Chapter 6: Enthalpy changes
Question Enthalpy, H / kJ mol–1
1 C lassify each process as exothermic or endothermic: Hreactants CH4(g) + 2O2(g)
a the burning of magnesium in air reactants
b the crystallisation of copper(II) sulfate from a
∆H = –890.3 kJ mol–1
saturated solution
c the thermal decomposition of magnesium nitrate Hproducts CO2(g) + 2H2O(l)
d the fermentation of glucose by yeast 0 products
e the evaporation of sea water. 0
Reaction pathway
Enthalpy changes and enthalpy profile
diagrams Figure 6.3 Enthalpy profile diagram for the combustion
of methane.
We call the energy exchange between a chemical reaction
and its surroundings at constant pressure the enthalpy the enthalpy of the products must be greater than the
change. Enthalpy is the total energy associated with enthalpy of the reactants. We can see from the enthalpy
the materials that react. The symbol for enthalpy is H. profile diagram for the thermal decomposition of calcium
We cannot measure enthalpy, but we can measure an carbonate (Figure 6.4) that Hproducts – Hreactants is positive.
enthalpy change when heat energy is exchanged with the
surroundings. We can write this as: CaCO3(s) CaO(s) + CO2(g) ΔH = +572 kJ mol–1
ΔH = Hproducts – Hreactants
The positive sign shows that the reaction is endothermic. 91
enthalpy enthalpy of enthalpy of
change products reactants Enthalpy, H / kJ mol–1
The symbol Δ is the upper case Greek letter ‘delta’. This Hproducts CaO(s) + CO2(g)
symbol is often used to mean a change in a quantity. products
For example, ΔT means a change in temperature and ΔH
means the enthalpy change. ∆H = +572 kJ mol–1
The units of enthalpy change are kilojoules per mole Hreactants CaCO3(s)
(kJ mol–1). 00 reactant(s)
We can draw enthalpy profile diagrams (also known Reaction pathway
as reaction pathway diagrams) to show enthalpy changes.
The enthalpy of the reactants and products is shown Figure 6.4 Enthalpy profile diagram for the decomposition of
on the y-axis. The x-axis shows the reaction pathway, calcium carbonate.
with reactants on the left and products on the right.
For an exothermic reaction, energy is released to the Question
surroundings. So the enthalpy of the reactants must be
greater than the enthalpy of the products. We can see
from the enthalpy profile diagram for the combustion of
metWhaencea(nFiinguclrued6e.3t)hitshiantfHorpmrodautcitosn– iHnrtehacetaenqtsuisatnioengaftoivre.
the reaction:
CH4(g) + 2O2(g) CO2(g) + 2H2O(l) 2 D raw enthalpy profile diagrams for:
ΔH = –890.3 kJ mol–1
a the combustion of sulfur to form sulfur dioxide
The negative sign shows that the reaction is exothermic. b the endothermic reaction
For an endothermic reaction, energy is absorbed from
H2O(g) + C(s) H2(g) + CO(g)
the surroundings by the chemicals in the reaction. So
Chapter 6: Enthalpy changes
The symbol for standard enthalpy change of combustion Standard enthalpy change of
is ΔH .— Oc Enthalpy changes of combustion are always solution, ΔHs—Ool
exothermic. The substances combusted can be either
elements or compounds. The standard enthalpy change of solution (ΔHs—Ool) is the
enthalpy change when one mole of solute is dissolved in
S(s) + O2(g) SO2(g) a solvent to form an infinitely dilute solution under
standard conditions.
ΔH — Oc [S(s)] = –296.8 kJ mol–1
CH4(g) + 2O2(g) CO2(g) + 2H2O(l) An infinitely dilute solution is one that does not produce
any further enthalpy change when more solvent is added.
ΔH — Oc [CH4(g)] = –890.3 kJ mol–1 An example is the addition of a small amount of solid
sodium hydroxide to a large amount of water.
Note that the first equation can be considered as either the
enthalpy change of combustion of sulfur or the enthalpy NaOH(s) + aq NaOH(aq)
change of formation of sulfur dioxide.
Question We use known amounts of solute and solvent with the
solvent in excess to make sure that all the solute dissolves.
3 C lassify each of the following reactions as ΔH — Or , ΔH —O
or ΔH — Oc : Standard enthalpy change of
f atomisation, ΔHa—Ot
a MgCO3(s) MgO(s) + CO2(g)
b C(graphite) + O2(g) CO2(g) The standard enthalpy change of atomisation, ΔH—aOt, is
the enthalpy change when one mole of gaseous atoms is
c HCl(g) + NH3(g) NH4Cl(s) formed from its element under standard conditions. 93
d H2(g) + _21 O2(g) H2O(l)
Standard enthalpy change of The standard enthalpy change of atomisation of hydrogen
neutralisation, ΔH —n O relates to the equation:
The standard enthalpy change of mneouletroafliwsaattioerni(sΔH —nO ) _ 21 H2(g) H(g) ΔH —aOt [ _12 H2(g) ]= +218 kJ mol–1
is the enthalpy change when one
formed by the reaction of an acid with an alkali under Standard enthalpy change of hydration
standard conditions. of an anhydrous salt
For example: The standard enthalpy change of hydration of an
anhydrous salt is the enthalpy change when one mole of
HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) a hydrated salt is formed from one mole of the anhydrous
ΔH — On = –57.1 kJ mol–1 salt under standard conditions.
For any acid–alkali reaction the ionic equation is: For example:
H+(aq) + OH–(aq) H2O(l) Na2S2O3(s) + 5H2O(l) Na2S2O3.5H2O(s)
ΔH —O = –55.0 kJ mol–1
The other ions in solution (Cl– and Na+) are spectator ions
and take no part in the reaction (see page 13).
This should not be confused with the standard enthalpy
change of hydration of gaseous ions to form aqueous ions
(see page 265 for this material, which is not required for AS
level).
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Measuring enthalpy changes
We can measure the enthalpy change of some reactions by
different techniques. These are outlined in the ‘Measuring
enthalpy changes’ box.
Measuring enthalpy changes As 1 cm3 of water weighs 1 g, we can substitute volume
of water in cm3 of water for mass of water in g in the
Calorimetry equation. Aqueous solutions of acids, alkalis and salts
We can measure the enthalpy change of some are assumed to be largely water.
reactions by a technique called calorimetry. The
apparatus used is called a calorimeter. A simple With solutions we make the assumptions that:
calorimeter can be a polystyrene drinking cup
(Figure 6.5), a vacuum flask or a metal can. ■■ 1 cm3 of solution has a mass of 1 g
■■ the solution has the same specific heat capacity as water.
thermometer (reading to 0.2 °C)
plastic lid Note: A rise in temperature is given a positive sign. So
the value of ΔH is negative for an exothermic reaction.
A fall in temperature is given a negative sign. So the
value of ΔH is positive for an endothermic reaction.
reaction polystyrene cup The enthalpy change of neutralisation by
94 mixture
experiment
We can find the enthalpy change of neutralisation
Figure 6.5 A polystyrene cup can act as a calorimeter for of sodium hydroxide with hydrochloric acid by mixing
finding some enthalpy changes. equal volumes of known equimolar concentrations of
acid and alkali together in a polystyrene cup. A typical
When carrying out experiments in calorimeters we procedure for the reaction above is as follows.
use known amounts of reactants and known volumes 1 Place 50 cm3 of 1.0 mol dm–3 hydrochloric acid in the cup
of liquids. We also measure the temperature change and record its temperature.
of the liquid in the calorimeter as the reaction occurs. 2 Add 50 cm3 of 1.0 mol dm–3 sodium hydroxide (at the
The thermometer should be accurate to 0.1 or 0.2 °C. same temperature) to the acid in the cup.
Calorimetry relies on the fact that it takes 4.18 J of 3 Stir the reaction mixture with the thermometer and
energy to increase the temperature of 1 g of water by record the highest temperature.
1 °C. The energy required to raise the temperature
of 1 g of a liquid by 1 °C is called the specific heat In this experiment most of the heat is transferred to
capacity, c, of the liquid. So, the specific heat the solution, as the polystyrene cup is a good insulator.
capacity of water is 4.18 J g–1 °C–1. Cooling of the warm solution is not a great problem:
the reaction is rapid so the maximum temperature is
The energy transferred as heat (the enthalpy change) reached before much cooling of the warm solution has
is given by the relationship: occurred. However, there are still heat losses to the air
and to the thermometer, which make the result less
exothermic than the data book value of –57.1 kJ mol–1.
ΔH = –mcΔT
where: Results and calculation
= 100 g (50 cm3 of acid
ΔH is the enthalpy change, in J mass of solution plus 50 cm3 of alkali and
m is the mass of water, in g
c is the specific heat capacity, in J g–1 °C–1 assuming that 1.0 cm3 of
ΔT is the temperature change, in °C
solution has a mass of 1.0 g)
Chapter 6: Enthalpy changes
Measuring enthalpy changes (continued)
specific heat capacity = 4.18 J g–1 °C–1 (assuming 7 Weigh the cup and its contents to calculate the mass of
that the heat capacity of sodium hydroxide which dissolved.
the solution is the same as
the heat capacity of water) Results and calculations
starting temperature of = 21.3 °C mass of polystyrene cup = 23.00 g
reactant solutions
final temperature of = 27.8 °C mass of polystyrene cup + water = 123.45 g
product solution
temperature rise = +6.5 °C mass of water = 100.45 g
use the relationship ΔH = –mcΔT mass of cup + water + sodium hydroxide = 124.95 g
heat energy released = –100 × 4.18 × 6.5 = –2717 J
mass of sodium hydroxide that dissolved = 1.50 g
initial temperature of water = 18.0 °C
At the start, the reaction mixture contained 50 cm3 final temperature of water = 21.6 °C
of 1.0 mol dm–3 hydrochloric acid and 50 cm3 of
1.0 mol dm–3 sodium hydroxide. The number of moles of temperature rise = +3.6 °C
each (and of the water formed) is calculated using
From the results, 1.50 g of sodium hydroxide dissolved
_ c_o_n_c_e_n_tr_a_t _io_n_1 _ ×0_ 0v_o0_ l_u_m _e_ _(i_n_c_m__3_) = _1_.10_0 _×0_ 50_0 _ = 0.050 moles in 100.45 cm3 (100.45 g) of water and produced a
temperature change of +3.6 °C.
So 2717 J of energy was released by 0.050 moles of acid.
( )enthalpy = – mass of × specific heat × temperature
Therefore for one mole of acid (forming one mole of
change water capacity (in change
water) the energy released was (in J) (in g) J g–1 °C–1) (in °C) 95
_ 02_.70_15_70_ == = –(100.45 × 4.18 × 3.6)
–54 340 J mol–1 = –1511.57 J = –1.5 kJ (to 2 significant figures)
–54 kJ mol–1 (to
The enthalpy change for 1.5 g sodium hydroxide is –1.5 kJ
2 significant figures). The enthalpy change for 1.0 mole of sodium hydroxide
(Mr = 40 g mol–1)
The negative sign shows that the reaction is – _14_.05_ × 1.5 kJ = –40 kJ
is exothermic.
Enthalpy change of solution by experiment ΔH —O = –40 kJ mol–1
The enthalpy change of solution of sodium sol
hydroxide can be found using a polystyrene cup as
a calorimeter. We use known amounts of solute and In this experiment we are assuming that the specific
solvent with the solvent in excess to make sure that heat capacity of the solution is the same as the
all the solute dissolves. specific heat capacity of water. The heat losses in this
The procedure is: experiment, however, are likely to be considerable
1 Weigh an empty polystyrene cup. because the sodium hydroxide takes some time to
2 Pour 100 cm3 of water into the cup and weigh the cup dissolve. This means that the reaction mixture has a
longer period of cooling.
and water.
3 Record the steady temperature of the water with a Finding the enthalpy change of combustion
thermometer reading to at least the nearest 0.2 °C. Experiment: the enthalpy change of combustion of
4 Add a few pellets of sodium hydroxide (corrosive!) propan-1-ol
which have been stored under dry conditions. We can find the enthalpy change of combustion by
5 Keep the mixture stirred continuously with a burning a known mass of substance and using the heat
released to raise the temperature of a known mass of
thermometer and record the temperature at fixed water. The apparatus used for this consists of a spirit
intervals, e.g. every 20 seconds. burner and a metal calorimeter (Figure 6.6).
6 Keep recording the temperature for 5 minutes after the
maximum temperature has been reached.
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Measuring enthalpy changes (continued)
thermometer 6 Remove the spirit burner, place the cap on it and
lid reweigh it.
metal Results and calculations
calorimeter
clamp water To find the standard enthalpy change of combustion we
need to know:
shield ■ the mass of fuel burnt
■ the temperature rise of the water
wick alkane or cap for ■ the mass of the water
spirit burner alcohol burner ■ the relative molecular mass of the fuel (propan-1-ol).
fuel
Figure 6.6 A simple apparatus used to find the enthalpy mass of water in calorimeter = 100 g
change of combustion of fuels.
mass of spirit burner and propan-1-ol at start = 86.27 g
mass of spirit burner and propan-1-ol at end = 86.06 g
The procedure is: mass of propan-1-ol burnt = 0.21 g
1 Weigh the spirit burner containing propan-1-ol. The initial temperature of water = 30.9 °C
final temperature of water = 20.2 °C
cap on the burner must be kept on when the burner temperature change of the water
is not lit to avoid evaporation of the fuel. = +10.7 °C
2 Pour 100 cm3 (100 g) of water into the calorimeter.
Using the relationship ΔH = –mcΔT (mass of water ×
For greater accuracy this should be weighed out. specific heat capacity of water × temperature change)
96 3 Stir the water and record its temperature with a energy released by burning 0.21 g propanol
thermometer reading to at least the nearest 0.1 °C. = –(100 × 4.18 × 10.7) = –4472.6 J
4 Place the spirit burner beneath the calorimeter, the mass of 1 mole of propan-1-ol, C3H7OH, is 60 g
remove the cap and light the wick. The length of the so for 60 g propan-1-ol the energy released is
wick should have been previously adjusted so that –4472.6 × _06_.2_01_ = –1 277 885.7 J mol–1 = –1300 kJ mol–1
the material of the wick does not burn and the flame (to 2 significant figures)
just touches the bottom of the calorimeter.
This is much less than the data book value of –2021 kJ mol–1,
5 Keep stirring the water with the thermometer until mainly due to heat losses to the surroundings.
there is a temperature rise of about 10 °C. Record
this temperature.
questions 5 E xplain why the enthalpy change of neutralisation of one
mole of sulfuric acid, H2SO4, is not the standard enthalpy
4 a Calculate the energy transferred when the change of neutralisation.
temperature of 75 cm3 of water rises from 23 °C
to 54 °C. 6 A student added 10 g (0.25 mol) of sodium hydroxide to
40 cm3 of water to make a concentrated solution. All the
b When 8 g of sodium chloride is dissolved in 40 cm3 sodium hydroxide dissolved. He measured the maximum
of water the temperature falls from 22 °C to 20.5 °C. temperature rise. He suggested that these results would
Calculate the energy absorbed by the solution when give an accurate value for the standard enthalpy change of
sodium chloride dissolves. solution. Give two reasons why he is incorrect.
c A student added 50 cm3 of sodium hydroxide to 50 cm3 7 A student calculated the standard enthalpy change of
of hydrochloric acid. Both solutions were at 18 °C to –co87m0b kuJs mtiooln–1o. Tf ehtehdaantoalbΔoHo —k Oc [vCa2lHue5OisH–]1b3y6c7a klJo mrimole–t1r.yExapslain
start with. When the solutions were mixed a reaction the difference between these values.
occurred. The temperature rose to 33 °C. Calculate the
energy released in this reaction.
