Pre-U STPM Chemistry Term 1 CC039342a

192 Chemistry Term 1 STPM CHAPTER 5 11 This means that, doubling the concentration of NO2 would increase the rate of reaction by 22 or 4 times. Similarly, if the concentration of NO2 is halved, then the rate would decrease to ( 1 2 )2 or 1 4 of its original value. 12 However, changing the concentration of CO has no effect on the rate of reaction. However, CO must be present for the reaction to occur. 13 The order of reaction with respect to a substance is the power to which the concentration of the substance is raised in an experimentally determined rate equation. Example 5.4 The rate equation for the following reaction: 2A + B + 2C → products is, Rate = k[A][B] 2 (a) What is the overall order of the reaction? (b) What is the unit for the rate constant, k? (c) The initial rate of reaction at a certain concentration of A, B and C, is x mol dm–3 s–1. What will the new rate be (at constant temperature) if, (i) the concentration of A is doubled, (ii) the concentration of A is doubled and the concentration of B halved, (iii) the concentration of C is tripled? Solution (a) Overall order = 1 + 2 = 3 (b) Unit for k = (Mol dm–3 s–1) (Mol dm–3)3 = mol–2 dm6 s–1 (c) Originally: x = k[A][B] 2 (i) New rate = k[2A][B] 2 = 2k[A][B] 2 = 2x mol dm–3 s–1 (ii) New rate = k[2A] 1 2 B 2 = 2 1 4 [A][B] 2 = 1 2 [A][B] 2 = 1 2 x mol dm–3 s–1 (iii) Since the reaction is zero order with respect to C, the rate remains the same (x mol dm–3 s–1). Definition of order of reaction

193 Chemistry Term 1 STPM CHAPTER 5 Quick Check 5.3 1 Consider the following reaction: A + B → product The reaction is first order with respect to A and second order with respect to B. (a) Write the rate equation for the reaction. (b) What is the unit of the rate constant? (c) The rate of reaction at temperature T is x mol dm–3 s–1. Find in terms of x, the rates of reaction at temperature T when (i) the concentration of A is halved. (ii) the concentration of B is halved. (iii) the concentration of A is doubled and the concentration of B tripled. 5.3 The Effect of Temperature on Reaction Kinetics 1 The relationship between the rate constant and temperature is an exponential one. It is given by the Arrhenius equation: k = Ae –Ea RT 2 Taking natural log throughout: ln k = ln A – Ea RT Or: log k = log A – Ea 2.303RT where k = Rate constant A = Constant Ea = Activation energy R = Universal gas constant T = Absolute temperature (K) 3 The constant A is known as the frequency factor. It composes two components: (a) The probability factor, which is the fraction of collisions with the correct orientation. (b) The collision frequency, which is the number of collisions per unit time. 4 The e –Ea RT term refers to the fraction of particles having energy equal to or greater than the activation energy. 5 From the above equation, it can be seen that the rate constant would increase (a) when the activation energy is decreased (by addition of suitable catalyst). R = 8.31 J mol–1 K–1 Probability factor and collision frequency Fraction of particles with energy  activation energy 2015/P1/Q19

194 Chemistry Term 1 STPM CHAPTER 5 (b) when the temperature increases. 6 A plot of ln k against 1 T gives a straight line with the slope equals to –Ea R . 7 A plot of log k against 1 T gives a straight line with the slope equals to – Ea 2.303R . 8 If the rate constant of a particular reaction is k1 at temperature T1 , and is k2 at temperature T2 , then, log k2 k1 = Ea 2.303R  1 T1 – 1 T2  Similarly, In k2 k1 = Ea R  1 T1 – 1 T2  Quick Check 5.4 1 The variation of the rate constant with temperature for a reaction is given in the table below. Temperature/°C 283 302 427 518 k 4.05  10–7 1.40  10–6 1.33  10–3 4.49  10–2 Plot a suitable graph to determine the activation energy in kJ mol–1. 5.4 The Role of Catalysts in Reactions 1 A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed by the reaction. 2 A catalyst reacts with the reactants to form an intermediate but is regenerated at the end of the reaction. 3 During the course of the reaction, the chemical nature of the catalyst might change, but remains unchanged chemically at the end of the reaction. 4 A catalyst alters, (a) the rate and rate constants of a reaction, (b) the mechanism of reaction, (c) the order of reaction, (d) the activation energy of the reaction. 2008/P2/Q1(b) Gradient = In k –Ea R 1 T Gradient = log k 1 T –Ea 2.303R Effects of adding a catalyst

195 Chemistry Term 1 STPM CHAPTER 5 5 A catalyst does not affect (a) the yield, (b) the overall stoichiometry, (c) the enthalpy change, of the reaction. 6 The rate of reaction is proportional to the amount of catalyst used. However, since they are regenerated they are usually used in small quantities because catalysts are expensive. 7 A catalyst alters the rate of reaction by allowing the reaction to proceed by an alternative pathway with lower activation energy. By doing so, more particles will have enough energy to overcome the lower activation energy and increases the number of effective collisions per second as illustrated by the Maxwell-Boltzmann distribution curve below: Fraction of molecule E Energy a ʼ Ea 8 The energy profile for a typical chemical reaction in the presence and in the absence of a catalyst is shown below. Activation energy without catalyst Activation energy with catalyst Reactants Products Energy Progress of reaction 9 The action of catalysts is very specific. A catalyst that catalyses one reaction might not catalyse another reaction. 10 A catalyst which is in the same physical state as the reactants is known as a homogeneous catalyst. 11 An example of homogeneous catalysis is the catalytic oxidation of atmospheric sulphur dioxide by atmospheric oxides of nitrogen. For example, 2SO2 (g) + O2 (g) NO2 (g) → 2SO3 (g) Ea = activation energy without catalyst Ea ′ = activation energy in the presence of a catalyst Exam Tips The more catalyst we use, the faster the rate of reaction. Info Chem It is easier to get to the top of a building by using the lift (catalysed) than using the stairs (uncatalysed).

196 Chemistry Term 1 STPM CHAPTER 5 12 Examples of homogeneous catalysis are: Reaction Catalyst S2 O8 2–(aq) + 2I– (aq) → 2SO4 2–(aq) + I2 (aq) Fe3+(aq) or Fe2+(aq) CH3 COOH(l) + C2 H5 OH(l) → CH3 COOC2 H5 (aq) + H2 O(l) H2 SO4 (aq) 2KClO3 (s) → 2KCl(s) + 3O2 (g) MnO2 (s) CH3 COCH3 (aq) + I2 (aq) → CH3 COCH2 I(aq) + HI(aq) H+(aq) 2MnO4 – (aq) + 5C2 O4 2–(aq) + 16H+(aq) → 2Mn2+(aq) + 10CO2 (g) + 8H2 O(l) Mn2+(aq) 13 A catalyst which is not in the same physical state as the reactants is known as a heterogeneous catalyst. 14 One example of heterogeneous catalysis is the oxidation of SO2 to SO3 using vanadium(V) oxide as catalyst at 450 °C. 2SO2 (g) + O2 (g) V2 O5 (s) → 2SO3 (g) 15 Examples of heterogeneous catalysis are: Reaction Catalyst H2 (g) + I2 (g) → 2HI(g) Pt(s) CH2==CH2 (g) + H2 (g) → CH3 CH3 Ni(s) N2 (g) + 3H2 (g) → 2NH3 (g) Fe(s) 4NH3 (g) + 5O2 (g) → 4NO(g) + 6H2 O(l) Pt(s) Homogeneous Catalysis 1 Homogeneous catalysis is usually explained using the intermediate product theory. 2 Consider the reaction between aqueous potassium iodide and potassium persulphate: 2KI(aq) + K2 S2 O8 (aq) → 2K2 SO4 (aq) + I2 (aq) The ionic equation for the reaction is: 2I– (aq) + S2 O8 2–(aq) → I2 (aq) + 2SO4 2–(aq) 3 The activation energy of this reaction is expected to be high because it involves collisions between negative charged particles which tend to repel one another. Hence, the rate is low. 4 However, the rate increases in the presence of a little aqueous iron(III) chloride as catalyst. 5 In the presence of Fe3+(aq) ions, the reaction does not involve direct collisions between the negatively charged I– and S2 O8 2– ions. 6 Instead, the Fe3+ ions collide with the I– ions to produce iodine, and Fe3+ is converted into Fe2+ (the intermediate product). 2Fe3+(aq) + 2I– (aq) → 2Fe2+(aq) + I2 (aq) I – I – S2 O8 2– Repulsion Attraction Attraction S2 O8 2– Fe3+ Fe2+ VIDEO Heterogeneous Catalyst