Chapter 6: Enthalpy changes
Hess’s law Enthalpy change of reaction
from enthalpy changes of
Conserving energy formation
The Law of Conservation of Energy states that ‘energy We can calculate the enthalpy change of reaction by using
cannot be created or destroyed’. This is called the First Law the type of enthalpy cycle shown in Figure 6.8.
of Thermodynamics.
reactants ∆Hr products
This law also applies to chemical reactions. The total
energy of the chemicals and their surroundings must ∆H1 ∆H2
remain constant. In 1840 Germain Hess applied the Law of elements
Conservation of Energy to enthalpy changes.
Figure 6.8 An enthalpy cycle for calculating an enthalpy
Hess’s law states that ‘the total enthalpy change in a change of reaction. The dashed line shows the indirect (two-
chemical reaction is independent of the route by which step) route.
the chemical reaction takes place as long as the initial and
final conditions are the same’.
Enthalpy cycles We use the enthalpy changes of formation of the reactants
and products to calculate the enthalpy change of the
We can illustrate Hess’s law by drawing enthalpy cycles reaction. we take note of the directions of the arrows to
(Hess cycles). In Figure 6.7, the reactants A and B combine find the one-stage (direct) and two-stage (indirect) routes.
directly to form C. This is the direct route. When we use Hess’s law we see that:
Two indirect routes are also shown. One other way ΔH2 = ΔH1 + ΔHr
of changing A + B to C is to convert A + B into different direct route indirect route
substances F + G (intermediates), which then combine to 97
form C. So ΔToHcr a=lcΔuHla2te–tΔheHe1nthalpy change of reaction using this
type of enthalpy cycle we use the following procedure:
F+G
indirect route 1 ■■ write the balanced equation at the top
■■ draw the cycle with elements at the bottom
A+B direct route C ■■ draw in all arrows, making sure they go in the
correct directions
■■ apply Hess’s law, taking into account the number of moles
of each reactant and product.
indirect route 2 Y+Z If there are 3 moles of cahparnogdeuocft,foe.rgm. 3aCtioOn2(bgy),3w. eAmlsoust
X multiply the enthalpy
remember that the standard enthalpy change of formation
of an element in its standard state is zero.
Figure 6.7 The enthalpy change is the same no matter which
route is followed. worked example
Hess’s law tells us that the enthalpy change of reaction
for the direct route is the same as for the indirect route. It 1 Calculate the standard enthalpy change for the
does not matter how many steps there are in the indirect reaction:
route. We can still use Hess’s law.
2NaHCO3(s) Na2CO3(s) + CO2(g) + H2O(l)
We can use Hess’s law to calculate enthalpy
changes that cannot be found by experiments using The relevant enthalpy changes of formation are:
calorimetry. For example, the enthalpy change of
formation of propane cannot be found by direct ΔH —O [NaHCO3(s)] = –950.8 kJ mol–1
experiment because hydrogen does not react with
carbon under standard conditions. f
ΔH —O [Na2CO3(s)] = –1130.7 kJ mol–1
f
ΔH —O [CO2(g)] = –393.5 kJ mol–1
f
ΔH —O [H2O(l)] = –285.8 kJ mol–1
f
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worked example (continued) Enthalpy change of formation
The enthalpy cycle is shown in Figure 6.9. from enthalpy changes of
combustion
2 NaHCO3(s) ∆Hr Na2CO3(s) + CO2(g) + H2O(l)
We can calculate the enthalpy change of formation of
2 ∆Hf NaHCO3(s) ∆H1 ∆H f [Na2CO3(s)] many compounds by using the type of enthalpy cycle
∆H2 +∆Hf [CO2(g)] shown in Figure 6.10.
+∆Hf [H2O(l)] elements + oxygen ∆Hf compound formed + oxygen
2Na(s) + 2C(graphite) + 3O2(g) + H2(g)
Figure 6.9 The enthalpy cycle for the decomposition of ∆H1 ∆H2
sodium hydrogencarbonate. The dashed line shows the combustion products
two-step route.
Using Hess’s law: Figure 6.10 An enthalpy cycle for calculating an enthalpy
change of formation from enthalpy changes of combustion.
The dashed line shows the two-step route.
ΔH2 = ΔH1 + ΔHr
ΔH —O [Na2CO3(s)] + ΔH —O [CO2(g)] + ΔH —O [H2O(l)] We use the enthalpy changes of combustion of the
f f f
= 2ΔH —O [NaHCO3(s)] + ΔH —O reactants and products to calculate the enthalpy change
f r
(–1130.7) + (–393.5) + (–285.8) = 2(–950.8) + ΔHr of formation. When we take note of the direction of
the arrows to find the one-stage (direct) and two-stage
–1810.0 = –1901.6 + ΔHr (indirect) routes and use Hess’s law we see that:
98 So ΔH —O = (–1810.0) – (–1901.6)
r
= +91.6 kJ mol–1 (for the equation shown) ΔH1 = ΔHf + ΔH2
Note: direct route indirect route
i the value for ΔH —O [NaHCO3(s)] is multiplied by 2 So ΔToHcf a=lcΔuHla1te–tΔhHe e2nthalpy change of formation using
this type of cycle:
f ■■ write the equation for enthalpy change of formation at the
because 2 moles of NaHCO3 appear in the equation
ii ΔthHe— O fva[Hlu2eOs(lf)o] raΔreHa— O fd[dNead2CtoOg3e(st)h],eΔr tHo— O fg[iCveO2Δ(Hg)2].aTnadke
care to account for the fact that some values may be
positive and some negative. top; add oxygen on both sides of the equation to balance
the combustion reactions
■■ draw the cycle with the combustion products at the bottom
Question ■■ draw in all arrows, making sure they go in the
correct directions
8 a Draw an enthalpy cycle to calculate ΔH —O for ■■ apply Hess’s law, taking into account the number of moles
the reaction of each reactant and product.
r
2Al(s) + Fe2O3(s) 2Fe(s) + Al2O3(s) worked example
b Calculate ΔH — O r using the following information: 2 Calculate the standard enthalpy change of formation
of ethane, C2H6.
ΔH —O [Fe2O3(s)] = –824.2 kJ mol–1
The relevant enthalpy changes of combustion are:
f
ΔH —O [Al2O3(s)] = –1675.7 kJ mol–1
f
C(graphite) + O2(g) CO2(g)
ΔH —O [C(graphite)] = –393.5 kJ mol–1
c
H2(g) + _21 O2(g) H2O(l)
ΔH —O [H2(g)] = –285.8 kJ mol–1
c
Chapter 6: Enthalpy changes
worked example (continued) anhydrous salt ∆Hhyd hydrated salt
C2H6(g) + 3_12 O2(g) 2CO2(g) + 3H2O(l) ∆H1 ∆H2
ΔH —O [C2H6(g)] = –1559.7 kJ mol–1 dilute aqueous
solution of salt
c
The enthalpy cycle is shown in Figure 6.11.
2C(graphite) + 3H2(g) + 321 O2(g) ∆Hf C2H6(g) + 3 1 O2(g) Figure 6.12 An enthalpy cycle to calculate the enthalpy
2 change of hydration of an anhydrous salt. The dashed line
shows the two-step route.
2∆Hc [C(graphite)] ∆H1 ∆H2 ∆Hc [C2H6(g)]
+ 3∆Hc [H2(g)] 2CO2(g) + 3H2O(l)
The enthalpy cycle for calculating the enthalpy of
Figure 6.11 The enthalpy cycle to find the enthalpy hydration of anhydrous sodium thiosulfate is shown in
change of formation of ethane using enthalpy changes of Figure 6.13.
combustion. The dashed line shows the two-step route.
Na2S2O3(s) + 5H2O(l ) ∆Hhyd Na2S2O3.5H2O(s)
Using Hess’s law: ∆H=s–o7l [.N6 ak2JSm2Ool3–(1s)] ∆H1 ∆H2 ∆Hs=ol +[N4a72.4S2kOJ m3.5oHl–21O(s)]
ΔH1 = ΔHf + ΔH2 Na2S2O3(aq)
2(–393.5) + 3(–285.8) = ΔHf + (–1559.7)
–1644.4 = ΔHf + (–1559.7) Figure 6.13 An enthalpy cycle to calculate the enthalpy
change when anhydrous sodium thiosulfate is hydrated. The
So ΔHf = –1644.4 – (–1559.7) = –84.7 kJ mol–1 dashed line shows the two-step route.
Question Using Hess’s law: 99
ΔH1 = ΔHhyd + ΔH2
9 a Draw an enthalpy cycle to calculate the enthalpy (–7.6) = ΔHhyd + (+47.4)
change of formation of ethanol, C2H5OH, using So ΔHhyd = (–7.6) – (+47.4) = –55.0 kJ mol–1
enthalpy changes of combustion.
b Calculate a value for ΔH —O [C2H5OH(l)] using the Question
following data: 10 Suggest why it is difficult to measure the enthalpy
f
change directly when an anhydrous salt is converted
ΔH —O [C(graphite)] = –393.5 kJ mol–1 to a hydrated salt.
c
ΔH —O [H2(g)] = –285.8 kJ mol–1
c
ΔH —O [C2H5OH(l)] = –1367.3 kJ mol–1
c
Calculating the enthalpy Bond energies and enthalpy
change of hydration of an changes
anhydrous salt
Hydrated salts such as hydrated copper(II) sulfate, Bond breaking and bond making
iCounSsO. I4t.i5sHv2eOry, contain water molecules surrounding their
difficult to measure the enthalpy change Enthalpy changes are due to the breaking and forming
when an anhydrous salt such as anhydrous sodium of bonds. Breaking bonds requires energy. The energy
thiosulfate becomes hydrated. is needed to overcome the attractive forces joining the
atoms together. Energy is released when new bonds are
Na2S2O3(s) + 5H2O(l) Na2S2O3.5H2O(s) formed. Bond breaking is endothermic and bond forming
is exothermic.
We can, however, use an enthalpy cycle to calculate this.
We use the standard enthalpy changes of solution to In a chemical reaction:
complete the enthalpy cycle (Figure 6.12).
■■ if the energy needed to break bonds is less than the energy
released when new bonds are formed, the reaction will
release energy and is exothermic.
Cambridge International AS Level Chemistry
■■ if the energy needed to break bonds is more than the Average bond energy
energy released when new bonds are formed, the reaction
will absorb energy and is endothermic. Bond energy is affected by other atoms in the molecule. The
O H bond in water has a slightly different bond energy
We can draw enthalpy level (reaction pathway) diagrams value to the O H bond in ethanol; in ethanol the oxygen is
to show these changes (Figure 6.14). In reality, not all the
bonds in a compound are broken and then re-formed connected to a carbon atom rather than another hydrogen
during a reaction. In most reactions only some of the bonds atom. The O H bond is in a different environment.
in the reactants are broken and then new bonds are formed Identical bonds in molecules with two (or more) types of
in a specific sequence. The minimum energy required to bond have different bond energies when we measure them.
break certain bonds in a compound to get a reaction to It takes more energy to break the first O H bond in water
start is called the activation energy (see page 141). than to break the second. For these reasons we use average
bond energies taken from a number of bonds of the same
type but in different environments.
Bond energy We cannot usually find the value of bond energies
The amount of energy needed to break a specific covalent directly so we have to use an enthalpy cycle. The average
bond is called the bond dissociation energy. We sometimes bond energy of the C H bond in methane can be found
call this the bond energy or bond enthalpy. The symbol using the enthalpy changes of atomisation of carbon and
for bond energy is E. We put the type of bond broken in hydrogen and the enthalpy change of combustion or
brackets after the symbol. So E(C H) refers to the bond formation of methane.
energy of a mole of single bonds between carbon and
hydrogen atoms. The enthalpy cycle for calculating the average C H
bond energy is shown in Figure 6.15. Using the enthalpy
The bond energy for double and triple bonds refers to a cycle shown in Figure 6.15, the average C H bond energy
mole of double or triple bonds. Two examples of equations can be found by dividing the value of ΔH on the diagram
relating to bond energies are: by four (because there are four C H bonds in a molecule
100 Br2(g) 2Br(g) E(Br Br) = +193 kJ mol–1 of methane).
O O(g) 2O(g) E(O O) = +498 kJ mol–1 CH4(g) ∆H C(g) + 4H(g)
The values of bond energies are always positive because ∆Hf [CH4(g)] ∆+H4a∆t H[(agtr[a12pHh2it(eg))]]
they refer to bonds being broken.
C(graphite) + 2H2(g)
When new bonds are formed the amount of energy
released is the same as the amount of energy absorbed Figure 6.15 An enthalpy cycle to find the average bond
when the same type of bond is broken. So, for the energy of the C H bond. The dashed line shows the
formation of oxygen molecules from oxygen atoms: two-step route.
2O(g) O2(g) E(O O) = –498 kJ mol–1
a b 2H + 2Br
C + 4H + 4O
bond bond forming
H H H + O O bond breaking
C O O breaking H H + Br Br
H Br energy
Energy H Br absorbed
Energy
H bond
forming
energy H O H
released H O H
O C O+
Reaction pathway Reaction pathway
Figure 6.14 a An energy level diagram showing bond breaking and bond forming for the combustion of methane (exothermic).
b An energy level diagram showing bond breaking and bond forming for the decomposition of hydrogen bromide (endothermic).
Chapter 6: Enthalpy changes
Question It is often easier to set out the calculation as a balance
sheet, as shown below.
11 U se the information in Figure 6.15 and the information Bonds broken ΔH1 (kJ) Bonds formed ΔH2 (kJ)
below to show that the average bond energy of the 6 × N H = 6 × 391
C H bond is 415.9 kJ mol–1. 1 × N NH
3 × H total = –2346
ΔH —O [CH4] = –74.8 kJ mol–1 = 1 × 945 = 945
= 3 × 436 = 1308
f
ΔH —aOt [_ 21 H 2] = +218 kJ mol–1
ΔH—aOt [C(graphite)] = +716.7 kJ mol–1
total = +2253
Note in these calculations that:
Calculating enthalpy changes ■■ one triple bond in nitrogen is broken
using bond energies ■■ three single bonds in hydrogen are broken
■■ six single N H bonds in hydrogen are formed (because each
We can use bond enthalpies to calculate the enthalpy
change of a reaction that we cannot measure directly. of the two ammonia molecules has three N H bonds)
For example, the reaction for the Haber process (see ■■ values for bond breaking are positive, as these are
page 129):
endothermic, and values for bond forming are negative, as
these are exothermic.
From the enthalpy cycle in Figure 6.16:
N2(g) + 3H2(g) 2NH3(g) ΔΔHHrr == eΔnHth1 a+lp yΔcHh2ange for bonds broken
+ enthalpy change for bonds
The enthalpy cycle for this reaction is shown in Figure ΔHr = 2253 + (–2346) = –93 kJ mol–1 formed
6.16. The relevant bond energies are:
Question 101
E(N N) = 945 kJ mol–1
E(H H) = 436 kJ mol–1
E(N H) = 391 kJ mol–1
N N(g) + 3H H(g) ∆Hr 2H N H(g) 12 T he equation for the combustion of ethanol is:
H C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l)
∆H1 ∆H2
a R ewrite this equation to show all the bonds in the
E(N N) 2× 3 × –E(N H) reactants and products.
+ 3 × E(H H)
b Use the following bond energies (in kJ mol–1) to
calculate a value for the standard enthalpy change
2N(g) + 6H(g) of this reaction:
HC))
Figure 6.16 The enthalpy cycle for ammonia synthesis. The E(C = +347
dashed line shows the two-step route. E(C = +410
E(C OO O))) === +336
E(O +496
E(C +805
E(O H) = +465
c The standard enthalpy change of combustion
of ethanol is –1367.3 kJ mol–1. Suggest why this
value differs from the value obtained using
bond energies.
Cambridge International AS Level Chemistry
Summary
■ When a chemical reaction occurs, energy is ■ The standard enthalpy changes of hydration and
transferred to or from the surroundings. solution can be defined in terms of one mole of a
specified compound reacting completely.
■ In an exothermic reaction, heat is given out to the
surroundings so the products have less energy than ■ The standard enthalpy change of neutralisation can
the reactants. In an endothermic reaction, heat is be defined in terms of one mole of water formed
absorbed from the surroundings so the products when hydrogen ions and hydroxide ions react.
have more energy than the reactants.