197 Chemistry Term 1 STPM CHAPTER 5 7 The Fe2+ ions then collide with S2 O8 2– ions to produce SO4 2– ion and Fe3+ is regenerated. 2Fe2+(aq) + S2 O8 2–(aq) → 2Fe3+(aq) + 2SO4 2–(aq) 8 The two steps can be combined as follows: 2Fe3+(aq) + 2I– (aq) → 2Fe2+(aq) + I2 (aq) 2Fe2+(aq) + S2 O8 2–(aq) → 2Fe3+(aq) + 2SO4 2–(aq) Overall reaction: S2 O8 2–(aq) + 2I– (aq) → 2SO4 2–(aq) + I2 (aq) 9 The catalysed steps are expected to have lower activation energies because they involve collisions between opposite charged particles. 10 Transition elements and their compounds are good homogeneous catalysts because they exhibit variation oxidation states. Energy Progress of reaction Catalysed reaction Uncatalysed reaction 11 Other examples of homogeneous catalysis are: Reaction Catalyst 2SO2 (g) + O2 (g) → 2SO3 (g) NO2 (g) CH3 COOH(I) + C2 H5 OH(I) → CH3 COOC2 H5 (I) + H2 O(l) H2 SO4 (aq) 2KCIO3 (s) → 2KCI(s) + 3O2 (g) MnO2 (s) 2MnO4 – (aq) + 5C2 O4 2–(aq) + 16H+(aq) → 2Mn2+(aq) + 10CO2 (g) + 8H2 O(l) Mn2+(aq) CH3 COCH3 (aq) + l2 (aq) → CH3 COCH2 l(aq) + HI(aq) H+(aq) Example 5.5 Using balanced equations, illustrate the action of the following catalyst in the reactions concerned. (a) 2SO2 (g) + O2 (g) → 2SO3 (g) [Catalyst: NO2 (g)] (b) 2KClO3 (s) → 2KCl(s) + 3O2 (g) [Catalyst: MnO2 (s)] Solution (a) 2SO2 (g) + 2NO2 (g) → 2SO3 (g) + 2NO(g) 2NO(g) + O2 (g) → 2NO2 (g) (b) 2KClO3 (s) + 4MnO2 (s) → 2KCl(s) + 2Mn2 O7 (s) 2Mn2 O7 (s) → 4MnO2 (s) + 3O2 (g) Transition elements or their compounds are good homogeneous catalyst. Exam Tips Exam Tips The sum of the individual steps must give the same overall stoichiometric equation. Exam Tips Exam Tips The SO3 formed dissolves in water to form acid rain. SO3 (g) + H2 O(ce) → H2 SO4 (aq)

198 Chemistry Term 1 STPM CHAPTER 5 Heterogeneous Catalysis 1 The adsorption theory is usually used to explain the action of heterogeneous catalyst. 2 One must differentiate between adsorption and absorption. 3 Adsorption occurs on the surface of a substance while absorption occurs in the body of a substance. 4 An example of heterogeneous catalysis is the reaction between hydrogen gas and iodine vapour at 450 °C in the presence of solid nickel as catalyst. H2 (g) + I2 (g) Ni(s) 2HI(g) 5 The H2 and I2 molecules get adsorbed on the surface of nickel. This is done by the formation of temporary bonds between the vacant or half-filled 3d orbitals of nickel. Surface of Ni H Ni H Ni I Ni I Ni H Ni H Ni I Ni I Ni 6 The adsorption serves three important purposes: (a) It brings the H2 and I2 molecules closer to one another. (b) It weakens the covalent bonds in the H2 and I2 molecules. (c) It holds the reactant molecules in the right orientation for new bonds to be formed. Bond in the process of forming H Ni H Ni I Ni I Ni H Ni H Ni I Ni I Ni Bond in the process of breaking 7 Eventually, the old bonds are broken completely and new bonds are formed. The HI molecules are then released from the nickel surface (this is called desorption) providing a new surface for the adsorption of other reactant molecules. H Ni H Ni I Ni I Ni H Ni H Ni I Ni I Ni 8 However, there are some heterogeneous catalysis that can be explained through the intermediate product theory. 9 One such example in the oxidation of sulphur dioxide by oxygen to form sulphur trioxide in the contact process. 2SO2 (g) + O2 (g) → 2SO3 (g) The reaction is catalysed by vanadium(V) oxide, V2 O5 (s). 10 The adsorption of SO2 onto the surface of vanadium(V) oxide facilitates the oxidation via the following reaction: SO2 (g) + V2 O5 (s) → SO3 (g) + V2 O4 (s) The vanadium is temporarily reduced to vanadium(IV) oxide. Adsorption Absorption

199 Chemistry Term 1 STPM CHAPTER 5 One of the products can act as catalyst for the reaction. 11 The vanadium(IV) oxide then reacts with oxygen that is adsorbed on its surface and gets converted back to vanadium(V) oxide. 2V2 O4 (s) + O2 (g) → 2V2 O5 (s) 12 Examples of heterogeneous catalysis are: Reaction Catalyst CH2 = CH2 (g) + H2 (g) → CH3 CH3 (g) Ni(s) or Pt(s) C2 H5 OH(g) → C2 H4 (g) + H2 O(l) Al2 O3 (s) or SiO2 (s) N2 (g) + 3H2 (g) → 2NH3 (g) Fe(s) 4NH3 (g) + 5O2 (g) → 4NO(g) + 6H2 O(l) Pt(s) or Rh(s) Autocatalysis 1 For most reactions, the rate decreases with time as the concentration of the reactants decreases. 2 However, there are some reactions where the rate increases initially then decreases thereafter. Such reactions are called autocatalytic reactions. 3 In autocatalytic reactions, one of the products acts as catalyst for the reaction. One example is the reaction between acidified potassium manganate(VII) and the ethanedioate (oxalate) ions at room temperature. 2MnO4 – (aq)+ 5C2 O4 2–(aq) + 16H+ (aq) → 2Mn2+(aq) + 10 CO2 (g) + 8H2 O(l) 4 When a few drops of potassium manganate(VII) is added to an acidified solution of sodium ethanedioate, it takes a while for the purple colour to disappear. 5 After that, addition of more aqueous potassium manganate(VII) see the immediate decolourisation of the potassium manganate(VII). 6 The reaction is slow in the beginning, but speeds up when Mn2+(aq) ions (which acts as a catalyst for the reaction) are formed. Rate increases as catalyst is produced Rate Time Rate decreases as concentration of reactants decreases 7 Another autocatalytic reaction is the reaction between propanone (or acetone) and iodine. CH3 —C—CH3 + I2 → CH3 —C—CH2 I + H+ + I– || || O O The H+ ions produced act as catalyst for the reaction. 2010/P1/Q9

200 Chemistry Term 1 STPM CHAPTER 5 Example 5.6 Sketch a concentration/time graph for an autocatalytic reaction. Explain the shape of your curve. Solution Concentration Time Reaction is slow as no catalyst is produced Reaction is fast as catalyst is produced Reaction is slow as little reactant is left Enzymes 1 Enzymes are substances that catalyse biological reaction in living organism. 2 Enzymes are protein molecules. They are very specific in their action. 3 Enzymes are very efficient catalyst compared to ‘inorganic catalyst’. 4 For example, the enzyme catalase lowers the activation energy for the decomposition of hydrogen peroxide from 85 kJ mol–1 to 20 kJ mol–1, while the inorganic catalyst, platinum, only lowers it to 55 kJ mol–1. 5 However, enzymes are very sensitive to changes in temperature and pH. Most enzymes function best within a small range of temperature (usually around body temperature of 37 °C) and pH. 6 For example, the optimum pH for pepsin is around 2.0 while the optimum pH for lipase is about 9.0. 7 If the temperature goes too high, or the pH gets too high or too low, the protein molecules would get denatured and the enzyme becomes ineffective. 8 The graphs below show the activity of enzymes at different temperatures and pH. pH Rate Temperature Rate 37 °C Optimum pH Enzymes are biological catalysts. Enzymes are sensitive to temperature and pH change. Exam Tips Exam Tips Denaturation of enzymes is due to disruption of their structures. Info Chem Decomposition of H2 O2 : 2H2 O2 → 2H2 O + O2 Info Chem Without enzymes, the biological reactions in our body will be too slow to sustain life.