■ Hess’s law states that ‘the total enthalpy change for
■ Energy changes in chemical reactions that lead to a chemical reaction is independent of the route by
heating or cooling are called enthalpy changes (ΔH). which the reaction takes place’.
■ Exothermic enthalpy changes are shown as negative ■ Hess’s law can be used to calculate enthalpy
values (–). changes for reactions that do not occur directly or
■ Endothermic enthalpy changes are shown as cannot be found by experiment.
positive values (+). ■ Hess’s law can be used to calculate the enthalpy
■ Standard enthalpy changes are compared under change of a reaction using the enthalpy changes of
formation of the reactants and products.
standard conditions of pressure, 105 Pa (100 kPa),
and temperature, 298 K (25 °C). ■ Hess’s law can be used to calculate the enthalpy
■ Enthalpy changes can be calculated experimentally change of formation of a compound using the
using the relationship: enthalpy changes of combustion of the reactants
and products.
enthalpy change = –mass of liquid × specific
102 heat capacity × temperature change ■ Bond breaking is endothermic; bond making is
ΔH = –mcΔT exothermic.
■ The standard enthalpy change of formation (ΔH —Of ) ■ Bond energy is a measure of the energy needed to
break a covalent bond.
is the enthalpy change when one mole of a ■ Average bond energies are often used because the
compound is formed from its elements under strength of a bond between two particular types of
standard conditions. atom is slightly different in different compounds.
■ The standard enthalpy change of combustion ■ Hess’s law can be used to calculate the enthalpy
(ΔH —Oc ) is the enthalpy change when one mole of change of a reaction using the average bond
a substance is burnt in excess oxygen under energies of the reactants and products.
standard conditions.
■ The standard enthalpy change of atomisation
(ΔH —aOt ) is the enthalpy change when one mole of
gaseous atoms is formed from the element in its
standard state under standard conditions.
Chapter 6: Enthalpy changes
End-of-chapter questions
1 Copper(II) nitrate decomposes on heating. The reaction is endothermic.
2Cu(NO3)2(s) 2CuO(s) + 4NO2(g) + O2(g)
a Draw an enthalpy level diagram (reaction profile diagram) for this reaction. [3]
b Draw an enthalpy cycle diagram to calculate the standard enthalpy change for this reaction, using [3]
enthalpy changes of formation. [3]
c Calculate the enthalpy change for this reaction using the following enthalpy changes of formation. [3]
[2]
ΔH —O [Cu(NO3)2(s)] = –302.9 kJ mol–1
f
ΔH —O [CuO(s)] = –157.3 kJ mol–1
f
ΔH —O [NO2(g)] = +33.2 kJ mol–1
f
d Copper(II) sulfate is soluble in water. A student dissolved 25.0g of copper(II) sulfate in 100cm3 of water in
a polystyrene beaker stirring all the time. The temperature of the water fell by 2.9°C.
i Calculate the enthalpy change of solution of copper(II) sulfate. (specific heat capacity of
water = 4.18Jg–1°C–1; relative molecular mass of copper(II) sulfate = 249.7 gmol–1)
ii Suggest one source of error in this experiment and explain how the error affects the results.
Total = 14
2 Propanone is a liquid. It has the structure HOH
HCCCH
HH 103
The equation for the complete combustion of propanone is:
CH3COCH3(l) + 4O2(g) 3CO2(g) + 3H2O(l)
a Use the following bond energies (in kJmol–1) to calculate a value for the standard enthalpy change of
this reaction:
E(C C) = +347 [4]
E(C H) = +413
[2]
E(O O) = +496
[2]
E(C O) = +805
E(O H) = +465 [3]
[2]
b Suggest why it would be more accurate to use bond energies that are not average bond energies in [1]
this calculation.
c The standard enthalpy change of combustion of propanone is –1816.5kJmol–1. Suggest why this value
differs from the value obtained using bond energies.
d The standard enthalpy change of formation of propanone is –248kJmol–1.
i Define the term standard enthalpy change of formation.
ii Write the equation that describes the standard enthalpy change of formation of propanone.
iii Explain why the enthalpy change of formation of propanone cannot be found by a single experiment.
Total = 14
Cambridge International AS Level Chemistry
3 240cm3 of ethane (C2H6) was burnt in a controlled way and found to raise the temperature of 100cm3 of
water by 33.5°C. (specific heat capacity of water = 4.18Jg–1K–1; 1mol of gas molecules occupies 24.0dm3
at r.t.p.)
a How many moles of ethane were burnt? [1]
b Calculate the heat change for the experiment. [2]
c Calculate the molar enthalpy change of combustion for ethane, as measured by this experiment. [2]
d Use the values below to calculate the standard molar enthalpy change for the complete combustion of ethane.
ΔH —O [CO2] = –394kJmol–1
f
ΔH —O [H2O] = –286kJmol–1
f
ΔH —O [C2H6] = –85 kJ mol–1 [4]
[2]
f
e Give possible reasons for the discrepancy between the two results.
Total = 11
4 a Define standard enthalpy change of combustion. [3]
b When red phosphorus burns in oxygen the enthalpy change is –2967kJmol–1. For white phosphorus the
enthalpy change is –2984kJmol–1. For both forms of phosphorus the reaction taking place is: [5]
[3]
P4(s) + 5O2(g) P4O10(s)
i Use this information to calculate the enthalpy change for the transformation: P4(white) P4(red)
ii Represent these changes on an enthalpy profile diagram.
Total = 11
104 5 a Define standard enthalpy change of formation. [3]
b Calculate the standard enthalpy change of formation of methane from the following standard enthalpy [4]
changes of combustion: [4]
carbon = –394kJmol–1
hydrogen = –286kJmol–1
methane = –891kJmol–1
c Calculate the standard enthalpy change of combustion of methane using the following bond energies:
E(C H) = +412kJmol–1
E(O O) = +496kJmol–1
E(C O) = +805kJmol–1
E(O H) = +463kJmol–1
Total = 11
6 a Define average bond enthalpy. [2]
b Use the average bond enthalpies that follow to calculate a value for the enthalpy change for the reaction:
[3]
H2 + I2 2HI [3]
E(H H) = +436kJmol–1
E(I I) = +151kJmol–1
E(H I) = +299kJmol–1
c Represent these changes on an enthalpy profile diagram.
Total = 8
Chapter 6: Enthalpy changes
7 a Define enthalpy change of solution. [3]
b Given the enthalpy changes ΔH1 and ΔH2 below, construct a Hess’s cycle that will enable you to find
the enthalpy change, ΔHr, for the reaction:
MgCl2(s) + 6H2O(l) MgCl2.6H2O(s) ΔHr
MgCl2(s) + aq MgCl2(aq) ΔH1
MgCl2.6H2O(s) + aq MgCl2(aq) ΔH2 [4]
Total = 7
8 a Define standard enthalpy change of reaction. [3]
b Given the enthalpy changes ΔH1 and ΔH2 below, construct a Hess’s cycle that will enable you to find
the enthalpy change, ΔHr, for the reaction:
MgCO3(s) MgO(s) + CO2(g) ΔHr
MgCO3(s) + 2HCl(aq) MgCl2(aq) + CO2(g) + H2O(l) ΔH1
MgO(s) + 2HCl(aq) H2O(l) + MgCl2(aq) ΔH2 [4]
Total = 7
9 In an experiment, a spirit burner is used to heat 250cm3 of water by burning methanol (CH3OH). 105
(Ar values: C = 12.0, H = 1.0, O = 16.0; specific heat capacity of water = 4.18Jg–1°C–1)
Results: [2]
starting temperature of water = 20.0°C [2]
starting mass of burner + fuel = 248.8g [2]
final temperature of water = 43.0°C
final mass of burner + fuel = 245.9g [3]
a How many joules of heat energy went into the water?
b How many moles of fuel were burnt?
c Calculate an experimental value for the enthalpy change of combustion of methanol from these results.
d Suggest three reasons why your answer is much smaller than the accepted standard enthalpy of
combustion of methanol.
Total = 9
106
chapter 7:
Redox reactions
Learning outcomes ■■ use changes in oxidation numbers to help balance
chemical equations.
You should be able to:
■■ calculate oxidation numbers of elements in
compounds and ions
■■ describe and explain redox processes in terms of
electron transfer and changes in oxidation number
Chapter 7: Redox reactions
Introduction
Some types of reactions can cost a lot of money due
to the damage they cause. Rusting is an oxidation
reaction that destroys about 20% of iron and steel
every year. Rust is hydrated iron(III) oxide. This forms
when iron reacts with oxygen in the presence of water.
Another costly example of oxidation is the reaction
between hydrogen and oxygen that is used to propel
some types of rockets into space. In this reaction, the
hydrogen is oxidised – but the oxygen is also reduced.
In fact, oxidation and reduction always take place
together, in what we call redox reactions.
Figure 7.1 A redox reaction is taking place when the fuel in
the Space Shuttle’s rockets burns.
What is a redox reaction? We can also define reduction as addition of hydrogen to 107
a compound and oxidation as removal of hydrogen from
A simple definition of oxidation is gain of oxygen by a compound. This is often seen in the reaction of organic
an element. For example, when magnesium reacts with compounds (see page 237).
oxygen, the magnesium combines with oxygen to form
magnesium oxide. Magnesium has been oxidised. There are two other ways of finding out whether or
not a substance has been oxidised or reduced during a
2Mg(s) + O2(g) 2MgO(s) chemical reaction:
A simple definition of reduction is loss of oxygen. When ■■ electron transfer
copper(II) oxide reacts with hydrogen, this is the equation ■■ changes in oxidation number.
for the reaction:
CuO(s) + H2(g) Cu(s) + H2O(l) Question
Copper(II) oxide loses its oxygen. Copper(II) oxide has 1 a In each of the following equations, state which
been reduced. reactant has been oxidised:
But if we look carefully at the copper oxide/hydrogen i PbO + H2 Pb + H2O
equation, we can see that oxidation is also taking place.
The hydrogen is gaining oxygen to form water. The ii CO + Ag2O 2Ag + CO2
hydrogen has been oxidised. We can see that reduction
and oxidation have taken place together. i ii 2Mg + CO2 2MgO + C
Oxidation and reduction always take place together. b In each of the following equations, state which
We call the reactions in which this happens redox reactant has been reduced:
reactions. Redox reactions are very important. For
example, one redox reaction – photosynthesis – provides i 5CO + I2O2 5CO2 + I2
food for the entire planet, and another one – respiration – i i 2H2S + SO2 3S + 2H2O
keeps you alive. both are redox reactions. iii CH2 CH2 + H2
CH3CH3
Cambridge International AS Level Chemistry
Redox and electron transfer ■■ Each copper(II) ion gains two electrons. The copper ions
have been reduced.
Half-equations
Cu2+ + 2e– Cu
We can extend our definition of redox to include reactions
involving ions. Balancing half-equations
Oxidation Is Loss of electrons. We can construct a balanced ionic equation from two half-
Reduction Is Gain of electrons. equations by balancing the numbers of electrons lost and
The initial letters shown in bold spell OIL RIG. This may gained and then adding the two half-equations together.
help you to remember these two definitions! The numbers of electrons lost and gained in a redox
reaction must be equal.
Sodium reacts with chlorine to form the ionic compound worked examples
sodium chloride.
1 Construct the balanced ionic equation for the reaction
2Na(s) + Cl2(g) 2NaCl(s) between nickel and iron(III) ions, Fe3+, from the half-
equations:
We can divide this reaction into two separate equations, Ni(s) Ni2+(aq) + 2e–
one showing oxidation and the other showing reduction.
We call these half-equations. Fe3+(aq) + e– Fe2+(aq)
When sodium reacts with chlorine: ■■ Each Ni atom loses two electrons when it is
oxidised. Each Fe3+ ion gains one electron when it is
■■ Each sodium atom loses one electron from its outer shell. reduced.
Oxidation is loss of electrons (OIL). The sodium atoms have
been oxidised. ■■ So two Fe3+ ions are needed to gain the two
electrons lost when each Ni2+ ion is formed
108 Na Na+ + e–
2Fe3+(aq) + 2e– 2Fe2+(aq)
This half-equation shows that sodium is oxidised. Ni(s) Ni2+(aq) + 2e–
It is also acceptable to write this half-equation as:
■■ The balanced ionic equation is:
Na – e– Na+ Ni(s) + 2Fe3+(aq) Ni2+(aq) + 2Fe2+(aq)
■■ Each chlorine atom gains one electron to complete its outer note how the electrons have cancelled out.
shell. Reduction is gain of electrons (RIG). The chlorine
atoms have been reduced. 2 Construct the balanced ionic equation for the reaction
othf eiopdriedseeinocneso(If–h) ywdirtohgmenaniognasna(Hte+)(.VUIIs) eiotnhse(fMonlloOw4–i)nign
Cl2 + 2e– 2Cl– two half-equations to help you:
This is a half-equation showing chlorine being reduced. i 2I–(aq) I2(aq) + 2e–
There are two chlorine atoms in a chlorine molecule, so
two electrons are gained. ii MnO4–(aq) + 8H+(aq) + 5e– Mn2+(aq) + 4H2O(l)
In another example iron reacts with copper(II) ions, ■■ When two iodide ions are oxidised, they lose two
Cu2+, in solution to form iron(II) ions, Fe2+, and copper. electrons. Each MnO4– ion gains five electrons when
it is reduced.
Fe(s) + Cu2+(aq) Fe2+(aq) + Cu(s)
■■ So we must multiply equation i by 5 and equation
ii by 2 to balance the number of electrons:
■■ Each iron atom loses two electrons to form an Fe2+ ion. The 10I–(aq) 5I2(aq) + 10e–
iron atoms have been oxidised.
Fe Fe2+ + 2e– 2MnO4–(aq) + 16H+(aq) + 10e– 2Mn2+(aq) + 8H2O(l)
It is also acceptable to write this half-equation as: ■■ The balanced ionic equation is:
Fe – 2e– Fe2+ 2MnO4–(aq) + 10I–(aq) + 16H+(aq)
2Mn2+(aq) + 5I2(aq) + 8H2O
Chapter 7: Redox reactions
Question 4 The sum of the oxidation numbers in a compound
is zero.
2 a Write two half-equations for the following
reactions. For each half-equation state whether 5 The sum of the oxidation numbers in an ion is equal to
oxidation or reduction is occurring. the charge on the ion.
i Cl2 + 2I– I2 + 2Cl– 6 In either a compound or an ion, the more
electronegative element is given the negative
i i 2Mg + O2 2MgO oxidation number.
i ii 4Fe + 3O2 2Fe2O3
b Zinc metal reacts with IO3– ions in acidic solution.
Construct a balanced ionic equation for this
reaction, using the two half-equations below:
2IO3– + 12H+ + 10e– I2 + 6H2O
Zn Zn2+ + 2e–
Oxidation numbers Figure 7.2 This is part of a ship’s hull. It is made of iron 109
protected by bars of magnesium metal. The magnesium
What are oxidation numbers? atoms (oxidation number = 0) are oxidised to Mg2+ ions
(oxidation number = +2) in preference to iron atoms changing
We can extend our definition of redox even further to include to Fe3+. This is called sacrificial protection.
oxidation and reduction in reactions involving covalent
compounds. We do this by using oxidation numbers Applying the oxidation number rules
(oxidation numbers are also called oxidation states). An
oxidation number is a number given to each atom or ion in a In the following examples we shall use ‘ox. no.’ as an
compound that shows us its degree of oxidation. Oxidation abbreviation for oxidation number.
numbers can be positive, negative or zero. The + or – sign must
always be included. Higher positive oxidation numbers mean Compounds of a metal with a non-metal
that an atom or ion is more oxidised. Higher negative oxidation The metal always has the positive ox. no. and the non-
numbers mean that an atom or ion is more reduced. metal has the negative ox. no. For example in sodium
oxidIfe,wNead2oOn, oNtak=no+w1 and O = –2.