201 Chemistry Term 1 STPM CHAPTER 5 5.5 Order of Reactions and Rate Constants The Initial Rate Method 1 Initial rate is the rate of reaction at time zero where the concentration of the reactants is maximum. 2 It can be obtained by drawing tangent to the concentration/time graph at time = 0. 3 For a general reaction: aA + bB → products In order to determine the order with respect to substance A, two or more experiments are conducted using varied amount of A but keeping the concentration of B constant throughout. Hence, whatever changes in the rates is due to A and not B. 4 To determine the order with respect to B, we keep the concentration of A constant while varying the concentration of B. Example 5.7 The data below is the result of the kinetic study of the following reaction: 2A + B → products Experiment [A]/mol dm–3 [B]/mol dm–3 Initial rate/mol dm–3 min–1 I 0.12 0.26 0.514 II 0.12 0.52 2.06 III 0.36 0.26 1.03 (a) Determine, (i) the order of reaction with respect to A and B, (ii) what is the overall order? (b) Write the rate equation for the reaction. (c) Using experiment (I), calculate the rate constant and state its unit. (d) Calculate the rate when the concentrations of A and B are 0.34 mol dm–3 and 0.52 mol dm–3 respectively. Solution (a) Let the rate equation be: Rate = k[A] x [B] y (i) Experiment II Experiment I : 2.06 0.514 =  0.52 0.26  y 4 = 2y ∴ y = 2 x yConcentration of A/mol dm–3 0 Time Initial rate = y x 2011/P2/Q6 2015/P1/Q12 2017/P1/Q9, Q11, Q20 2011/P1/Q9 2016/P1/Q11 2018/P1/Q9

202 Chemistry Term 1 STPM CHAPTER 5 Experiment III Experiment I : 1.03 0.514 =  0.36 0.12  y 2 = 2x ∴ x = 1 (ii) Overall order = (2 + 1) = 3 (b) Rate = k[A][B]2 (c) 0.514 = k(0.12)(0.26)2 k = 63.4 mol–2 dm6 min–1 (d) Rate = 63.4  (0.34)(0.52)2 = 5.83 mol dm–3 min–1 Quick Check 5.5 1 The data below refers to the reaction at 300 K. 2NO(g) + O2 (g) → 2NO2 (g) Experiment [NO]/mol dm–3 [O2 ]/mol dm–3 Relative initial rate I 0.58 0.29 1.0 II 1.16 0.29 4.0 III 1.16 0.58 8.0 Determine the rate equation for the reaction at 300 K. 2 Consider the data below regarding the reaction: A + 2B → products Experiment [A]/mol dm–3 [B]/mol dm–3 Initial rate/mol dm–3 s–1 I 1.0 1.0 1.44 × 10–2 II 1.0 8.0 9.12 × 10–1 III 4.0 1.0 5.88 × 10–2 Determine the rate equation for the reaction. 3 Chlorine dioxide disproportionates in alkaline solution according to the equation: 2ClO2 (aq) + 2OH– (aq) → ClO3 – (aq) + ClO2 – (aq) + H2 O(l) The kinetic results of the above reaction are given in the table below: Experiment [CIO2 ]/mol dm–3 [OH– ]/mol dm–3 Initial rate/M s–1 I 0.10 0.184 0.032 II 0.14 0.184 0.063 III 0.14 0.368 0.126 (a) Determine the oxidation number of chlorine in (i) ClO2 , (ii) ClO3 – and (iii) ClO2 – . (b) Determine the order of reaction with respect to (i) ClO2 and (ii) OH– . (c) Calculate the rate constant and state its unit. (d) Calculate the rate of reaction when the concentrations of ClO2 and OH– are 0.23 mol dm–3 and 0.37 mol dm–3 respectively.

203 Chemistry Term 1 STPM CHAPTER 5 Graphical Method (Integrated Rate Equation Method) First Order Reaction 1 The differential equation for a first order reaction is: A → products Rate = k[A] 1 – dC dt = kC (where C is the concentration of A at time t) – dC C = k dt Integrating the equation: C t ∫ dC C = – ∫ kt C0 0 ln C – ln C0 = –kt (where C0 is the concentration of A at time ‘zero’) ∴ In C = –kt + In C0 Or: log C = – kt 2.303 + log C0 Or: In C0 C = kt 2 A plot of ln C against t gives a straight line with the slope equals to –k. t Gradient = –k In C 3 Similarly, a plot of ln C0 C against t would give a straight line with the slope equal to k. 4 From the integrated rate equation we can obtain another important quantity of the reaction, which is the half-life, t 1 2 . The half-life is defined as the time taken for the initial amount (concentration, mass, moles etc) of a reactant to decrease to half its value. 5 The half-life can be determined graphically from the concentrationtime graph as shown on the right: 6 From the first order equation : ln C = –kt + ln C0 When t = t 1 2 , C = C0 2 Substituting into the above equation: ln C0 2 = –kt 1 2 + ln C0 Definition of half-life Gradient = k t In C0 C t C0 C0 2 [A] t 2 1

204 Chemistry Term 1 STPM CHAPTER 5 Rearranging: t 1 2 = ln 2 k 7 The expression above shows that the half-life of a first order reaction is independent of the initial amount of the reactant. 8 The concentration/time graph for a typical first order reaction is shown below: 1.0 0.5 0 2nd half life 3rd half life 1st half life Time/minute 5 10 15 20 25 [A] The half-life for the reaction is 7.0 minutes. The time taken for the amount of A to decrease from 1.0 to 0.5, and from 0.5 to 0.25, and from 0.25 to 0.125 is the same. Example 5.8 The half-life of a first order reaction is 15 minutes. (a) Calculate the time taken for the original amount of A to 1 16 decrease to its original amount. (b) Calculate the rate constant. Solution (a) 1 → 1 2 → 1 4 → 1 8 → 1 16 The time required = 4 half-lives = 4  15 = 60 minutes (b) Using the formula: t 1 2 = ln 2 k k = ln 2 15 = 4.62  10–2 min–1

205 Chemistry Term 1 STPM CHAPTER 5 Example 5.9 The following data is obtained for a reactant A in a particular reaction. Time/min 0.0 10.0 20.0 30.0 40.0 [A]/mol dm–3 1.00 0.94 0.88 0.83 0.78 Plot a graph of log[A] against time. (a) Use your graph to determine the rate constant. (b) Calculate the half-life. Solution Time/min 0 10 20 30 40 log [A] 0 –0.027 –0.056 –0.081 –0.11 –10 –12 10 20 30 40 50 Time/min –8 –6 –4 –2 Iog [A] (× 10–2) Gradient = = 11.0 × 10–2 40 – – 2.75 × 10–3 (a) Gradient of graph = – k 2.303 = – 2.75  10–3 ∴ k = 6.33  10–3 min–1 (b) Half-life = 0.693 k = 0.693 6.33 × 10–3 = 100.9 min Second Order Reaction 1 The differential equation for a second order reaction is: A → products Rate = k[A] 2 – dC dt = kC2 – dC C2 = k dt

206 Chemistry Term 1 STPM CHAPTER 5 Integrating the equation: C t ∫ dC C2 = – ∫ k dt C0 0 – 1 C + 1 C0 = – kt ∴ 1 C = kt + 1 C0 2 For a second order reaction, a plot of 1 C against t gives a straight line with the slope equals to k. 3 The half-life for the second order reaction is obtained as follows: 1 C = kt + 1 C0 When t = t 1 2 , C = C0 2 Thus, 1 C0 2 = kt 1 2 + 1 C0 t 1 2 = 1 kC0 4 The half-life of a second order reaction is inversely proportional to the initial concentration of the reactant. If the initial amount is high, the half-life will be shorter and vice versa. 1.0 0.5 0 100 200 300 400 1st half life 2nd half life 3rd half life Time/s [A] Gradient = k t 1 C The higher the concentration, the shorter the half-life. From the graph: 1st half-life = 5 min 2nd half-life = 10 min 3rd half-life = 20 min 2014/P1/Q17

207 Chemistry Term 1 STPM CHAPTER 5 Example 5.10 The data below refers to the following reaction: A → products Plot a graph of 1 [A] against time. (a) Use your graph to determine the rate constant. (b) Calculate the half-life of the reaction when the concentration of A is (i) 1.00 mol dm–3, (ii) 0.50 mol dm–3. Solution Time/s 0 40 80 120 160 1 [A] ( 10–1)/mol–1 dm3 63 11.0 15.6 20.2 24.9 8 20 0 40 60 80 100 120 140 160 10 12 14 16 18 20 22 24 26 t/s Gradient = = (22.6 – 10.0) ×10–1 140 – 32 1.17 × 10–2 (× 10–2)/mol–1 dm3 [A] 1 (a) Gradient = k = 1.17  10–2 mol–1 dm3 s–1 (b) Using the formula: t 1 2 = 1 kC0 (i) When C0 = 1.00 mol dm–3 t 1 2 = 1 (1.17  10–2) (1) = 85.5 s (ii) When C0 = 0.50 mol dm–3 t 1 2 = 1 (1.17  10–2) (0.5) = 170.9 s

208 Chemistry Term 1 STPM CHAPTER 5 Zero Order Reaction 1 The differential equation for a zero order reaction is: A → products Rate = k[A] 0 Or: Rate = k – dC dt = k – dC = k dt Integrating the equation: C t ∫dC = – ∫ k dt C0 0 C – C0 = – kt ∴ C = – kt + C0 2 For a zero order reaction, a plot of C against t gives a straight line with a slope equals to –k. 3 The half-life for a zero order reaction is: t 1 2 = C0 2k 4 The half-life of a zero order reaction is directly proportional to the initial concentration of the reactant. That means the half-life is longer if the initial amount of reactant is higher. 0 0.5 1.0 2 4 6 8 10 12 14 Time/min 1st half life 2nd half life 3rd half life [A] The higher the concentration, the longer the half-life. From the graph: 1st half-life = 6.5 min 2nd half-life = 3.25 min 3rd half-life = 1.63 min Gradient = k t C

209 Chemistry Term 1 STPM CHAPTER 5 5 The table below summarises the integrated rate equation and the half-life for zero-, first- and second-order reactions. Order Rate equation Integrated rate equation Half-life, t 1 2 0 Rate = k C = – kt + C0 C0 2k 1 Rate = k[A] ln C = – kt + ln C0 ln C0 C = kt In 2 k 2 Rate = k[A] 2 1 C = kt + 1 C0 1 kC0 6 The concentration/time graph for the zero, first and second order reactions are shown in the plot below for comparison. Zero order First order Second order Time Concentration Rate/Concentration Graph 1 For a first order reaction, a plot of rate against [concentration] gives a straight line passing through the origin with the gradient equal to the rate constant. Gradient = k [A] Rate 2 For a second order reaction, a plot of rate against [concentration]2 gives a straight line passing through the origin with the gradient equal to the rate constant. Gradient = k [A] 2 Rate 2009/P1/Q9 2008/P1/Q8 Info Chem For a second order reaction: [A] Rate

210 Chemistry Term 1 STPM CHAPTER 5 3 For a zero order reaction, a plot of rate against [concentration] gives a line parallel to the concentration axis. The intercept on the rate axis gives the rate constant. k [A] Rate Quick Check 5.6 1 The table below refers to the reaction: X → Y + Z Time/min 0 10 20 30 40 [X]/mol dm–3 1.00 0.75 0.56 0.45 0.38 (a) Plot a suitable graph to determine the order of reaction with respect to X. (b) Calculate the rate constant for the reaction. 2 The data below refers to the hydrolysis of sucrose. Time/min 0 20 50 65 150 [Sucrose]/mol dm–3 0.38 0.27 0.18 0.15 0.087 (a) Plot a suitable graph to determine the order of reaction with respect to X. (b) Calculate the rate constant for the reaction. 3 Consider the data below: Time/min 0.0 5.0 10.0 15.0 20.0 25.0 [A]/mol dm–3 1.00 0.88 0.75 0.62 0.50 0.37 Plot a graph to determine the order with respect to A. 4 The decomposition of dinitrogen pentoxide follows a second order kinetic. 2N2 O5 (g) → 2N2 O4 (g) + O2 (g) [N2 O5 ]/mol dm–3 1.00 1.49 1.90 2.13 Initial rate/mol dm–3 s–1 1.89 2.45 2.90 3.15 (a) Plot a suitable graph to show that the reaction is first order with respect to N2 O5 . (b) Determine the rate constant. 5 The decomposition of hydrogen peroxide is first order with respect to [H2 O2 ]. 2H2 O2 (aq) → 2H2 O(l) + O2 (g) Sketch the graph you would obtain for the following: (a) Rate against [H2 O2 ] (b) Rate against [H2 O2 ] 2 (c) [H2 O2 ] against time

211 Chemistry Term 1 STPM CHAPTER 5 Rate Equation and Reaction Mechanism 1 One-step reactions are called elementary reactions. Very few chemical reactions occur in only one step. 2 Most chemical reactions take place in several consecutive steps. These sequence of steps is called the reaction mechanism. 3 Reaction mechanism is usually proposed on the basis of experimental results which are provided by the rate equation. 4 The sum of the individual steps in the reaction mechanism gives the overall stoichiometric equation for the reaction. 5 In a multiple steps reaction, one of the steps will be the slowest (i.e. the one that has the highest activation energy). This step forms the ‘bottle neck’ of the reaction. It is called the rate-determining step. 6 The overall rate of reaction is then dependent on the ratedetermining step. The rate equation would contain only the species that are involved in the rate-determining step. 7 For example, A + B + C → products Assuming the reaction mechanism is: A + B Slow X X + C Fast products 8 The overall rate of reaction depends only on the rate of formation of the intermediate, X. Hence, the rate equation is given by: Rate = k[A][B] 9 Increasing the concentration of C does not affect the rate of reaction because C cannot take part in the reaction until ‘X’ is formed. 10 The energy profile for such a reaction is as shown below: Energy Reaction pathway A + B + C Products high Ea low Ea ‘X’ + C Alkaline Hydrolysis of 2-methyl-2-indopropane 1 The equation of hydrolysis is: CH3 CH3 | | CH3 —C—CH3 + OH– → CH3 —C—CH3 + I– | | I OH Or: (CH3 ) 3 C—I + OH– → (CH3 ) 3 C—OH + I– Elementary reactions Exam Tips Sum of all the individual steps must give the overall balanced stoichiometric equation. Exam Tips Exam Tips The rate-determining step is the step with the highest activation energy. 'X' is the intermediate. 2008/P2/Q1(b)(i) 2011/P1/Q9 2013/P1/Q9

212 Chemistry Term 1 STPM CHAPTER 5 This is known as a SN1 mechanism. Energy Reaction pathway (CH3 ) 3 COH + I– (CH3 ) 3 C+ + OH– + I– (CH3 ) 3 Cl + OH– 2 Kinetic studies of the reaction show that the reaction is first order with respect to (CH3 )3 CI and zero order with respect to OH– . The rate equation is given by: Rate = k[(CH3 )3 CI] 3 This shows that the reaction must involve more than one step. The rate determining step would involve only (CH3 )3 CI. 4 The proposed mechanism is as follows: (CH3 )3 C—I Slow (CH3 )3 C+ + I– (CH3 )3 C+ + OH– Fast (CH3 )3 C—OH 5 The proposed mechanism is consistent with the rate equation. Furthermore, the first step would require high activation energy because it involves the breaking of covalent bonds. 6 The second step is expected to be fast as it involves the combination of opposite charged particles. 7 The energy profile is as shown in the diagram. Alkaline Hydrolysis of 1-iodobutane 1 The equation for the hydrolysis is: CH3 CH2 CH2 CH2 I + OH– → CH3 CH2 CH2 CH2 OH + I– 2 Kinetic studies show that the reaction is first order with respect to both 1-iodobutane and OH– ions. The rate equation is given by, Rate = k[CH3 CH2 CH2 CH2 I][OH– ] 3 In this case, the species appearing in the rate equation are the same as those in the stoichiometric equation. This is an example of a one step (or elementary) reaction. 4 The proposed mechanism is: Direct collision between CH3 CH2 CH2 CH2 I and OH– to form an intermediate (or an activated complex). H H CH3 CH2 CH2 CH2 I + OH– Slow HO.......C.......I | CH2 | CH2 | CH3 Activated complex 5 In the activated complex, the carbon-oxygen bond is in the process of forming, while the carbon-iodine bond is in the process of breaking. 6 The activated complex then decomposes rapidly to yield the product. CH3 CH2 CH2 CH2 (OH)I– Fast CH3 CH2 CH2 CH2 OH + I– – – This is a SN2 mechanism.