Oxidation number rules the ox. no. of one of the atoms, we
can often work it out using the invariable ox. nos. in rule 2.
We can deduce the oxidation number of any atom or For example in sodium sulfide:
ion by using oxidation number rules. It is important to
note that an oxidation number refers to a single atom in ■■ ox. no. of each Na atom = +1
a compound. ■■ for two sodium atoms = +2
1 The oxidation number of any uncombined element is ■■ Na2S has no overall charge, so the total ox. no. is zero
zero. For example, the oxidation number of each atom (rule 4)
in S8, Cl2 and Zn is zero. ■■ ox. no. of S = –2.
2 In compounds many atoms or ions have fixed
oxidation numbers Compounds of a non-metal with a non-metal
– Group 1 elements are always +1
– Group 2 elements are always +2 In compounds containing two different non-metals, the
– fluorine is always –1 sign of the ox. no. depends on the electronegativity of each
– hydrogen is +1 (except in metal hydrides such as atom (see page 157). The most electronegative element is
given the negative sign (rule 6).
NaH, where it is –1)
– oxygen is –2 (except in peroxides, where it is –1, and
in F2O, where it is +2).
3 The oxidation number of an element in a monatomic
ion is always the same as the charge. For example,
Cl– is –1, Al3+ is +3.
Cambridge International AS Level Chemistry
Sulfur dioxide, SO2 Redox and oxidation number
■■ ox. no. of each O atom = –2
■■ for two oxygen atoms = 2 × (–2) = –4 We can define oxidation and reduction in terms of the
■■ SO2 has no charge, so the total ox. no. is zero (rule 4) oxidation number changes of particular atoms during
■■ ox. no. of S = +4 a reaction.
Iodine trichloride, ICl3 Oxidation is an increase of oxidation number.
■■ chlorine is more electronegative than iodine, so chlorine is – Reduction is a decrease in oxidation number.
and iodine is + For example, when tin reacts with nitric acid, the
■■ ox. no. of each Cl atom = –1 oxidation numbers of each atom of tin and nitrogen
■■ for three chlorine atoms = 3 × (–1) = –3 change as shown below.
■■ ICl3 has no charge, so the total ox. no. is zero (rule 4)
■■ ox. no. of I = +3 reduction
Hydrazine, N2H4 Sn + 4HNO3 → SnO2 + 4NO2 + 2H2O
■■ nitrogen is more electronegative than hydrogen, so nitrogen oxidation numbers 0 +5 +4 +4
is – and hydrogen is + oxidation
■■ ox. no. of each H atom = +1 (rule 2)
■■ for four hydrogen atoms = 4 × (+1) = +4
■■ N2H4 has no charge, so the total ox. no. is zero (rule 4)
■■ ox. no. of two N atoms = –4
■■ ox. no. of each N atom = –2
Compound ions
Compound ions are ions with two or more different atoms.
110 Examples are the sulfate ion, oSuOt4t2h–e, and the nitrate ion,
nNoOt 3k–n. Wowe. use rule 5 to work ox. no. that we do
Nitrate ion, NO3–
■■ ox. no. of each O atom = –2
■■ for three oxygen atoms = 3 × (–2) = –6
■■ NO3– has a charge of 1–, so the total ox. no. of N and O atoms
is –1 (rule 5)
■■ ox. no. of the nitrogen atom plus ox. no. of the three oxygen
atoms (–6) = –1
■■ ox. no. of N = +5 Figure 7.3 Copper reacts with silver nitrate to form silver and
copper(II) nitrate. The ox. no. of each copper atom has increased
by two. The ox. no. of each silver ion decreases by one.
Question Each tin atom (Sn) has increased in ox. no. by +4: tin
has been oxidised. Each nitrogen atom has decreased in
3 S tate the ox. no. of the bold atoms in these ox. no. by –1: nitrogen has been reduced. The ox. no. of
compounds or ions: each oxygen atom is unchanged at –2. The ox. no. of each
hydrogen atom is unchanged at +1. Oxygen and hydrogen
a P2O5 e NH3 are neither oxidised nor reduced.
b SO42– f ClO2–
c H2S g CaCO3 In this reaction nitric acid is acting as an oxidising agent:
d Al2Cl6
■■ oxidising agents increase the ox. no. of another atom
■■ an atom in the oxidising agent decreases in ox. no.
■■ the oxidising agent is the substance which gets reduced – it
gains electrons.
Chapter 7: Redox reactions
In this reaction tin is acting as a reducing agent: Nitrate ions
■■ reducing agents decrease the ox. no. of another atom Sodium, nitrogen and oxygen can form two different
■■ an atom in the reducing agent increases in ox. no. cnoom. opfosuonddiusmNais+N+1Oa2–ndantdheNoax+.NnOo.3–of(Foixgyugreen7.i4s)–. 2T.hSeooixt.is
■■ the reducing agent is the substance that gets oxidised – it the ox. no. of nitrogen that varies.
loses electrons.
Question ■■ The ox. no. of N in the NO2– ion is +3. So NaNO2 is sodium
nitrate(III).
■■ The ox. no. of N in the NO3– ion is +5. So NaNO3 is sodium
nitrate(V).
4 a Deduce the change in ox. no. for the bold atoms or
ions in each of the following equations. In each Note that the ox. no. comes after the ion it refers to.
case, state whether oxidation or reduction has Ions containing oxygen and one other element have the
taken place.
ending -ate (but hydroxide ions, OH–, are an exception to
i 2I– + Br2 I2 + 2Br– this rule). For example, ions containing chlorine and
oxygen are chlorates and ions containing sulfur and
i i (NH4)2Cr2O7 N2 + 4H2O + Cr2O3 oxygen are sulfates.
i ii A s2O3 + 2I2 + 2H2O As2O5 + 2H+ + 4I– The names of inorganic acids containing oxygen end
in –ic. The Roman number goes directly after the ion that
i v 2KMnO4 + 16HCl contains the oxygen and another element.
2MnCl2 + 2KCl + 5Cl2 + 8H2O
■■ pHh3PoOsp3hisopruhsoissp+h3o. ric(III) acid because the ox. no. of
b Identify the reducing agent in each of the equations ■■ HClO4 is chloric(VII) acid because the ox. no. of chlorine is +7.
above.
Naming compounds a – 111
Na+ O
We sometimes use Roman numbers, in brackets, to name N
compounds. We use these systematic names to distinguish
different compounds made of the same elements. For O
example, there are two types of iron chloride. We show the
difference by naming them iron(II) chloride and iron(III) b–
chloride. The numbers in brackets are the oxidation OO
numbers of the iron.
Na+ N
■■ In iron(II) chloride, the ox. no. of the iron is +2. The
compound contains Fe2+ ions. The formula is FeCl2. O
■■ In iron(III) chloride, the ox. no. of the iron is +3. The Figure 7.4 a One formula unit of ‘sodium nitrate(III)’ and
compound contains Fe3+ ions. The formula is FeCl3. b one formula unit of ‘sodium nitrate(V)’.
Salts of the common acids are usually named without
We can also use oxidation numbers to distinguish between including the ox. no. of the non-metal ion. For example,
non-metal atoms in molecules and ions. aMNnogdt(eNKaO2lSs3Oo)2t4hiissamtpwoagteandsesosiuinumomt ssnutailtftraeattethenenoootxtp.montaoag.snsoiefustmihuemsmunlefiattatrela(tiVfe(iIVt).)
has only one oxidation state.
Oxides of nitrogen
There are several oxides of nitrogen, including N2O, NO
aonx.dnNo.Oo2f. We distinguish between these according to the
the nitrogen atom. (The ox. no. of oxygen is
generally –2.)
■■ The ox. no. of N in N2O is +1. So this compound is
nitrogen(I) oxide.
■■ The ox. no. of N in NO is +2. So this compound is
nitrogen(II) oxide.
■■ Tnhiteroogxe.nn(oIV. )oof xNidine.NO2 is +4. So this compound is
Cambridge International AS Level Chemistry
Question Balancing chemical equations
using oxidation numbers
5 G ive the full systematic names of the following:
a Na2SO3 e F eSO4 We can use oxidation numbers to balance equations
b Na2SO4 f C u2O involving redox reactions. This method is especially
c Fe(NO3)2 g H 2SO3 useful where compound ions such as nitrate(V) or
d Fe(NO3)3 h M n2O7 manganate(VII) are involved.
From name to formula worked examples
You can work out the formula of a compound from 4 Copper(II) oxide (CuO) reacts with ammonia (NH3) to
its name. form copper, nitrogen (N2) and water.
Step 1 Write the unbalanced equation and identify
the atoms which change in ox. no. (shown here in red).
worked example CuO + NH3 Cu + N2 + H2O
+2 –2 –3 +1 0 0 +1 –2
3 Each formula unit of sodium chlorate(V) contains Step 2 Deduce the ox. no. changes.
one sodium ion. What is the formula of sodium
chlorate(V)? ox. no. change = +3
We know that:
■■ sodium has an ox. no. of +1 CuO + NH3 → Cu + N2 + H2O
■■ oxygen has an ox. no. of –2 +2 –3 00
112 ■■ the ox. no. of chlorine is +5
■■ the chlorate(V) ion has a charge of 1– (to balance
the 1+ charge of the sodium). ox. no. change = –2
Step 3 Balance the ox. no. changes.
We can work out the formula of the chlorate(V) ion
from the oxidation numbers of oxygen and chlorine ox. no. change = 2 × (+3) = +6
(let n be the number of oxygen atoms):
ox. no.(Cl) + ox. no.(O) = –1
+5 n × (–2) = –1
n = 3
3CuO + 2NH3 → 3Cu + N2 + H2O
So the chlorate(V) ion is ClO3– and sodium chlorate(V)
is NaClO3. 3 × (+2) 2 × (–3) 0 0
ox. no. change = 3 × (–2) = –6
Question The change in ox. nos. are –2 for the copper and +3
for the nitrogen. To balance the ox. no. changes, we
6 G ive the formulae of: need to multiply the copper by 3 and the nitrogen in
a sodium chlorate(I) the ammonia by 2. The total ox. no. changes are then
b iron(III) oxide balanced (–6 and +6). Note that we do not multiply
c potassium nitrate(III) the N2 by 2 because there are already two atoms of
d phosphorus(III) chloride. nitrogen present. Once these ratios have been fixed
you must not change them.
Step 4 Balance the atoms.
There are sbiaxlhayndcreodgwenithatsoimx osninththeeri2gNhHt (3aosn3Hth2eOl)e. Tfth. is
These are
also balances the number of oxygen atoms. The final
equation is
3CuO + 2NH3 3Cu + N2 + 3H2O
Chapter 7: Redox reactions
worked examples (continued) Question
5 Manganate(VII) ions ((MHn+)Oto4–f)orremacMt wn2i+thioFnes2,+Fioe3n+sioinns 7 U se the oxidation number method to balance these
the presence of acid equations.
and water. a H2SO4 + HI S + I2 + H2O
b HBr + H2SO4 Br2 + SO2 + H2O
Step 1 Write the unbalanced equation and identify c V3+ + I2 + H2O VO2+ + I– + H+
the atoms that change in ox. no.
MnO4– + Fe2+ + H+ Mn2+ + Fe3+ + H2O
+7 –2 +2 +1
+2 +3 +1 –2
Step 2 Deduce the ox. no. changes. Summary
ox. no. change = –5
■ Redox reactions can be explained in terms of:
MnO4– + Fe2+ + H+ → Mn2+ + Fe3+ + H2O – increase in oxidation number (oxidation
+7 +2 +2 +3 state), which is oxidation
– decrease in oxidation number, which is
ox. no. change = +1 reduction.
Step 3 Balance the ox. no. changes. ■ Oxidation numbers can be used to balance
ox. no. change = 1 × (–5) = –5 equations.
MnO4– + 5Fe2+ + H+ → Mn2+ + 5Fe3+ + H2O ■ Redox reactions can be explained in terms
of electron loss (oxidation) or electron gain
(reduction).
+7 +2 +2 +3
113
ox. no. change = 5 × (+1) = +5
Step 4 Balance the charges.
Initially ignore the hydrogen ions, as these will be used
to balance the charges.
■ The total charge on the other reactants is:
(1–)(from MnO4–) + (5 × 2+)(from 5Fe2+) = 9+
■ The total charge on the products is:
(2+)(from Mn2+) + (5 × 3+)(from 5Fe3+) = 17+
■ To balance the charges we need 8 H+ ions on
the left.
MnO4– + 5Fe2+ + 8H+ Mn2+ + 5Fe3+ + H2O
Step 5 Balance the hydrogen atoms in the water.
MnO4– + 5Fe2+ + 8H+ Mn2+ + 5Fe3+ + 4H2O
cambridge International AS Level chemistry
End-of-chapter questions
1 In the industrial production of nitric acid the following changes take place to the nitrogen.
stage 1 stage 2 stage 3 stage 4
N2 NH3 NO NO2 HNO3
a Give the oxidation number of the nitrogen atom in each molecule. [5]
b For each stage, state whether oxidation or reduction has taken place. In each case explain your answer. [2]
c Give the full systematic name for NO2. [1]
d Nitric acid, HNO3, reacts with red phosphorus.
[5]
P + 5HNO3 H3PO4 + 5NO2 + H2O [1]
By referring to oxidation number changes, explain why this is a redox reaction.
e Explain why nitric acid can be described as an oxidising agent in this reaction.
Total = 14
2 Calcium reacts with cold water to form calcium hydroxide, Ca(OH)2, and hydrogen, H2.
a State the oxidation number of calcium in:
i calcium metal [1]
ii calcium hydroxide. [1]
b State the oxidation number of hydrogen in:
i water [1]
ii hydrogen gas. [1]
114 c Write two half-equations for the reaction between water and calcium hydroxide to show:
i the change from calcium to calcium ions [1]
ii the change from water to hydroxide ions and hydrogen. [1]
d In which one of the half-equations in part c is a reduction occurring? Give a reason for your answer. [1]
e Write a balanced equation for the reaction of calcium with water. [1]
f Explain the role played by water in this reaction. [1]
Total = 9
3 The unbalanced equation for the reaction of sulfur dioxide with bromine is shown below.
SO2 + Br2 + H2O SO42– + Br– + H+
a State the oxidation number of sulfur in:
i SO2 [1]
ii SO42– [1]
b State the oxidation number of bromine in:
i Br2 [1]
ii Br– [1]
[1]
c Identify the reducing agent in this reaction. Give a reason for your answer.
[1]
d State the change in oxidation number for: [1]
[2]
i each S atom
ii each bromine atom.
e Construct a balanced equation for this reaction.
Total = 9
Chapter 7: Redox reactions
4 Aluminium reacts with hydrochloric acid to form aluminium chloride, AlCl3, and hydrogen. This is a redox reaction. [3]
a Explain in term of electrons, what is meant by a redox reaction. [1]
b i Write a half-equation to show aluminium changing to aluminium ions. [1]
ii Write a second half-equation to show what happens to the hydrogen ions from the acid. [1]
iii What is the change in oxidation number when a hydrogen ion turns into a hydrogen atom? [1]
c Construct a balanced ionic equation for the reaction between aluminium atoms and hydrogen ions.
Total = 7
5 Iodine, I2, reacts with thiosulfate ions, S2O32– to form iodide ions, I–, and tetrathionate ions, S4O62–.
I2 + 2S2O32– 2I– + S4O62–
a State the oxidation number of each sulfur atom in:
i a S2O32– ion [1]
ii a S4O62– ion. [1]
b Explain in terms of electron transfer why the conversion of iodine to iodide ions is a reduction reaction. [1]
c When a salt containing iodide ions is warmed with concentrated sulfuric acid and MnO2, iodine is evolved.