213 Chemistry Term 1 STPM CHAPTER 5 SUMMARY SUMMARY 7 The energy profile takes the following form. Energy Reaction pathway CH3 CH2 CH2 CH2 I + OH– CH3 CH2 CH2 CH2 OH + I– H H HO C I CH2 CH2 CH3 – Example 5.11 Nitrogen monoxide reacts with chlorine in the gaseous phase according to the equation: 2NO(g) + Cl2 (g) → 2NOCl(g) The experimental rate equation is: Rate = k[NO][Cl2 ] Propose a mechanism for the reaction. Solution NO + Cl2 Slow NOCl2 NOCl2 + NO Fast 2NOCl Exam Tips The energy profile for a SN1 mechanism has two 'humps', while that for a SN2 mechanism has one 'hump'. 1 The rate of reaction is defined as the rate of change of the concentration of a substance with time. 2 The instantaneous rate is obtained by drawing tangents to the concentration/ time graph at a particular moment. 3 Activation energy is the minimum energy that has to be overcome before a reaction can occur. 4 Effective collisions are collisions that have enough energy to overcome the activation energy, and also in the correct orientation. 5 The factors that affect the rates of reactions are: (a) Concentration (c) Temperature (b) Pressure (d) Catalyst 6 Increasing temperature increases the fraction of particles having energy equal to or greater than the activation energy. 7 A catalyst increases the rate of reaction by providing an alternative pathway, with lower activation energy, for the reaction to occur. 8 Homogeneous catalyst is a catalyst that is in the same physical state as the reactants.

214 Chemistry Term 1 STPM CHAPTER 5 1 Which of the following statements about the rate constant for a first order reaction is not correct? A It has a unit of s–1. B It is a constant at constant temperature. C It does not change even if a catalyst is added. D It is independent of the concentrating of the reactants. 2 Which of the following rate equations is correct for the following reaction? 3O2 (g) → 2O3 (g) 9 Heterogeneous catalyst is a catalyst that is in a different physical state as the reactants. 10 Autocatalysis is a reaction where one of the products formed can act as a catalyst for the reaction. 11 Enzymes are catalysts that catalyse biological reactions in living organism. 12 The order of reaction with respect to a substance is the power to which the molar concentration of that reactant is raised in an experimentally determined rate equation. 13 Half-life is the time taken for the amount of a reactant to decrease to half of its original quantity. 14 Order Rate equation Integrated rate equation Half-life, t 1 2 0 Rate = k C = –kt + C0 C0 2k 1 Rate = k[A] ln C = –kt + ln C0 ln C0 C = kt In 2 k 2 Rate = k[A] 2 1 C = kt + 1 kC0 1 kC0 15 The Arrhenius equation is given by: k = Ae –Ea RT 16 The rate constant of a reaction increases with temperature irrespective of whether the reaction is endothermic or exothermic. 17 The rate constant is dependent on temperature and the presence of a catalyst. 18 Reaction mechanism describes the sequence of steps leading to the formation of products. 19 The rate-determining step is the step with the highest activation energy (or the slowest step) in the reaction mechanism. 20 The rate equation only involves species in the rate-determining step. 21 Change Effect on rate of reaction Effect on rate constant Increasing concentration Increases No effect Decreasing concentration Decreases No effect Increasing pressure Increases No effect Decreasing pressure Decreases No effect Increasing temperature Increases Increases Decreasing temperature Decreases Decreases Addition of catalyst Increases Increases Objective Questions STPM PRACTICE 5 A ∆[O2 ] ∆t = ∆[O3 ] ∆t B – ∆[O2 ] ∆t = ∆[O3 ] ∆t C – ∆[O2 ] ∆t = ∆[O3 ]2 ∆t D – ∆[O2 ] ∆t = 3 2 ∆[O3 ]2 ∆t

215 Chemistry Term 1 STPM CHAPTER 5 3 The Maxwell distribution curves for a gaseous reaction under different conditions are shown below. Energy Number of molecules X Y Which of the following is true? A The curve X is obtained at a higher temperature B The curve X is obtained at a higher pressure C The total area of curve X is the same as curve Y D Curve Y is one with the presence of a catalyst. 4 In acidic solution, bromate(V) ions oxidise bromide ions to bromine. BrO3 – + 6H+ + 5Br– → 3Br2 + 3H2 O Given the following data: Experiment [BrO3 – ]/ mol dm–3 [Br– ]/ mol dm–3 [H+]/ mol dm–3 Relative rate 1 0.40 0.28 0.031 1 2 0.60 0.28 0.031 1.5 3 0.60 0.56 0.031 3 4 0.40 0.28 0.062 4 Determine the overall order of the reaction. A 2 B 3 C 4 D Insufficient data 5 Which of the following statements regarding catalyst is not true? A A catalyst alters the mechanism of a reaction. B A catalyst does not take part chemically in the reaction. C The amount of catalyst used affects the rate of a reaction. D The same amount of products would be obtained from a particular reaction irrespective of whether a catalyst is present or not. 6 In general, as the temperature increases, A the rate will increase if the reaction is endothermic. B the rate will decrease if the reaction is exothermic. C the rate will increase regardless of whether the reaction is endothermic or exothermic. D the rate will remain the same if the reaction is zero order. 7 The acid catalysed reaction between propanone and iodine is: CH3 COCH3 + I2 H+ CH2 ICOCH3 + HI The result of an experiment is shown below: Time [I2]/mol dm–3 From this, it can be concluded that A the acid is not involved in the reaction. B it is a third order reaction overall. C the rate of reaction does not depend on the concentration of iodine. D the rate-determining step involves only iodine. 8 The mechanism for the reaction: P + 2Q → PQ2 is shown below: P + Q → PQ slow PQ + Q → PQ2 fast Which of the following rate equations is consistent with the above mechanism? A rate = k[PQ] B rate = k[PQ][Q] C rate = k[P][Q] D rate = k[PQ2 ] 9 Combustion of ethanol in pure oxygen is faster than in air because A oxygen is a catalyst for the combustion process. B oxygen lowers the activation energy. C the concentration of oxygen is higher. D nitrogen from the air hinders the combustion process.

216 Chemistry Term 1 STPM CHAPTER 5 10 The reaction between iodine and propanone in the presence of dilute sulphuric acid is given by the equation below: I 2 (aq) + CH3 COCH3 (aq) → H+(aq) + I– (aq) + CH2 ICOCH3 The reaction is zero order with respect to iodine. It can be deduced that A the reaction is first order with respect to propanone. B sulphuric acid acts as a catalyst. C iodine is involved in the rate-determining step. D the rate equation is: rate = k[CH3 COCH2 ][I2 ][H+] 11 Consider the reaction: 4NH3 (g) + 7O2 (g) → 4NO2 (g) + 6H2 O(g) Which of the following species in the above reaction appears or disappears the fastest? A NH3 C NO2 B O2 D H2 O 12 The initial rate of decomposition of a compound X was measured at two initial concentration of X with the following results. [X]/mol dm–3 0.20 0.30 Initial rate/mol dm–3 min–1 0.68 1.52 What is the order of reaction with respect to substance X? A Zero C 1.5 B 1.0 D 2 13 The rate of a chemical reaction is roughly doubled by increasing the temperature by 10 °C. This is because A the bond energy of the particles is halved by raising the temperature by 10 °C. B increasing the temperature by 10 °C doubles the rate of collision. C increasing the temperature by 10 °C doubles the number of particles having energy greater than the activation energy. D increasing the temperature by 10 °C doubles the average kinetic energy of the particles. 14 Consider the reaction: 2A → B + 2C Time In[A ] The gradient of the line is equal to A k B k–1 C –k D ln k 15 Which of the following is necessarily true of an overall second order reaction? A The half-life is a constant. B The rate of reaction would increase four folds if the concentration of the reactants is doubled. C The reaction occurs in two steps. D It must involve two substances. 16 A reaction is represented by equation below. aA + bB → cC ΔH = 0.00 The rate of reaction with respect to A is x mol dm–3 s–1. Which of the following statement is true of the reaction? A The rate of reaction is independent of temperature B The rate of reaction with respect to B is bx mol dm–3 s–1 C The activation energy of the forward reaction is the same as the activation energy of the reverse reaction D Catalyst does not affect the activation energy of the reaction 17 A catalyst A increases the kinetic energy of the reactant particles. B increases the fraction of particles having energy greater than the activation energy. C increases the rate of collisions between the particles. D increases the intensity of the collisions between the particles.

217 Chemistry Term 1 STPM CHAPTER 5 18 Which of the following statements correctly explains why a small increase in temperature usually leads to a large increase in the rate of a gaseous reaction? A The number of collisions increases at higher temperature. B The activation energy decreases at higher temperature. C The number of collisions with energy greater than the activation energy increases. D The gas occupies a larger volume at higher temperature. 19 Consider the reaction below. 2P + 2Q → R The rate of reaction increases eight times when the concentrations of P and Q are doubled. Which of the following rate law is not correct? A Rate = k[R][P][Q] C Rate = k[P]3 B Rate = k[P][Q]2 D Rate = k[Q]3 20 Which of the following statements is not true? A The rate of a chemical reaction does not depend on the amount of catalyst used. B The rate of exothermic reaction increases with increasing temperature. C All endothermic reactions require heating. D All exothermic reactions can occur spontaneously. 21 In a first order reaction, it takes 8.2 minutes for the concentration of the reactant to decrease from 0.085 mol dm–3 to 0.055 mol dm–3. The half-life of the reaction is A 6.2 minutes C 13.0 minutes B 8.2 minutes D Insufficient data 22 The half-life of a first order reaction 2.3 × 103 s. Which statement is true about the reaction? A The activation energy for the reaction is high B The rate constant is 3.0 × 10–4 s–1 C The rate is 3.0 × 10–8 mol dm–3 s–1 when the concentration is 0.010 mol dm–3 D The rate constant is independent of temperature 23 The graphs of initial rate versus concentration for the reaction W + X → products are shown below: Rate [W] Rate [X] What is the order of reaction with respect to W and X? W X A 1 1 B 1 2 C 0 2 D 2 1 24 The rate equation for the gaseous reaction between X and Y is: Rate = k[X]2 [Y] When 0.20 mol gas X and 0.20 mol of gas Y are mixed in a 2.0 dm3 flask at 350 °C, the initial rate is 5.60 × 10–3 mol dm–3 s–1. Which of the following statement(s) is/are correct of the reaction? I The rate determining step is trimolecular. II The numerical value for the rate constant at 350 °C is 5.60. III The rate of reaction is 1.12 × 10–2 mol dm–3 s–1 at 700 °C. A I only B II only C I and II D I, II and III 25 Consider the following equation: X + Y + Z → Products In a kinetic study of the above reaction, the following graphs are obtained. Concentration Time [Z] [Y] [X] What is the order of reaction with respect to X, Y and Z? X Y Z A 1 1 1 B 1 2 0 C 2 1 0 D 0 1 2

218 Chemistry Term 1 STPM CHAPTER 5 26 The rate constant for a reaction at 30 °C is 5.60 × 10–4 s–1. What is the half-life of this reaction? A 1237 s B 2209 s C 7487 s D Insufficient data 27 Propanone reacts with iodine according to the equation: CH3 COCH3 + I2 → CH2 ICOCH3 + HI The reaction is catalysed by dilute sulphuric acid. The rate equation is: Rate = k[CH3 COCH3 ][H+] Which statement(s) is/are correct? I Increasing the concentration of dilute sulphuric acid increases the rate of reaction. II The rate determining step involves the reaction of propanone and iodine. III The unit of the rate constant is s–1. IV Dilute sulphuric acid is consumed in the reaction. A I only C II and III B I and IV D II and IV 28 Aqueous potassium iodide reacts with hydrogen peroxide in the presence of dilute sulphuric acid according to the following mechanism: H2 O2 + I– → H2 O + IO– Fast IO– + H+ → HOI Slow HOI + H+ + I– → I2 + H2 O Fast Which statement(s) is/are true of the reaction? I The rate equation is: Rate = k[IO– ][H+]. II The reaction is first order with respect to H2 O2 , I– and H+. III Dilute sulphuric acid acts as a catalyst. A I only C I and III B II only D I, II and III 29 The rate constant of an exothermic reaction is 4.50 × 10–2 s–1 at 35 °C. Which statement is true of the reaction? A It is a second order reaction. B The half-life for the reaction is 15.4 s. C The rate constant at 50 °C is less than 4.50 × 10–2 s–1. D The reaction is non-reversible. 30 The dissociation of dinitrogen pentoxide is as follows: 2N2 O5 (g) → 4NO2 (g) + O2 (g) The mechanism for the reaction is: N2 O5 Slow 2NO2 + O N2 O5 + O Fast 2NO2 + O2 What is the unit of the rate constant for the decomposition? A s–1 C mol–1 dm3 s–1 B mol dm–3 s–1 D mol–2 dm6 s–1 31 Which statement about endothermic reactions is true? A They cannot occur at room conditions. B The rate of reaction decreases with increasing temperature. C The activation energy of the forward reaction is lower than that of the reverse reaction. D The rate of reaction increases with the addition of a catalyst. 32 The initial rate of reaction between P and Q at different concentration is given in the table below: [P]/mol dm–3 [Q]/mol dm–3 Initial rate/mol dm–3 s–1 x y r 2x 2y 4r 2x 4y 16r What is the initial rate if the concentration of P and Q is 3x and 3y respectively? A r B 4r C 9r D Cannot be calculated because the rate constant is not given. 33 Which statement about catalyst is not true? A The rate of reaction increases with the amount of catalyst used. B The catalyst lowers the activation energy for the forward reaction more than the rate of the reverse reaction. C The yield of the reaction does not depend on the amount of catalyst used. D The chemical nature of the catalyst before and after the reaction is the same.

219 Chemistry Term 1 STPM CHAPTER 5 Structured and Essay Questions 1 Ethyl ethanoate undergoes hydrolysis in the presence of dilute sulphuric acid according to the equation: CH3 COOC2 H5 (l) + H2 O(l) → CH3 COOH(aq) + C2 H5 OH(aq) An investigation of the hydrolysis in an excess of acid at 300 K gives the following information. Time/min 0 10 40 50 80 [Ester]/mol dm–3 1.0 0.72 0.24 0.18 0.06 (a) What is the function of dilute sulphuric acid? (b) Why must it be in excess? (c) Plot a graph of [ester] against time. Use your graph to determine the half-life of the reaction at initial [ester] of (i) 1.0 mol dm–3 (ii) 0.8 mol dm–3 (d) What is the order of reaction with respect to hydrogen peroxide? 2 Nitrogen monoxide burns in oxygen to produce nitrogen dioxide. 2NO(g) + O2 2NO2 (g) The reaction is first order with respect to oxygen and second order with respect to nitrogen dioxide. (a) Write the rate law for the reaction. (b) Given that it is a two-step reaction, write the mechanism for the reaction. (c) In a particular experiment, the initial rate of reaction is 9.6 × 10–3 mol dm–3 s–1 when the concentrations of nitrogen monoxide and oxygen are 0.80 mol dm–3 and 0.26 mol dm–3 respectively. (i) Calculate the rate constant (ii) Explain the effect on the reaction rate when [O2 ] is doubled and [NO] is decreased four times. 3 (a) The Arrhenius equation is given as k = Ae –Ea RT . Name the terms represented by the symbols, k, A, Ea and R. (b) The table below shows the variation of the rate constant of dinitrogen tetroxide with temperature. N2 O4 (g) → 2NO2 (g) Temperature/°C 32 40 50 60 k/s–1 0.182 0.466 1.35 3.31 (i) Plot a suitable graph to obtain the value of the activation energy. (ii) State one way of reducing the activation energy of the reaction. (iii) What is the order of reaction? Explain how you arrived at your answer. 4 The mechanism for the reaction between hydrogen peroxide and potassium iodide in the presence of dilute sulphuric acid is as follows: H2 O2 + I– → H2 O + IO– Fast IO– + H+ → HOI Slow HOI + H+ + I– → I2 + H2 O Fast (a) Write the equation for the overall reaction. (b) What is the role of dilute sulphuric acid in the reaction? (c) Derive the rate equation for the reaction. 5 The decomposition of hydrogen peroxide can be accelerated by adding a small amount of dilute hydrochloric acid. The mechanism for the catalysed reaction is: 2H+ + 2Cl– + H2 O2 → Cl2 + 2H2 O Cl2 + H2 O2 → 2Cl– + 2H+ + O2