2I– + MnO2 + 6H+ + 2SO42– I2 + Mn2+ + 2HSO4– + 2H2O
i State the systematic name for MnO2. [1]
ii What is the oxidation number of S in the SO42– ion? [1]
iii Which reactant gets oxidised in this reaction? Explain your answer by using oxidation numbers. [1]
iv Which substance is the oxidising agent? Explain your answer. [1]
Total = 7 115
6 The compound KBrO3 decomposes when heated.
2KBrO3 2KBr + 3O2
a State the oxidation numbers of bromine in:
i KBrO3 [1]
ii KBr. [1]
[3]
b Explain using oxidation numbers why this reaction is a redox reaction. [1]
c State the systematic name of KBrO3. [1]
d When KBrO3 reacts with hydrazine, N2H4, nitrogen gas is evolved. [2]
[3]
2KBrO3 + 3N2H4 2KBr + 3N2 + 6H2O Total = 12
i What is the oxidation number change of the bromine atom when KBrO3 is converted to KBr?
ii What is the oxidation number change for each nitrogen atom when N2H4 is converted to N2?
iii Use your answers to i and ii to explain why 2 moles of KBrO3 react with 3 moles of N2H4.
116
chapter 8:
equilibrium
Learning outcomes ■■ calculate:
You should be able to: – the value of equilibrium constants in terms of
concentrations or partial pressures
■■ explain what is meant by a reversible reaction and
dynamic equilibrium – the quantities of substances present at
equilibrium
■■ state Le Chatelier’s principle and apply it to deduce
qualitatively the effect of changes in temperature, ■■ describe and explain the conditions used in the
concentration or pressure on a system at equilibrium Haber process and the Contact process
■■ state whether changes in temperature, ■■ show understanding of, and use, the Brønsted–
concentration or pressure or the presence of a Lowry theory of acids and bases
catalyst affect the value of the equilibrium constant
for a reaction ■■ explain qualitatively the differences in behaviour
between strong and weak acids and bases and the
■■ deduce expressions for equilibrium constants in pH values of their aqueous solutions in terms of the
terms of concentrations, Kc, and partial pressure, Kp extent of dissociation.
Reversible reactions and A reaction in which the products can react to re-form the
equilibrium original reactants is called a reversible reaction. In this
case heating and adding water are not being carried out
Reversible reactions at the same time. However, there is a type of chemical
reaction in which the forward reaction and the backward
Some reactions can be reversed. For example, when reaction take place at the same time.
blue, hydrated copper(II) sulfate is heated, it loses its
water of crystallisation and changes to white, anhydrous In many chemical reactions the reactants are not
copper(II) sulfate. used up completely. Some products are formed but the
maximum theoretical yield is not obtained. A mixture
CuSO4.5H2O(s) CuSO4(s) + 5H2O(l) of products and reactants is formed. The products react
together to re-form reactants at the same time as the
hydrated copper(II) anhydrous reactants are forming products. This type of reversible
sulfate copper(II) sulfate reaction is called an equilibrium reaction. We show that
equilibrium reactions are reversible by the sign .
This is called the forward reaction.
When water is added to anhydrous copper(II) sulfate, For example, consider the reaction between hydrogen
and iodine carried out in a sealed glass tube at 400 °C:
the reaction is reversed.
CuSO4(s) + 5H2O(l) CuSO4.5H2O(s) H2(g) + I2(g) 2HI(g)
This is called the backward (or reverse) reaction. Molecules of hydrogen iodide are breaking down to
We can show these two reactions in the same equation hydrogen and iodine at the same rate as hydrogen
and iodine molecules are reacting together to form
by using two arrows. hydrogen iodide.
CuSO4.5H2O(s) CuSO4(s) + 5H2O(l)
hydrogen atom iodine atom
Cambridge International AS Level Chemistry
When fizzy drinks are made, carbon dioxide gas is Figure 8.4 shows what happens when 5.00 mol of
dissolved in the drink under pressure. When you take the hydrogen molecules and 5.00 mol of iodine molecules
lid off a bottle of fizzy drink, bubbles of carbon dioxide react at 500 K in a vessel of volume 1 dm3. As time
suddenly appear. When you put the lid back on, the passes, the purple colour of the iodine vapour fades
bubbles stop. This is because of the equilibrium until equilibrium is reached. At equilibrium the mixture
contains 0.68 mol of iodine, 0.68 mol of hydrogen and
CO2(g) CO2(aq) 8.64 mol of hydrogen iodide.
The forward reaction happens during manufacture and the Concentration of reagent / 10–3 mol dm–3 10.0 equilibrium reached 8.64
backward reaction happens on opening. 9.0 0.68
8.0 line shows increase in
Characteristics of equilibrium 7.0 concentration of hydrogen
6.0 iodide as the equilibrium
An equilibrium reaction has four particular features under 5.0 is established
constant conditions: 4.0
■■ it is dynamic 3.0 line shows decrease in
■■ the forward and reverse reactions occur at the same rate 2.0 concentration of both
■■ the concentrations of reactants and products remain 1.0 hydrogen and iodine as the
0.0 equilibrium is established
constant at equilibrium
■■ it requires a closed system. 0 Time
1 It is dynamic
The phrase dynamic equilibrium means that the
molecules or ions of reactants and products are Figure 8.4 The changes in the concentrations of reagents
as 5.00 mol of each of hydrogen and iodine react to form
an equilibrium mixture with hydrogen iodide in a vessel of
continuously reacting. Reactants are continuously being
118 changed to products and products are continuously being volume 1 dm3.
changed back to reactants.
Figure 8.5 shows that the same equilibrium can be
2 The forward and backward reactions occur achieved when 10.00 mol of hydrogen iodide molecules
at the same rate decompose to iodine and hydrogen iodide. You can see
that the same equilibrium concentrations of all three
At equilibrium the rate of the forward reaction equals molecules are achieved.
the rate of the backward reaction. Molecules or ions
of reactants are becoming products, and those in the
products are becoming reactants, at the same rate. Concentration of reagent / 10–3 mol dm–3 10.0 equilibrium reached
9.0 8.64
0.68
3 The concentrations of reactants and 8.0 line shows decrease in
products remain constant at equilibrium
7.0 concentration of hydrogen
The concentrations remain constant because, at iodide as the equilibrium
equilibrium, the rates of the forward and backward 6.0 is established
reactions are equal. The equilibrium can be approached 5.0
from two directions. For example, in the reaction
4.0 line shows increase in
H2(g) + I2(g) 2HI(g) 3.0 concentration of both hydrogen
2.0 and iodine as the equilibrium
1.0 is established
We can start by either: 0.0 Time
0
■■ using a mixture of colourless hydrogen gas and purple Figure 8.5 The changes in the concentrations of reagents
iodine vapour, or as 10.00 mol of hydrogen iodide react to form an equilibrium
mixture with hydrogen and iodine gases in a vessel of 1 m3.
■■ using only colourless hydrogen iodide gas.
Chapter 8: Equilibrium
Question Question 119
1 These questions relate to the information in Figure 8.5. 2 A beaker contains saturated aqueous sodium
a W hy are the concentrations of iodine and hydrogen
chloride solution in contact with undissolved solid
at equilibrium the same? sodium chloride. Sodium ions and chloride ions are
b D escribe how the depth of colour of the reaction constantly moving from solid to solution and from
solution to solid.
mixture changes as time progresses. a i Explain why this is a closed system.
c E xplain why there must be 8.64 mol of hydrogen ii E xplain why the concentration of the saturated
iodide molecules in the equilibrium mixture if sodium chloride solution does not change,
0.68 mol of iodine are present. even though ions are still moving into the
solution from the solid.
4 Equilibrium requires a closed system b B romine is a reddish-brown liquid that vaporises
at room temperature. Some liquid bromine is put
A closed system is one in which none of the reactants or in a closed jar. The colour of the bromine vapour
products escapes from the reaction mixture. In an open above the liquid gets darker and darker until the
system some matter is lost to the surroundings. Figure depth of colour remains constant. Bromine liquid
8.6 shows the difference between a closed system and an still remains in the jar. Explain what is happening in
open system when calcium carbonate is heated at a high terms of changes in concentration of the bromine
temperature in a strong container. molecules in the vapour.
Many chemical reactions can be studied without Changing the position of
placing them in closed containers. They can reach equilibrium
equilibrium in open flasks if the reaction takes place
entirely in solution and no gas is lost. Position of equilibrium
a The position of equilibrium refers to the relative amounts of
products and reactants present in an equilibrium mixture.
CaCO3(s) CaO(s) + CO2(g) ■■ If a system in equilibrium is disturbed (e.g. by a change
CaCO3(s) CaO(s) CO2(g)
in temperature) and the concentration of products is
b increased relative to the reactants, we say that the position
of equilibrium has shifted to the right.
CaCO3(s) → CaO(s) + CO2(g) ■■ If the concentration of products is decreased relative to
Figure 8.6 a A closed system. No carbon dioxide escapes. the reactants, we say that the position of equilibrium has
The calcium carbonate is in equilibrium with calcium shifted to the left.
oxide and carbon dioxide. b An open system. The calcium
carbonate is continually decomposing as the carbon dioxide is Le Chatelier’s principle
lost. The reaction eventually goes to completion.
Changes in both concentration and temperature affect
the position of equilibrium. When any of the reactants or
products are gases, changes in pressure may also affect the
position of equilibrium. French chemist Henri Le Chatelier
(1850–1936) observed how these factors affect the position
of equilibrium. He put forward a general rule, known as
Le Chatelier’s principle:
If one or more factors that affect an equilibrium is
changed, the position of equilibrium shifts in the direction
that reduces (opposes) the change.
Cambridge International AS Level Chemistry
We can predict the effect of changing concentration Question
and pressure by referring to the stoichiometric equation
for the reaction. We can predict the effect of changing 3 a Use this reaction:
the temperature by referring to the enthalpy change of
the reaction. CH3COOH(l) + C2H5OH(l)
CH3COOC2H5(l) + H2O(l)
How does change in concentration affect
the position of equilibrium? Explain what happens to the position of
equilibrium when:
When the concentration of one or more of the reactants
is increased: i more CH3COOC2H5(l) is added
■■ the system is no longer in equilibrium ii some C2H5OH(l) is removed.
■■ the position of equilibrium moves to the right to reduce the b Use this reaction:
effect of the increase in concentration of reactant Ce4+(aq) + Fe2+(aq) Ce3+(aq) + Fe3+(aq)
■■ more products are formed until equilibrium is restored.
Explain what happens to the position of
When the concentration of one or more of the products equilibrium when:
is increased:
■■ the system is no longer in equilibrium i the concentration of Fe2+(aq) ions is increased
■■ the position of equilibrium moves to the left to reduce the ii water is added to the equilibrium mixture.
effect of the increase in concentration of product
■■ more reactants are formed until equilibrium is restored.
For example, look at the reaction:
120 CH3COOH(l) + C2H5OH(l) CH3COOC2H5(l) + H2O(l)
ethanoic acid ethanol ethyl ethanoate water
What happens when we add more ethanol?
■■ The concentration of ethanol is increased.
■■ According to Le Chatelier’s principle, some of the ethanol
must be removed to reduce the concentration of the
added ethanol.
■■ The position of equilibrium shifts to the right.
■■ More ethanol reacts with ethanoic acid and more ethyl
ethanoate and water are formed.
What happens when we add more water? Figure 8.7 Stalactites and stalagmites are formed as a result
of water passing through rocks containing calcium carbonate.
■■ The concentration of water is increased. The solution running through these rocks contains water,
■■ According to Le Chatelier’s principle, some of the water dissolved carbon dioxide and calcium hydrogencarbonate:
must be removed to reduce the concentration of the CaCO3(s) + H2O(l) + CO2(aq) Ca(HCO3)2(aq)
added water.
■■ The position of equilibrium shifts to the left. When droplets of this mixture are formed on the roof of the
■■ So more water reacts with ethyl ethanoate and more cave, some of the carbon dioxide in the droplets escapes
ethanoic acid and ethanol are formed. into the air. The position of equilibrium shifts to the left and
calcium carbonate is deposited.
What happens when we remove some water?
■■ The concentration of water is decreased.
■■ According to Le Chatelier’s principle, some water must be
added to increase its concentration.
■■ The position of equilibrium shifts to the right.
■■ So more ethanoic acid reacts with ethanol and more water
and ethyl ethanoate are formed.
The effect of pressure on the position a Initial pressure
of equilibrium
X +Y Z 13 molecules
Change in pressure only affects reactions where gases b Pressure is increased contributing to
are reactants or products. The molecules or ions in solids the pressure.
and liquids are packed closely together and cannot be
compressed very easily. In gases, the molecules are far More molecules
apart (Figure 8.8). of Z are formed,
reducing the total
number of molecules
from 13 to 11.
gas higher pressure
means higher
concentration
For example, consider the reaction:
2SO2(g) + O2(g) 2SO3(g)
lower pressure means There are three moles of gas molecules on the left of the
lower concentration equation and two on the right.
What happens when we increase the pressure?
Figure 8.8 Pressure has a considerable effect on the ■■ The molecules are closer together, because the pressure
concentration of gases. is higher.
The pressure of a gas is caused by the molecules hitting the ■■ According to Le Chatelier’s principle, the reaction must
walls of the container. Each molecule in a mixture of gases shift in the direction that reduces the number of molecules
contributes towards the total pressure. So, at constant of gas.
temperature, the more gas molecules there are in a given
volume, the higher the pressure. ■■ The position of equilibrium shifts to the right.
■■ So more SO2 reacts with O2 to form SO3.
Figure 8.9 shows what happens when we increase the
pressure on the reaction represented by: What happens when we decrease the pressure?
X(g) + Y(g) Z(g) ■■ The molecules are further apart, because the pressure
is lower.
1 mol 1 mol 1 mol
■■ According to Le Chatelier’s principle, the reaction must
In this reaction there are two moles of gas on the left shift in the direction that increases the number of molecules
and one on the right. When the pressure is increased at of gas.
constant temperature:
■■ The position of equilibrium shifts to the left.
■■ the molecules are closer together, because the pressure ■■ dSoecmomorpeoSsOit2ioannodfOS2Om3 moloecleucluesleasr.e formed by the
has increased
Table 8.1 summarises the effect of changes in pressure on
■■ the position of equilibrium shifts to minimise this increase two other gas reactions.
■■ it shifts in the direction of fewer gas molecules (in the note that:
direction that opposes the increase in pressure) ■■ if there are equal numbers of molecules of gas on each side
■■ more product, Z, is formed from X and Y until equilibrium of the equation, the position of equilibrium is not affected
by a change in pressure
is re-established.
■■ in a reaction involving gases and solids (or liquids), it is only
the molecules of gases that count when determining how
pressure affects the position of equilibrium.
Cambridge International AS Level Chemistry
Change in Fewer molecules of More molecules of position of equilibrium shifts to the right. We can explain
pressure gas on right gas on right this using Le Chatelier’s principle:
N2(g) + 3H2(g) N2O4(g) 2NO2
pressure ■■ an increase in temperature increases the energy of
increase 2NH3(g) equilibrium position the surroundings
pressure equilibrium position shifts towards reactants:
decrease shifts towards products: more N2O4 forms ■■ according to Le Chatelier’s principle, the reaction will go in
more NH3 forms equilibrium position the direction that opposes the increase in energy
equilibrium position shifts towards products:
shifts towards reactants: more NO2 forms ■■ so the reaction will go in the direction in which energy is
more N2 and H2 form absorbed, which is the endothermic reaction
Table 8.1 The effect of changes in pressure on gas reactions. ■■ the position of equilibrium shifts to the right, producing
more H2 and I2.
If an endothermic reaction is favoured by an increase in
temperature, an exothermic reaction must be favoured by
a decrease in temperature:
Question ■■ a decrease in temperature decreases the energy of
the surroundings
4 a Predict the effect of increasing the pressure on the
following reactions: ■■ according to Le Chatelier’s principle, the reaction will go in
the direction that opposes the decrease in energy
i N2O4(g) 2NO2(g)
■■ so the reaction will go in the direction in which energy is
ii CaCO3(s) CaO(s) + CO2(g) released, which is the exothermic reaction.
b P redict the effect of decreasing the pressure on Table 8.3 summarises the effect of temperature changes
the reaction: on the position of equilibrium for endothermic and
exothermic reactions.
2NO2(g) 2NO(g) + O2(g)
122 Temperature Endothermic Exothermic reaction
The effect of temperature on the position change reaction 2SO2(g) + O2(2gS)O3(g)
of equilibrium Temperature 2HI H2 + I2 position of equilibrium
increase position of equilibrium shifts towards
The decomposition of hydrogen iodide is an endothermic shifts towards products:
reaction.
2HI H2 + I2 ΔHr = +9.6 kJ mol–1 more H2 and I2 formed reactants: more SO2
position of equilibrium and O2 formed
The effect of temperature on the equilibrium concentration of shifts towards reactant:
hydrogen iodide and hydrogen at equilibrium for the forward Temperature more HI formed position of equilibrium
decrease shifts towards product:
more SO3 formed
reaction is shown in Table 8.2. Table 8.3 Effect of temperature on endothermic and
exothermic reactions.
Temperature / °C Equilibrium Equilibrium Question
concentration of concentration
HI / mol dm–3 omfoHl 2d(mor–3I2) / 5 a Predict the effect of increasing the temperature on
the reaction:
25 0.934 0.033
230 0.864 0.068 H2(g) + CO2(g) H2O(g) + CO(g)
430 0.786 0.107
490 0.773 0.114 ΔHr = +41.2 kJ mol–1
b In the reaction
Table 8.2 Effect of temperature on the decomposition of Ag2CO3(s) Ag2O(s) + CO2(g)
hydrogen iodide.
increasing the temperature increases the amount
You can see from Table 8.2 that, as the temperature of carbon dioxide formed at constant pressure. Is
increases, the concentration of product increases. The this reaction exothermic or endothermic? Explain
your answer.
Chapter 8: Equilibrium
Do catalysts have any effect on the You can see that this expression gives an approximately
position of equilibrium? constant value close to about 160 whatever the starting
concentrations of H2, I2 and HI.
A catalyst is a substance that increases the rate of a We call this constant the etqhuatilcibonricuemntrcaotniosntasnhta, vKec.bTehene
chemical reaction. Catalysts speed up the time taken to subscript ‘c’ refers to the fact
reach equilibrium, but they have no effect on the used in the calculations.
position of equilibrium once this is reached. This is There is a simple relationship that links Kc to the
because they increase the rate of the forward and reverse equilibrium concentrations of reactants and products
reactions equally. and the stoichiometry of the equation. This is called an
equilibrium expression.
Equilibrium expressions and the For a general reaction:
equilibrium constant, Kc
mA + nB pC + qD
Equilibrium expressions
(where m, n, p and q are the number
When hydrogen reacts with iodine in a closed tube at of moles in the equation)
500 K, the following equilibrium is set up:
H2 + I2 2HI concentration of product D
Table 8.4 shows the relationship between the equilibrium Kc = [C]p [D]q number of moles of product D
claosntcceonlutrmatniornesfeorftHo 2t,hIe2caonndceHnIt.rTathieonsq, iunarmeoblr damck–e3t,soifn the [A]m [B]n number of moles of reactant B
the substance inside the brackets. The results are obtained
as follows: concentration of reactant B
■■ several tubes are set up with different starting 123
concentrations of hydrogen and iodine
worked examples
■■ the contents of the tubes are allowed to reach equilibrium
at 500 K 1 Write an expression for Kc:
■■ the concentrations of hydrogen, iodine and hydrogen iodide
at equilibrium are determined.
The last column in Table 8.4 shows the number we get N2(g) + 3H2(g) 2NH3(g)
bpyaratricruanlagrinwgayth. We ceognectetnhtirsateixopnrseossfiHon2,bIy2 and HI in a Kc = _[ _N[_N_2_]H_ [_H3__]22_]_ 3_
taking the square
of the concentration of hydrogen iodide and dividing it by
the concentrations of hydrogen and iodine at equilibrium. 2 Write an expression for Kc:
So for the first line of data in Table 8.4: 2SO2(g) + O2(g) 2SO3(g)
_ [_H_[_H_2_]I_ []_I2_2 _]_ = _ ( _0_._6_8__×_(_8_1._60_ 4–_3_)×_(_ 0_1_.06_–_8_3)_×2__1_0__–_3_)_ Kc = _[ _S_O[_S_2_O]_2_3 [_]O_2_ 2_]_
= 161
Ceqounicleibnrtiruamti o/ nmoofl Hdm2 a–t3 Ceqounicleibnrtiruamti o/ nmoofl Id2mat–3 Concentration of HI at _[_H[_H_2_]I_ []_I2_2_ ]_
0.68 × 10–3 0.68 × 10–3 equilibrium / mol dm–3 161
0.50 × 10–3 0.50 × 10–3 159
1.10 × 10–3 2.00 × 10–3 8.64 × 10–3 161
2.50 × 10–3 0.65 × 10–3 6.30 × 10–3 160
18.8 × 10–3
16.1 × 10–3
Table 8.4 The relationship between the equilibrium concentrations of H2, I2 and HI in the reaction H2 + I2 2HI.
Cambridge International AS Level Chemistry
In equilibrium expressions involving a solid, we ignore the Some examples of equilibrium
solid. This is because its concentration remains constant, calculations
however much solid is present. For example:
Ag+(aq) + Fe2+(aq) Ag(s) + Fe3+(aq) worked examples
The equilibrium expression for this reaction is: 5 In this calculation we are given the number of moles
of each of the reactants and products at equilibrium
Kc = _ [ A__g_+_( a[_Fq_ e)_]3 _+[ F(_ae_q2_+)_](a_q_)_]_ together with the volume of the reaction mixture.
What are the units of Kc? Ethanol reacts with ethanoic acid to form ethyl
ethanoate and water.
In the equilibrium expression each figure within a
square bracket represents the concentration in mol dm–3. C H3COOH(l) + C2H5OH(l) CH3COOC2H5(l) + H2O(l)
The units of Kexcptrheesrseifoonr.e depend on the form of the
equilibrium ethanoic acid ethanol ethyl ethanoate water
worked examples 500 cm3 of the reaction mixture at equilibrium
contained 0.235 mol of ethanoic acid and 0.0350 mol of
3 State the units of Kc for the reaction: ethanol together with 0.182 mol of ethyl ethanoate and
0.182 mol of water. Use this data to calculate a value of
H2 + I2 2HI Kc for this reaction.
124 Kc = _[ H_[_H2_]I [_]I2_2 _] Step 1 Write out the balanced chemical equation with
Units of Kc = _ (( mm__oo_ll_ dd_mm__––33_))_×× _ ((_mm__oo_ll dd__mm_––_33_)) the concentrations beneath each substance.
CH3COOH(l) + C2H5OH(l)
0_._23_55_0×_01_ 0_0_0 _0._03_55_0×_01_ 0_0_0
0.470 mol dm–3 0.070 mol dm–3
CH3COOC2H5(l) + H2O(l)
_0._18_25_0×_01_ 0_0_0 _0._18_25_0×_01_ 0_0_0
The units of mol dm–3 cancel, so Kc has no units. 0.364 mol dm–3 0.364 mol dm–3
4 State the units of Kc for the reaction: Step 2 Write the equilibrium constant for this reaction
in terms of concentrations.
2SO2(g) + O2(g) 2SO3(g) Kc = _ [ [C_CH_H_33C_C_OO_OO_H_C _ ]2 _[HC_52_]H [_H5_O2_OH_]_]
Units of Kc = _ ( m__o_l_ d_m(_m_–o3_ )l_ ×d_ m_(m_–_o3)_l ×_d_m( m_–_o3_)l_ ×d _m(_m_–_o3)_l d__m_–_3_) Step 3 Substitute the equilibrium concentrations into
the expression
= _ (_m_o__l1 d_ m__–_3_) = dm3 mol–1
Kc = _ (( 00_.._34_67_04_))_×× _ ((_00_..0_37_60_4)_) = 4.03
(to 3 significant figures)
Question Step 4 Add the correct units by referring back to the
equilibrium expression:
6 Write equilibrium expressions for the following
reactions and state the units of Kc. _ (( mm__oo_ll_ dd_mm__––33_))_×× _ ((_mm__oo_ll dd__mm_––_33_))
a CO(g) + 2H2(g) CH3OH(g) The units of mol dm–3 cancel, so Kc has no units.
Therefore Kc = 4.03.
b 4HCl(g) + O2(g) 2H2O(g) + 2Cl2(g)
Note: if there are equal numbers of moles on the top
and bottom of the equilibrium expression, you can
use moles rather than concentration in mol dm–3 in
the calculation. In all other cases, if volumes are given,
the concentrations must be calculated before they are
substituted into the equilibrium expression.
Chapter 8: Equilibrium
worked examples (continued) worked examples (continued)
6 In this example we are only given the initial 7 In this example we are given the initial and equilibrium
concentrations of the reactants and the equilibrium concentrations of the reactants but not the products.
concentration of the product.
Ethyl ethanoate is hydrolysed by water:
Propanone reacts with hydrogen cyanide as follows:
CH3COOC2H5 + H2O CH3COOH + C2H5OH
C H3COCH3 + HCN CH3C(OH)(CN)CH3 ethyl ethanoate water ethanoic acid ethanol
propanone hydrogen cyanide product 0.1000 mol of ethyl ethanoate are added to 0.1000 mol
of water. A little acid catalyst is added and the
A mixture of 0.0500 mol dm–3 propanone and mixture made up to 1 dm3 with an inert solvent. At
0.0500 mol dm–3 hydrogen cyanide is left to reach equilibrium 0.0654 mol of water are present. Calculate
equilibrium at room temperature. At equilibrium Kc for this reaction.
the concentration of the product is 0.0233 mol dm–3.
Calculate Kc for this reaction. Step 1 Write out the balanced chemical equation with
all the data underneath.
Step 1 Write out the balanced chemical equation with
all the data underneath: CH3COOC2H5 + H2O
CH3COCH3 + HCN CH3C(OH)(CN)CH3 CH3COOH + C2H5OH
initial conc. 0.1000 0.1000 0 0
initial conc. 0.0500 0.0500 0 mol dm–3 mol dm–3
mol dm–3 mol dm–3
conc. at 0.0654
Conc. at to be to be equilibrium mol dm–3
0.0233 mol dm–3
equilibrium calculated calculated
Step 2 Calculate the equilibrium concentrations of the Step 2 Calculate the unknown concentrations:
reactants. – the chemical equation shows that 1 mole of
The chemical equation shows that for every mole of CH3COOC2H5 reacts with 1 mole water, so the 125
product formed, 1 tmheoleeqoufilCibHr3iCuOmCcHo3nacnednt1ramtioolnesoafrHeCaNs 0e.q0u6i5li4b rmiuoml dcmon–3c(eanstwraetisotnarotfeCdHw3iCthOOthCe2Hsa5miseailnsoitial
are consumed. So concentrations of ethyl ethanoate and water)
follows: – the amount of water used in forming the products is
(0.1000 – 0.0654) = 0.0346 mol dm–3.
CH3COCH3; 0.0500 – 0.0233 = 0.0267 mol dm–3 The chemical equation shows that 1 mole of water
HCN; 0.0500 – 0.0233 = 0.0267 mol dm–3 formed 1 mole of ethanoic acid and 1 mole of ethanol.
So the concentrations of both the products at
Step 3 Write the equilibrium constant for this reaction equilibrium is 0.0346 mol dm–3
in terms of concentrations:
Kc =_ [[ C_C_HH_33_CC_(OO__CH_H ) _(3C_] _N[H_)C_C_HN_3]_] CH3COOC2H5 + H2O
Step 4 Substitute the equilibrium concentrations into CH3COOH + C2H5OH
the expression conc. at 0.0654 0.0654 0.0346 0.0346
equilibrium
mol dm–3
Kc = _ ( 0_._0_2_6(_07_.)0_×2_ 3(_03_.)_0 _2_6_7_) = 32.7
(to 3 significant figures) Step 3 Write the equilibrium constant for this reaction
in terms of concentrations:
Step 5 Add the correct units by referring back to the Kc = _ [[ C_CH_H_33C_C_OO_OO_H_C _] 2 [_HC_52_]H [_H5_O2_OH_]_]
equilibrium expression. Step 4 Substitute the equilibrium concentrations into
_ ( m__o_l_ d(_mm_ o_–l3_ )d_( mm__o–_3l )_d_m_–_3_) = _m_o_l_ 1d_ m__–_3 = dm3 mol–1
the expression:
So Kc = 32.7 dm3 mol–1
Kc = _ (( 00_.._00_36_44_65_))_×× _ ((_00_..0_03_6_44_65_)) = 0.280
(to 3 significant figures)
Cambridge International AS Level Chemistry
worked examples (continued) alters the position of equilibrium. It is shifted in the
direction that results in fewer gas molecules being formed.
Step 5 Add the correct units by referring back to the However, if all other conditions remain constant, the value
equilibrium expression. of Kc does not change when the pressure is altered.
_ (( mm__oo_ll_ dd_mm__––33_))_×× _ ((_mm__oo_ll dd__mm_––_33_))
Kc and temperature changes
The units of mol dm–3 cancel, so Kc has no units.
Therefore Kc = 0.280. We have seen on page 122 that for an endothermic
reaction, an increase in temperature shifts the reaction in
the direction of more products.
So for the endothermic reaction 2HI H2 + I2:
■■ the concentrations of H2 and I2 increase as the temperature
increases
Question
7 Calculate the value of Kbcefloowr t:he following reaction ■■ the concentration of HI falls as the temperature increases.
using the information
Look at how these changes affect the equilibrium expression:
H2(g) + CO2(g) H2O(g) + CO(g) Kc = _[ H_[H_2_]I _][I2_ 2_]
initial concentration of H2(g) = 10.00 mol dm–3. We see that the equilibrium constant must increase with
increasing temperature. This is because [H2] and [I2] are
initial concentration of CO2(g) = 10.00 mol dm–3. increasing and [HI] is decreasing. Table 8.5 shows how the
value of Kc for this reaction changes with temperature.
equilibrium concentration of CO(g) = 9.47 mol dm–3.
Kc and concentration changes rveaalucteaonftsKocr Temperature / K Kc (no units)
126 If all other conditions remain constant, the 300 1.26 × 10–3
does not change when the concentration of
products is altered.
Take the example of the decomposition of 500 6.25 × 10–3
hydrogen iodide. 1000 18.5 × 10–3
2HI H2 + I2 Table 8.5 Variation of Kc for the reaction 2HI H2 + I2
with temperature.
The equilibrium constant at 500 K for this reaction is
6.25 × 10–3. For an exothermic reaction, an increase in temperature
shifts the reaction in favour of more reactants.
Kc = _[ H_[H_2_]I _][I2_ 2_] = 6.25 × 10–3
When more hydrogen iodide is added to the equilibrium Now look at the exothermic reaction:
mixture, the equilibrium is disturbed.
■■ The ratio of concentrations of products to reactants in the 2SO2 + O2 2SO3
equilibrium expression decreases. ■■ tTehmepcoernacteunrteriantciorenassoefsS. O2 and O2 increase as the
■■ To restore equilibrium, both [H2] and [I2] increase and [HI] ■■ The concentration of SO3 falls as the temperature increases.
decreases. How do these changes affect the equilibrium expression?
■■ Equilibrium is restored when the values of the
Kc = _[ S_O_[_S2_O]2_ 3[_]O_2 _2]_
concentrations in the equilibrium expression are such that
the value of Kc is once again 6.25 × 10–3. We see that the equilibrium constant must decrease with
increasing temperature. This is because [SO2] and [O2] are
Kc and pressure changes increasing and [SO3] is decreasing.
Where there are different numbers of gas molecules on
each side of a chemical equation, a change in pressure
Chapter 8: Equilibrium
Question Question
8 a Deduce the effect of increase in temperature on 9 The reaction below was carried out at a pressure of
the value of Kc for the reaction: 10.00 × 104 Pa and at constant temperature.