220 Chemistry Term 1 STPM CHAPTER 5 (a) Write an overall equation for the above exothermic reaction. (b) Sketch the energy profile for the catalysed and uncatalysed reaction. (c) Other than catalyst and concentration, state how you would increase the rate of decomposition of hydrogen peroxide. 6 Sodium bromate(V), NaBrO3 , reacts with sodium bromide in acidic solution according to the equation: BrO3 – + 6H+ + 5Br– → 3Br2 + 3H2 O A kinetic study of the reaction gives the following results. Experiment [BrO3 – ]/mol dm–3 [Br– ]/mol dm–3 [H+]/mol dm–3 Initial rate/mol dm–3 s–1 I 0.40 0.28 0.031 2.50 × 10–3 II 0.60 0.28 0.031 3.75 × 10–3 III 0.60 0.56 0.031 7.50 × 10–3 IV 0.40 0.28 0.062 1.00 × 10–2 (a) Determine the order of reaction with respect to BrO3 – , Br– and H+. (b) Calculate the rate constant, k, for the reaction. 7 Phosphorous(V) chloride undergoes thermal decomposition as follows: PCl5 (g) → PCl3 (g) + Cl2 (g) The graph below refers to the above reaction at temperature T. 100 200 300 400 500 600 700 1.0 0.8 0.6 0.4 0.2 0 Concentration/mol dm–3 Time/min (a) Determine the half-life of the decomposition corresponding to the initial concentration of PCl5 . Concentration of PCl5 /mol dm-3 1.0 0.6 Half-life/min (b) Write the rate equation for the reaction. (c) How would the half-life change if the reaction is carried out at a high temperature? Explain your answer.

CHAPTER CHEMICAL EQUILIBRIA 6 Equilibrium Law and Equilibrium Constant • Heterogeneous system • Relationship between Kc and Kp • Equilibrium constant and rate constant Factors Affecting Equilibrium System • Concentration • Pressure • Temperature • Catalyst • Noble gas Equilibrium and Temperature Chemical Equilibrium Reaction Quotient and Equilibrium Industrial Processes • Haber • Contact • Ostwald Chemical Equilibria Students should be able to: Chemical equilibria • describe a reversible reaction and dynamic equilibrium in terms of forward and backward reactions; • state mass action law from stoichiometric equation; • deduce expressions for equilibrium constants in terms of concentrations, Kc , and partial pressures, Kp, for homogeneous and heterogeneous systems; • calculate the values of the equilibrium constants in terms of concentrations or partial pressures from given data; • calculate the quantities present at equilibrium from given data; • apply the concept of dynamic chemical equilibrium to explain how the concentration of stratospheric ozone is affected by the photodissociation of NO2, O2 and O3 to form reactive oxygen radicals; • state the Le Chatelier’s principle and use it to discuss the effect of catalysts, changes in concentration, pressure or temperature on a system at equilibrium in the following examples: (a) the synthesis of hydrogen iodide, (b) the dissociation of dinitrogen tetroxide, (c) the hydrolysis of simple esters, (d) the Contact process, (e) the Haber process, (f) the Ostwald process; • explain the effect of temperature on equilibrium constant from the equation ln K = – ∆H RT + C Learning earning Outcomes Concept Map

222 Chemistry Term 1 STPM CHAPTER 6 6.1 Reversible Reaction and Dynamic Equilibrium 1 When aqueous sodium hydroxide reacts with aqueous hydrochloric acid, sodium chloride and water are formed. NaOH(aq) + HCl(aq) → NaCl(aq) + H2 O(l) 2 However, when sodium chloride is added to water, it does not produce sodium hydroxide and hydrochloric acid. Such a reaction is known as an irreversible reaction and is represented by a single arrow ‘→’. 3 Other irreversible reactions include: 2H2 O2 (aq) → 2H2 O(l) + O2 (g) Zn(s) + H2 SO4 (aq) → ZnSO4 (aq) + H2 (g) C2 H5 OH(l) + 3O2 (g) → 2CO2 (g) + 3H2 O(l) 4 The concentration/time graph of an irreversible reaction is shown in the diagram: A → B The concentration of the reactant, A, decreases with time while the concentration of the product, B, increases with time. Eventually when all A has been converted to B, the reaction stops completely. Beginning End A B 5 When hydrogen gas and iodine are heated in a closed container in the presence of nickel (as catalyst), hydrogen iodide is formed. H2 (g) + I2 (g) ∆ 2HI(g) 6 When hydrogen iodide is placed in a closed container containing nickel and the content heated, hydrogen iodide will decompose to form hydrogen and iodine. 2HI(g) ∆ H2 (g) + I2 (g) 7 This shows that the above reaction can occur in both directions, and is represented by the ‘’ sign. H2 (g) + I2 (g)  2HI(g) Such a reaction is called a reversible reaction. 8 For a reversible reaction: A + B  C + D Beginning ‘End’ A and B A, B, C and D 9 The reaction proceeding from the left to the right is called the forward reaction. The reaction that proceeds from the right to the left is called the reverse reaction. Concentration Time End of reaction [A] [B] Forward reaction Reverse reaction INFO Reversible Reaction and Dynamic Equilibrium

223 Chemistry Term 1 STPM CHAPTER 6 10 Other examples of reversible reactions are: CH3 COOH(l) + C2 H5 OH(l)  CH3 COOC2 H5 (l) + H2 O(l) N2 (g) + 3H2 (g)  2NH3 (g) PCl5 (g)  PCl3 (g) + Cl2 (g) 2NO2 (g)  N2 O4 (g) 2SO2 (g) + O2 (g)  2SO3 (g) 11 Equilibrium involving gases must be carried out in a closed container to prevent the gases from escaping. 12 When solid calcium carbonate is heated in an open container, all the calcium carbonate would decompose to form calcium oxide and carbon dioxide. CaCO3 (s) ∆ CaO(s) + CO2 (g) No equilibrium is achieved because the carbon dioxide produced escapes into the atmosphere. 13 However, if solid calcium carbonate is heated in a closed container, the following equilibrium would be achieved. CaCO3 (s)  CaO(s) + CO2 (g) At first, calcium carbonate decomposes to calcium oxide and carbon dioxide. Then some of the calcium oxide produced will react with the ‘trapped’ carbon dioxide to produce calcium carbonate. 14 At the end of the reaction, the container would contain a mixture of solid calcium carbonate, solid calcium oxide and gaseous carbon dioxide. The amount of substances present would not change with time as long as the conditions remain unchanged. 15 The concentration/time graph for a reversible reaction is shown in the diagram. C  D The concentration of the reactant, C, decreases with time, while the concentration of the product, D, increases with time. However, the concentration of C does not drop to ‘zero’ as some of the C formed would react to produce D. 16 After time t 1 , the concentration of C and D remains unchanged. The system is said to have achieved dynamic equilibrium. 17 At time t 1 , the rate of the forward reaction becomes equal to the rate of the reverse reaction. After this, for every molecule of C that gets converted to D, a molecule of D gets converted back to C at the same time. As a result, the concentration of C and D remains constant as long as there is no change to the reaction conditions. The system has achieved dynamic equilibrium. Heat CO2 (g) CaCO3 (s) CaCO3 (s) + CaO(s) Concentration Time [C] [D] t 1