2NO2(g) + O2(g) 2NO(g) N2(g) + O2(g) 2NO(g)
ΔHr = –115 kJ mol–1 The partial pressures of nitrogen and oxygen are both
b E xplain why increasing the concentration of 4.85 × 104 Pa.
oxygen in this reaction does not affect the value Calculate the partial pressure of the nitrogen(II) oxide,
of Kc. NO(g), at equilibrium.
Equilibria in gas reactions: the Equilibrium expressions involving partial
equilibrium constant, Kp pressures
Partial pressure We write equilibrium expressions in terms of partial
pressures in a similar way to equilibrium expressions in
For reactions involving mixtures of gases, it is easier to terms of concentrations. But there are some differences:
measure the pressure than to measure concentrations.
The total pressure in a mixture of gases is due to each ■■ we use p for partial pressure
molecule bombarding the walls of the container. At
constant temperature, each gas in the mixture contributes ■■ the reactants and products are written as subscripts after
to the total pressure in proportion to the number of moles the p
present (Figure 8.10). The pressure exerted by any one gas
in the mixture is called its partial pressure. ■■ the number of moles of particular reactants or products is
written as a power after the p
■■ square brackets are not used
■■ we give the equilibrium tceornmstsaonftptahretisaylmprbeoslsKupre(tsh).e 127
equilibrium constant in
For example, the equilibrium expression for the reaction:
N2(g) + 3H2(g) 2NH3(g)
gas A partial partial total is written as KP = _ p_N_p2_2×N_H_p_33 H__2
gas B pressure pressure pressure
= 8 Pa = 3 Pa = 11 Pa What are the units of Kp?
the volumes of the containers are the same dTehpeeunnditosnotfhperefossrumreoafrtehepaesqcuaillsi,bPriau.mTheexupnreistssioofnK. p
Figure 8.10 Each gas in this mixture contributes to the
pressure in proportion to the number of moles present.
The total pressure of a gas equals the sum of the partial worked examples
pressures of the individual gases.
8 For the reaction:
ptotal = pA + pB + pC …
N2O4(g) 2NO2(g)
where pA, pB, pC are the partial pressures of the individual
gases in the mixture. The equilibrium expression is:
Kp = _ PP_N 2N_ 2O _O2_4
The units are _ P_a_P×_a_P _a_ = Pa
Cambridge International AS Level Chemistry
worked examples (continued) worked examples (continued)
9 For the reaction: Step 3 Add the correct units.
2SO2(g) + O2(g) 2SO3(g) The units are _ P_a_P×_a_P×_a_P_×a_ P_a_ = _ P1_a_ = Pa–1
The equilibrium expression is:
Kp = _ P_S 2_ OP_2S 2_× O _3P_ _O_2 Kp = 9.1 × 10–6 Pa–1
The units are _ P_a_P×_a_P×_a_P_×a_ P_a_ = _ P1_a_ = Pa–1 11 In this calculation we are given the partial pressure
of two of the three gases in the mixture as well as the
Although the standard unit of pressure is the pascal, total pressure.
many chemists in industry use the atmosphere as the
unit of pressure. 1 atmosphere = 1.01 × 105 Pa. Using Nitrogen reacts with hydrogen to form ammonia.
‘atmospheres’ as units simplifies calculations because
the numbers used are not as large. N2(g) + 3H2(g) 2NH3(g)
Question The pressure exerted by this mixture of hydrogen,
nitrogen and ammonia at constant temperature is
10 Deduce the units of Kp for the following reactions: 2.000 × 107 Pa. Under these conditions, the partial
pressure of nitrogen is 1.490 × 107 Pa and the partial
a PCl5(g) PCl3(g) + Cl2(g) pressure of hydrogen is 0.400 × 107 Pa. Calculate the
value of Kp for this reaction.
b N2(g) + 3H2(g) 2NH3(g)
Step 1 Calculate the partial pressure of ammonia.
128 c 3Fe(s) + 4H2O(g) Fe3O4(s) + 4H2(g) We know that the total pressure is the sum of the
Calculations using partial pressures partial pressures.
Ptotal = PN2 + PH2 + PNH3
worked examples
2.000 × 107 = (1.490 × 107) + (0.400 × 107) + PNH3
So partial pressure of NH3 = 0.110 × 107 Pa
Step2 Write the equilibrium expression for the
reaction in terms of partial pressures.
Kp = _ P_NP_2_N 2× _ HP_3 _H 3 _2
10 In this example we are given the partial pressure of Step 3 Substitute the equilibrium concentrations into
each gas in the mixture. the expression.
In the reaction Kp = _ ( 1_._4_9_0_×_(_01_.01 _71_)0_×_× (_01_.04_70_)02__×_1_0_7_)_3
2SO2(g) + O2(g) 2SO3(g)
the equilibrium partial pressures at constant Step 4 Add the correct units.
tSeOm3 p=e8r.a0t×ur1e06a Prea.SCOa2l=cu1l.a0te× 106 Pa, O2 = 7.0 × 106 Pa, The units are _ P _a_×__PP_aa_×_× _PP_aa_ _×_P__a = _ P_1a_ _2 = Pa–2
reaction. the value of Kp for this
Kp = 1.27 × 10–15 Pa–2
Step 1 Write the equilibrium expression for the
reaction in terms of partial pressures.
Kp = _ P_2 S_ OP_2S2 _× O_3P_ _O_2
Step 2 Substitute the equilibrium concentrations into
the expression.
Kp = _ ( 1_._0_×_(_180_.0 _6)_×2_ ×1_07_6._0)2_×_1_0_6_ = 9.1 × 10–6 Pa–1
Chapter 8: Equilibrium
Question ■■ What happens if we decrease the temperature?
– A decrease in temperature decreases the energy of
11 The information below gives the data for the reaction the surroundings.
of hydrogen with iodine at 500 °C. – The reaction will go in the direction in which energy
is released.
H2(g) + I2(g) 2HI(g) – Energy is released in the exothermic reaction, in
which the position of equilibrium favours ammonia
The table shows the initial partial pressures and the production.
partial pressures at equilibrium of hydrogen, iodine – This shifts the position of equilibrium to the right. The
and hydrogen iodide. The total pressure was constant value of Kp increases.
throughout the experiment.
■■ What happens if we remove ammonia by condensing it to a
Partial pressures / Pa liquid? We can do this because ammonia has a much higher
boiling point than hydrogen and nitrogen.
hydrogen iodine hydrogen – The position of equilibrium shifts to the right to replace
iodide the ammonia that has been removed.
– More ammonia is formed from hydrogen and nitrogen to
Initially 7.27 × 106 4.22 × 106 0 keep the value of Kp constant.
At equilibrium
3.41 × 106 7.72 × 106
a D educe the partial pressure of the iodine Equilibrium and the production of
at equilibrium. sulfuric acid
b C alculate the value of Kp for this reaction, including The synthesis of sulfuric acid is carried out by the Contact
the units. process. The main equilibrium reaction involved is:
Equilibria and the chemical 2SO2(g) + O2(g) 2SO3(g) ΔHr = –197 kJ mol–1 129
industry
We can use Le Chatelier’s principle to show how to get the
An understanding of equilibrium is important in the best yield of sulfur trioxide.
chemical industry. Equilibrium reactions are involved
in some of the stages in the large-scale production of ■■ What happens when we increase the pressure?
ammonia, sulfuric acid and many other chemicals. – When we increase the pressure, the reaction goes in the
direction that results in fewer molecules of gas being
Equilibrium and ammonia production formed, to reduce the pressure.
– There are three molecules of gas on the left-hand side
The synthesis of ammonia is carried out by the Haber and two on the right-hand side, so the equilibrium shifts
process. The equilibrium reaction involved is: towards the right.
N2(g) + 3H2(g) 2NH3(g) ΔHr = –92 kJ mol–1
We can use Le Chatelier’s principle to show how to get the However, in practice, the reaction is carried out at just
best yield of ammonia. At high temperatures, when the above atmospheric pressure. This is because the value of Kp
reaction is faster, the position of equilibrium is to the left is very high. The position of equilibrium is far over to the
because the reaction is exothermic (ΔH is negative). right even at atmospheric pressure. Very high pressure is
unnecessary, and is not used as it is expensive.
■■ What happens if we increase the pressure?
■■ What happens if we decrease the temperature?
– When we increase the pressure, the reaction goes in – Decreasing the temperature shifts the position of
the direction that results in fewer molecules of gas equilibrium to the right.
being formed. – A decrease in temperature decreases the energy of the
surroundings so the reaction will go in the direction in
– The equilibrium shifts in the direction that reduces which energy is released.
the pressure. – This is the exothermic reaction, in which the position of
equilibrium favours SO3 production. The value of
– In this case there are four molecules of gas on the Kp increases.
left-hand side and two on the right-hand side. So the
equilibrium shifts towards the right.
– The yield of ammonia increases.
Cambridge International AS Level Chemistry
dASOolte3hsiosnuorgethmaftofhevecetdStObh3yeiaesbqasuboislroibbrirbnieugdmitinisniag9nc8oi%fnictsaiunnultflouyurbiscepcaracouicdsee.stsh, ethis a b
position of equilibrium is already far over to the right.
Question
12 The Haber process for the synthesis of ammonia may
operate at a temperature of 450 °C and pressure of
1.50 × 107 Pa using an iron catalyst. Figure 8.11 a The sour taste of lemons is due to citric acid
and that of vinegar is due to ethanoic acid. b Washing soda
is a base used to soften water prior to washing clothes. A
N2(g) + 3H2(g) 2NH3(g) solution of washing soda feels soapy.
ΔHr = –92 kJ mol–1
a S uggest why the temperature of more than 450 °C
is not used even though the rate of reaction would
be faster. Name of acid Formula Ions formed in water
b S uggest why the reaction is carried out at a high hydrochloric acid HCl H+ + Cl–
pressure rather than at normal atmospheric nitric acid HNO3 H+ + NO3–
pressure. Explain your answer.
c E xplain why the removal of ammonia as soon sulfuric acid H2SO4 2H+ + SO42–
as it is formed is an important part of this
industrial process.
d W hen the ammonia has been removed, ethanoic acid CH3COOH CH3COO– + H+
130 why doesn’t it decompose back to nitrogen benzoic acid C6H5COOH C6H5COO– + H+
and hydrogen?
Table 8.6 Formulae and ions of some common acids.
Acid–base equilibria A better definition of an acid is a substance that releases
hydrogen ions when it dissolves in water. For example:
Some simple definitions of acids
and bases HCl(g) + aq H+(aq) + Cl–(aq)
A very simple definition of an acid is that it is a substance The formulae for a number of bases are given in
which neutralises a base. A salt and water are formed. Table 8.7. Many metal oxides or hydroxides are bases.
Some bases dissolve in water to form hydroxide ions
2HCl(aq) + CaO(s) CaCl2(aq) + H2O(l) in solution. A base that is soluble in water is called an
alkali. For example:
acid base
The equation above also shows us a very simple definition NaOH(g) + aq Na+(aq) + OH–(aq)
of a base. A base is a substance that neutralises an acid.
Some alkalis are formed by the reaction of a base with
If we look at the formulae for a number of acids in water. When ammonia gas dissolves in water, some of
Table 8.6, we see that they all contain hydrogen atoms. the ammonia molecules react with water molecules.
When the acid dissolves in water, it ionises and forms Hydroxide ions are released in this reaction.
hydrogen ions. Note that in organic acids such as
carboxylic acids (see page 231) only some of the hydrogen NH3(g) + H2O(l) NH4+(aq) + OH–(aq)
atoms are capable of forming ions.
Aqueous ammonia is therefore an alkali. We can also see
from the equation above that the ammonia has accepted a
hydrogen ion to become NH4+. So a wider definition of a
base is a substance that accepts hydrogen ions.
Chapter 8: Equilibrium
Name of base Formula A Brønsted–Lowry acid is a proton donor.
calcium oxide CaO A Brønsted–Lowry base is a proton acceptor.
copper(II) oxide
sodium hydroxide CuO When hydrochloric acid is formed, hydrogen chloride gas
calcium hydroxide dissolves in water and reacts to form hydroxonium ions,
ammonia NaOH wHa3Oter+,iasnindvcohlvloerdidine ions (Figure 8.12). You can see that the
the reaction.
Ca(OH)2
NH3
Table 8.7 The formulae of some common bases.
HCl(g) + H2O(l) H3O+(aq) + Cl–(aq)
Question Hydrochloric acid is an acid because it donates a proton
to water. This means that water is acting as a Brønsted–
13 a Write an equation to show potassium hydroxide Lowry base. The water is accepting a proton.
dissolving in water.
H+ donated
b W rite an equation for liquid nitric acid dissolving
in water.
c Write ionic equations for: HCl(g) + Hba2sOe (l) H3O+(aq) + Cl–(aq)
i t he reaction in aqueous solution between
acid
sodium hydroxide and nitric acid
ii t he reaction in aqueous solution between Water can also act as an acid. When ammonia reacts with
water, it accepts a proton from the water and becomes an
potassium hydroxide and hydrochloric acid. NH4+ ion (Figure 8.13).
The Brønsted–Lowry theory of acids H+ donated 131
and bases
The definitions of acids and bases given above are limited
to reactions taking place in water. In 1923 the Danish NH3(g) + H2O(l) NH4+(aq) + OH–(aq)
chemist J. Brønsted and the English chemist T. Lowry base acid
suggested a more general definition of acids and bases.
This definition is based on the idea that in an acid–base Substances like water, which can act as either acids or
reaction, a proton is transferred from an acid to a base (a bases, are described as amphoteric.
proton is a hydrogen ion, H+).
HCl(g) + H2O(l) H3O+(aq) + Cl–(aq)
H +
proton –
H Cl
Cl O H OH
HH
the proton, H+, is donated …forming the …and leaving a
by the hydrogen chloride positive hydroxonium ion… negative chloride ion
and accepted by the water…
Figure 8.12 An acid is a proton donor. Hydrogen chloride is the acid in this reaction. A base is a proton acceptor. Water is the base
in this reaction. Remember that a proton is a hydrogen ion, H+.
Cambridge International AS Level Chemistry
NH3(g) + H2O(l) NH4+(aq) + OH–(aq)
H +
–
H HO HNH O
HN
HH HH
the proton, H+, is donated …forming …and the
by the water and accepted the positive negative
by the ammonia… ammonium ion… hydroxide ion
Figure 8.13 Water is the proton donor (it is the acid); ammonia is the proton acceptor (it is the base).
Brønsted–Lowry acids and bases do not have to involve Conjugate acids and conjugate bases
aqueous solutions. For example, when chloric(VII) acid
(HClO4) reacts with ethanoic acid (CH3COOH) in an inert In a reaction at equilibrium, products are being converted
solvent, the following equilibrium is set up: to reactants at the same rate as reactants are being
converted to products. The reverse reaction can also be
considered in terms of the Brønsted–Lowry theory of
H+ donated
acids and bases.
Consider the reaction:
HClO4 + CH3COOH ClO4– + CH3COOH2+
132 acid base
NH3(g) + H2O(l) NH4+(aq) + OH–(aq)
aIpnrpotrthooitnsonrteoaaCccctHieop3nCtoHOr.COlHO.4CisHt3hCeOacOidHbiesctahuesebaitseisbdeocnauatsienigt a In the reverse reaction, the aNcHtin4+giaosnadnoancaitdesana dprOoHto–nis
is to the OH– ion. So NH4+ is
acting as a base.
When an acid or base reacts with water, an equilibrium
mixture is formed. For acids such as hydrochloric acid, H+ donated
the position of equilibrium is almost entirely in favour of NH3(g) + H2O(l)
the products. But for ammonia the position of equilibrium acid base
favours the reactants. The equations can be written to NH4+(aq) + OH–(aq)
show this. For example: If a reactant is linked to a product by the transfer of a
proton we call this pair a conjugate pair. Consider the
HCl(g) + aq H+(aq) + Cl–(aq)
A forward arrow is used as this reaction goes to following reaction:
completion. conjugate pair
NH3(g) + H2O(l) NH4+(aq) + OH–(aq)
An equilibrium arrow is used as this reaction does not go HCl(g) + H2O(l) H3O+(aq) + Cl–(aq)
to completion. acid base acid base
Question conjugate pair
14 Identify which reactants are acids and which are bases Looking at the forward reaction:
in the following reactions: ■■ Cl– is the conjugate base of the acid HCl
■■ H3O+ is the conjugate acid of the base H2O.
a NH4+ + H2O NH3 + H3O+
b HCOOH + HClO2 HCOOH2+ + ClO2–
Chapter 8: Equilibrium
Looking at the reverse reaction: Strong and weak acids and bases
■■ HCl is the conjugate acid of the base Cl–
■■ H2O is the conjugate base of the acid H3O+. Strong acids
In a conjugate pair, the acid has one proton more.
When hydrogen chloride dissolves in water to form a solution
The conjugate pairs for the equilibrium between of concentration 0.1 mol dm–3, it ionises almost completely.
ammonia and water to form ammonium ions and We say that the acid is almost completely dissociated.
hydroxide ions are:
HCl(g) + H2O(l) H3O+(l) (aq) + Cl–(aq)
conjugate pair
The position of equilibrium is so far over to the right that
we can show this as an irreversible reaction. The pH of
this solution is pH 1. The pH of a solution depends on the
NH3(g) + H2O(l) NH4+(aq) + OH–(aq) tchoencceonntcreantitorantioofnhoydf rhoyxdornoxiuomniuiomnsi,oHns3,Oth+.eTlhoewherigthheerpH.
base acid acid base The low pH shows that there is a high concentration of
hydroxonium ions in solution.
conjugate pair
Acids that dissociate almost completely in solution are
The idea of conjugate acids and bases is sometimes called called strong acids.
the acid–1 base–1, acid–2 base–2 concept.
Question The mineral acids, hydrochloric acid, sulfuric acid and
nitric acid, are all strong acids.
15 a Identify the acid and the base on the right-hand
side of these equilibria. Weak acids
i HClO2 + HCOOH ClO2– + HCOOH2+ When ethanoic acid dissolves in water to form a solution 133
of concentration 0.1 mol dm–3, it is only slightly ionised.
ii H2S + H2O HS– + H3O+ There are many more molecules of ethanoic acid in
solution than ethanoate ions and hydroxonium ions. We
b I dentify the acid on the right-hand side of this say that the acid is partially dissociated.
equation which is conjugate with the base on the
left-hand side.
CH3NH2 + H2O CH3NH3+ + OH– CH3COOH(l) + H2O(l) CH3COO–(aq) + H3O+(aq)
ethanoic acid ethanoate ion hydroxonium
ion
The position of equilibrium is well over to the left. The
pH of this solution is pH 2.9. The pH is much higher
compared with a solution of hydrochloric acid of the
same concentration. This is because the concentration of
hydroxonium ions in solution is far lower.
Acids that are only partially dissociated in solution are
called weak acids.
Figure 8.14 Many foods have high quantities of sugar in them. Weak acids include most organic acids, hydrocyanic acid
The sugar is converted to acid by bacteria in your mouth. This (HCN), hydrogen sulfide and ‘carbonic acid’.
acid can attack the enamel on your teeth. By chewing sugar-
free gum, more saliva is produced. Saliva is slightly alkaline. It Although we sometimes talk about the weak acid,
neutralises the acid. carbonic acid, you will never see a bottle of it. The acid is
really an equilibrium mixture of carbon dioxide dissolved
in water. The following equilibrium is set up:
CO2(g) + H2O(l) HCO3–(aq) + H+(aq)
Cambridge International AS Level Chemistry
The amount of CveOr2ytshmatalfloarms Hs 2uCnOdi3ssioonciiasetesdrecaadribloy.nic questions
acid, H2CO3, is
16 Nitric acid is a strong acid but chloric(I) acid, HClO, is a
Strong bases weak acid.
When sodium hydroxide dissolves in water to form a E xplain the difference between a strong and a
a solution of concentration 0.1 mol dm–3, it ionises weak acid.
completely.
b W rite equations showing the ionisation of each of
NaOH(s) + aq Na+(aq) + OH–(aq) these acids in water.
The position of equilibrium is far over to the right. c S uggest relative pH values for 0.1 mol dm–3
The solution formed is highly alkaline due to the high aqueous solutions of:
concentration of hydroxide ions present. The pH of this
solution is pH 13. i chloric(I) acid
ii nitric acid.
Bases that dissociate almost completely in solution are d Hydrazine, N2H4, is a weak base.
called strong bases. i W rite a chemical equation to show the
The Group 1 metal hydroxides are strong bases. equilibrium reaction of hydrazine with water.
ii S tate the relative concentrations (high or low)
Weak bases
of the N2H4 molecules and the products.
When ammonia dissolves and reacts in water to form a 17 a The pH of a solution depends on the hydrogen
solution of concentration 0.1 mol dm–3, it is only slightly
ionised. There are many more molecules of ammonia in ion (hydroxonium ion) concentration. Which
134 solution than ammonium ions and hydroxide ions. concentration of ethanoic acid in Table 8.8 has
the highest concentration of hydrogen ions in
NH3(g) + H2O(l) NH4+(aq) + OH–(aq) solution?
b W hich acid or alkali in Table 8.8 has the highest
The position of equilibrium is well over to the left. The concentration of hydroxide ions?
pH of this solution is pH 11.1. The pH is much lower
compared with a solution of sodium hydroxide of the c E xplain why a solution of 0.1 mol dm–3 ethanoic
same concentration. This is because the concentration of acid has a lower electrical conductivity than a
hydroxide ions in solution is far lower. solution of 0.1 mol dm–3 hydrochloric acid.
d B oth hydrochloric acid and ethanoic acid
react with magnesium. The rate of reaction of
1.0 mol dm–3 hydrochloric acid with magnesium
is much faster than the rate of reaction of
1.0 mol dm–3 ethanoic acid. Explain why.
Bases which dissociate to only a small extent in solution
are called weak bases.
Ammonia, amines (see page 220) and some hydroxides of
transition metals are weak bases.
Table 8.8 compares the pH values of some typical
strong and weak acids and bases.
Acid or base pH of 1.0 mol dm–3 solution pH of 0.1 mol dm–3 solution pH of 0.01 mol dm–3 solution
hydrochloric acid (strong acid) 0.4 1.4 2.4
ethanoic acid (weak acid) 2.4 2.9 3.4
sodium hydroxide (strong base) 14.4 13.4 12.4
ammonia (weak base) 11.6 11.1 10.6
Table 8.8 pH values of some typical strong and weak acids and bases.
Chapter 8: Equilibrium
Summary ■ The quantities of reactants and products present at
equilibrium can be calculated from the equilibrium
■ A reversible reaction is one in which the products expression and a value of Kc (or Kp), given
can be changed back to reactants. appropriate data.
■ Chemical equilibrium is dynamic because the ■ A change in temperature affects the value of the
backward and forward reactions are both occurring equilibrium constant for a reaction but changes in
at the same time. concentration, pressure or the presence of a catalyst
do not affect the value of the equilibrium constant.
■ A chemical equilibrium is reached when the rates of
the forward and reverse reactions are equal. ■ The conditions used in the Haber process and the
Contact process are chosen so that a good yield of
■ Le Chatelier’s principle states that when the product is made.
conditions in a chemical equilibrium change, the
position of equilibrium shifts to oppose the change. ■ The Brønsted–Lowry theory of acids and bases
states that acids are proton donors and bases are
■ Changes in temperature, pressure and proton acceptors.
concentration of reactants and products affect the
position of equilibrium. ■ Strong acids and bases are completely ionised in
aqueous solution whereas weak acids and bases are
■ For an equilibrium reaction, there is a relationship only slightly ionised.
between the concentrations of the reactants and
products which is given by the equilibrium constant K. ■ Strong and weak acids and bases can be distinguished
by the pH values of their aqueous solutions.
■ Equilibrium constants in terms of concentrations,
Kc, and partial pressure, Kp, can be deduced from
appropriate data.
135
End-of-chapter questions
1 The reaction
2SO2(g) + O2(g) 2SO3(g)
reaches dynamic equilibrium in a closed vessel. The forward reaction is exothermic. The reaction is catalysed
by V2O5.
a Explain the term dynamic equilibrium. [2]
b What will happen to the position of equilibrium when:
[1]
i some sulfur trioxide, SO3, is removed from the vessel? [1]
ii the pressure in the vessel is lowered? [1]
iii more V2O5 is added? [1]
iv the temperature of the vessel is increased? [2]
c State Le Chatelier’s principle.
d Use Le Chatelier’s principle to explain what will happen to the position of equilibrium in the reaction
H2(g) + CO2(g) H2O(g) + CO(g)
when the concentration of hydrogen is increased. [5]
Total = 13
cambridge International as level chemistry
2 Hydrogen, iodine and hydrogen iodide are in equilibrium in a sealed tube at constant temperature.
The equation for the reaction is:
H2 + I2 2HI(g) ΔHr = –96kJmol–1
The partial pressures of each gas are shown in the table below.
Gas Partial pressure / Pa
H2 2.330 × 106
I2 0.925 × 106
HI 10.200 × 106
a Explain the meaning of the term partial pressure. [2]
b Calculate the total pressure of the three gases in this mixture. [1]
c Write an equilibrium expression for this reaction in terms of partial pressures. [1]
d Calculate a value for Kp for this reaction, including the units. [1]
e Use Le Chatelier’s principle to explain what happens to the position of equilibrium in this reaction when:
136 i the temperature is increased [5]
ii some iodine is removed. [5]
Total = 15
3 The equilibrium between three substances, a, B and c is shown below.
A(g) + B(g) C(g)
Initially there were 0.1mol of a and 0.2mol of B in the reaction mixture. a and B reacted together to produce an
equilibrium mixture containing 0.04mol of c. The total volume of the mixture was 2.00dm3.
a Calculate the number of moles of a and B at equilibrium. [2]
b Calculate the concentrations of a, B and c at equilibrium. [3]
c i Write the equilibrium expression for Kc. [1]
ii Calculate the value of Kc and give the units. [2]
Total = 8
Chapter 8: Equilibrium
4 Gaseous hydrogen and gaseous iodine react together to form hydrogen iodide.
H2 + I2 2HI
a The graph shows how the amount of hydrogen iodide varies with time in a 1.00dm3 container.
The initial amounts of hydrogen and iodine were 1.00mol H2 and 1.00mol I2.
2.0
Amount of HI / moles 1.5
1.0
0.5
0.0
0123456
Time / arbitrary units
Draw a similar graph to show how the number of moles of hydrogen varies with time. [5] 137
b Calculate the number of moles of iodine present at equilibrium. [1]
c i Write the equilibrium expression for Kc for the reaction between gaseous hydrogen and iodine. [1]
[2]
ii Calculate the value of Kc and give the units.
Total = 9
5 a Describe three characteristic features of chemical equilibrium. [3]
b When 1mol of N2O4 gas is allowed to come to equilibrium with NO2 gas under standard conditions,
only 20% of the N2O4 is converted to NO2. [1]
[4]
N2O4 2NO2 ΔHr = +58kJmol–1 [2]
[2]
i Give the equilibrium expression for this reaction.
ii Calculate the value of Kc for the reaction. Assume that the volume of the reaction mixture is 1dm3.
c Explain the effect on Kc of an increase in:
i pressure
ii temperature.
Total = 12
6 This question is about the following reaction:
CH3COOH(l) + C2H5OH(l) CH3COOC2H5(l) + H2O(l)
ethanoic acid ethanol ethyl ethanoate water
9.20g of ethanol are mixed with 12.00g of ethanoic acid in an inert solvent. The total volume of solution [2]
is 250cm3. The mixture is left to equilibrate for several days. At equilibrium 70% of the reactants are [2]
converted to products.
a What is the concentration of each reactant at the start?
b What is the concentration of each reactant at equilibrium?
cambridge International as level chemistry
c What is the concentration of each product at equilibrium? [2]
d i Write the equilibrium expression for this reaction. [1]
[1]
ii Calculate the value of Kc for the reaction. [1]
iii Explain why there are no units for Kc for this reaction. [1]
e What will happen to the numerical value of Kc if 100cm3 of water is added to the equilibrium mixture?
f What will happen to the yield of ethyl ethanoate if 100cm3 of water is added to the equilibrium [2]
mixture? Explain your answer.
Total = 12
7 a Hydrogen chloride and ammonia both ionise in water:
HCl + H2O H3O+ + Cl– equation 1
NH3 + H2O NH4+ + OH– equation 2
i State the name of the ion H3O+. [1]
ii Identify the acid and the base on the left-hand side of each equation. [2]
[5]
iii By referring to equation 1 and equation 2, explain why water is described as being amphoteric.
b When dissolved in an organic solvent, hydrogen chloride reacts with hydrogen iodide as follows:
HCl + HI H2Cl+ + I–
i Use the Brønsted–Lowry theory of acids and bases to explain which reactant is the acid and which [2]
reactant is the base.
138 ii Identify which of the products is the conjugate acid and which is the conjugate base of the
substances you have identified in part b i. [1]
c Hydrochloric acid is a strong acid but ethanoic acid, CH3COOH, is a weak acid. [2]
i Explain the difference between a strong acid and a weak acid.
ii Suggest a value of the pH for a 0.1moldm–3 solution of ethanoic acid in water. [1]
iii Write a chemical equation to show the reaction when ethanoic acid donates a proton to water. [2]
Total = 16
8 This question is about the reaction:
N2(g) + 3H2(g) 2NH3(g) ΔHr = –92kJmol–1
120.0mol of hydrogen gas are mixed with 40.0mol of nitrogen gas then pressurised. The mixture of gases
is passed at constant pressure over an iron catalyst at 450°C until the mixture reaches equilibrium.
The total volume of the mixture is 1.0dm3. 20% of the reactants are converted to ammonia.
a How many moles of nitrogen and hydrogen remain at equilibrium? [2]
b How many moles of ammonia are formed? [1]
c Write an equilibrium expression for Kc. [1]
d Calculate a value for Kc, including units. [2]
e What will happen to the numerical value of Kc when the pressure is raised? [1]
f What will happen to the numerical value of Kc when the temperature is raised? [1]
Total = 8
Chapter 8: Equilibrium
9 Ethanol can be manufactured by reacting ethene, C2H4, with steam.
C2H4(g) + H2O(g) C2H5OH(g)
a Write the equilibrium expression in terms of partial pressures, Kp, for this reaction. [1]
b State the units of Kp for this reaction. [1]
c The reaction is at equilibrium at 290°C and 7.00 × 106Pa pressure. Under these conditions the partial
[1]
pressure of ethene is 1.50 × 106Pa and the partial pressure of steam is 4.20 × 106Pa. [1]
[1]
i Calculate the partial pressure of ethanol.
ii Calculate the value of the equilibrium constant, Kp, under these conditions. [3]
d The reaction is carried out in a closed system. Explain the meaning of the term closed system.
e Use Le Chatelier’s principle to explain what will happen to the position of equilibrium in this reaction
when the pressure is increased.
f The results in the table below show the effect of temperature on the percentage of ethene converted
to ethanol at constant pressure. Use this information to deduce the sign of the enthalpy change for this
reaction. Explain your answer.
Temperature / °C % of ethene converted 139
260 40
290 38
320 36
[4]
Total = 12
140
Chapter 9:
Rates of reaction
Learning outcomes – temperature (in terms of both the Boltzmann
distribution and collision frequency)
You should be able to:
– catalysts (in terms of changing a reaction’s
■■ explain and use the terms mechanism, lowering the activation energy and
– rate of reaction the Boltzmann distribution)
– activation energy, including reference to the
Boltzmann distribution ■■ explain that catalysts can be homogeneous or
– catalysis heterogeneous and describe enzymes as biological
catalysts (proteins) that may have specificity.
■■ explain qualitatively how the following a ect the rate
of a chemical reaction:
– concentration (in terms of collision frequency)
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