224 Chemistry Term 1 STPM CHAPTER 6 18 If we plot the rate for the forward and reverse reaction for a reversible reaction, the following graph is obtained. The rate of the forward reaction decreases with time as the concentration of C decreases. The rate of the reverse reaction increases with time as more and more D is formed. 19 Dynamic equilibrium is defined as the situation where the rate of the forward reaction becomes equal to the rate of the reverse reaction (of a reversible reaction), and the amount of substances present do not vary with time as long as the conditions remain constant. 6.2 Law of Mass Action and Equilibrium Constant 1 The reaction between ethanoic acid and ethanol to form ester is a reversible reaction: CH3 COOH(l) + C2 H5 OH(l)  CH3 COOC2 H5 (l) + H2 O(l) 2 Begining with 1 mole of ethanoic acid and 1 mole of ethanol, how much ester would be produced when the system achieved equilibrium? 3 Similarly, if we begin with 2 moles of ethanoic acid and 1 mole of ethanol, how much ester would be present at equilibrium? 4 The answer is given by the equilibrium law which states that when a reversible reaction has achieved dynamic equilibrium, the ratio of the molar concentrations of the products to the molar concentration of the reactants, both raised to their respective stoichiometric coefficients, is a constant at constant temperature. 5 For a general reversible reaction: aA + bB  cC + dD [C]c [D]d [A]a [B]b = constant (where [ ] = equilibrium molar concentration) The constant is known as the equilibrium constant, Kc . 6 For a gaseous system, the concentration of a gas in a mixture is proportional to its partial pressure. Hence, the equilibrium constant can also be expressed in terms of partial pressure of the gases present at equilibrium. aA(g) + bB(g)  cC(g) + dD(g) Forward rate = Reverse rate Definition of dynamic equilibrium Statement of the equilibrium law Exam Tips Exam Tips Concentration of a gas is directly proportional to its pressure (or partial pressure). The equilibrium law is also known as the 'law of mass action'. Rate Time Dynamic equilibrium Forward reaction Reverse reaction t 1 2008/P1/Q11, Q12 2009/P1/Q10 2011/P1/Q13 2008/P2/Q6(a) 2010/P1/Q11 2009/P2/Q6(a) 2015/P1/Q14

225 Chemistry Term 1 STPM CHAPTER 6 [PC]c [PD]d [PA]a [PB]b = constant (where P is the partial pressure of the gas at equilibrium) The equilibrium constant is given the symbol of K p . 7 The partial pressure of a gas A in a mixture of gases is also given by: PA = XAPT (XA = mole fraction of A, PT = total pressure) ∴ K p is also given by the following expression: KP = [XC]c [XD]d [XA]a [XB]b (PT)∆n [Where ∆n = (c + d) – (a + b)] 6.3 Deduce Equilibrium Constant, K c and K p Homogeneous System 1 Consider the following equilibrium system: 2SO2 (g) + O2 (g)  2SO3 (g) 2 The equilibrium constant, Kc is given by: Kc = [SO3 ]2 [SO2 ]2 [O2 ] 3 The equilibrium constant, Kp is given by: Kp = P2 (SO3 ) P2 (SO2 ) P(O2 ) Heterogeneous System 1 Consider the following equilibrium in a closed container: CaCO3 (s)  CaO(s) + CO2 (g) Applying the equilibrium law, we get [CaO(s)] [CO2 (g)] [CaCO3 (s)] = constant 2 It was found that the equilibrium concentration of carbon dioxide (or its partial pressure) does not depend on the amount of CaCO3 or CaO present, as long as some of each is present at equilibrium. 3 Hence, the equilibrium constant, Kc and K p for the above heterogeneous system can be written as: Kc = [CO2 ] K p = P(CO2 ) Exam Tips Exam Tips Solid is not included in the expression for equilibrium constant. 2014/P1/Q20(a)

226 Chemistry Term 1 STPM CHAPTER 6 Exam Tips Exam Tips For the law of mass action, the order of reaction is taken to be the same as the stoichiometric coefficient of the species concerned. 4 The equilibrium constant for the following system: Ag(s) + Fe3+(aq)  Ag+(aq) + Fe2+(aq) is given by: Kc = [Fe2+] [Ag2+] [Fe3+] Relationship between Kc and Kp 1 Consider a gaseous equilibrium system below: aA(g) + bB(g)  cC(g) + dD(g) KP = (PC)c (PD)d (PA)a (PB)b 2 For a gas: PV = nRT ∴ P = n V RT But n V = concentration PA = [A]RT 3 Substituting the last expression into the expression for KP : KP = [C]c (RT)c [D]d (RT)d [A]a (RT)a [B]b (RT)b = [C]c [D]d [A]a [B]b (RT) (c + d) – (a + b) K p = Kc (RT) ∆n [where ∆n = (c + d) – (a + b)] 4 For gaseous system where the number of moles of gases is the same on both sides of the equation, that is ∆n = 0, then K p = Kc . Relationship between Equilibrium Constant and Rate Constant 1 Consider the general reaction: aA + bB  cC + dD 2 The rate of the forward reaction is given by: R(forward) = k1 [A] a [B] b (where k1 = rate constant for the forward reaction) The rate of the reverse reaction is given by: R(reverse) = k–1[C] c [D] d (where k–1 = rate constant for the reverse reaction) 2016/P1/Q12

227 Chemistry Term 1 STPM CHAPTER 6 At equilibrium: R(forward) = R(reverse) k1 [A] a [B] b = k–1[C] c [D] d [C]c [D]d [A]a [B]b = k1 k–1 ∴ k1 k–1 = Kc 3 Since the rate constant of a reaction is affected by temperature (and presence of a catalyst), consequently the equilibrium constant is only affected by changes in temperature. Example 6.1 Write the expression of Kc and K p (where appropriate) for the following equilibrium system. State the unit for the equilibrium constants. (a) CH3 COOH(l) + C2 H5 OH(l)  CH3 COOC2 H5 (l) + H2 O(l) (b) PCl5 (g)  PCl3 (g) + Cl2 (g) (c) NH2 COONH4 (s)  CO2 (g) + 2NH3 (g) (d) H2 (g) + I2 (g)  2HI(g) Solution (a) Kc = [CH3 COOC2 H5 ][H2 O] [CH3 COOH][C2 H5 OH] Kc has no unit. (b) Kc = [PCl3 ][Cl2 ] [PCl5 ] Unit of Kc = mol dm–3 K p = P(PCl3 ) P(Cl2 ) P(PCl5 ) Unit of K p = atm or Pa P(PCl5 ) (c) Kc = [CO2 ][NH3 ] 2 Unit of Kc = mol3 dm–9 K p = P[CO2 ]P 2 [NH3 ] Unit of K p = atm3 or Pa3 (d) Kc = [HI]2 [H2 ][I2 ] Kc has no unit. K p = P2 (HI) P(H2 ) P(I2 ) K p has no unit. Info Chem The values of equilibrium constants depend on temperature only.

228 Chemistry Term 1 STPM CHAPTER 6 6.4 Calculation Involving Equilibrium Constant Example 6.2 1 mol of glacial ethanoic acid is mixed with 1 mol of absolute ethanol in a closed flask and allowed to achieve equilibrium at room temperature. The equilibrium mixture was found to contain 0.67 mol of ethyl ethanoate. Calculate Kc for the following equilibrium at room temperature. CH3 COOH(l) + C2 H5 OH(l)  CH3 COOC2 H5 (l) + H2 O(l) Solution CH3 COOH C2 H5 OH CH3 COOC2 H5 H2 O Initial/mol 1.0 1.0 0 0 Final/mol 1 – 0.67 = 0.33 1 – 0.67 = 0.33 0.67 0.67 Kc = [CH3 COOC2 H5 ][H2 O] [CH3 COOH][C2 H5 OH] = (0.67)2 (0.33)2 = 4.12 Example 6.3 The equilibrium constant for the following reaction is 4.12 at 298 K. CH3 COOH(l) + C2 H5 OH(l)  CH3 COOC2 H5 (l) + H2 O(l) Calculate the mass of ethyl ethanoate produced from a mixture containing 2 mol of ethanoic acid and 2 mol of ethanol. Solution Let the amount of ethyl ethanoate produced be x mol. CH3 COOH C2 H5 OH CH3 COOC2 H5 H2 O Initial/mol 2.0 2.0 0 0 Final/mol 2 – x 2 – x x x Kc = [CH3 COOC2 H5 ][H2 O] [CH3 COOH][C2 H5 OH] 4.1 = (x)(x) (2 – x)(2 – x) ∴ x = 1.34 mol Mass of ethyl ethanoate produced = 1.34  88 g = 117.9 g 2013/P1/Q17 2016/P1/Q20(a)(i) 2014/P1/Q11, Q20(b)(i) 2018/P1/Q20(a)

229 Chemistry Term 1 STPM CHAPTER 6 Example 6.4 1 mole of nitrogen and 3 moles of hydrogen were allowed to achieve equilibrium at 450 °C and a total pressure of 20000 kPa. The equilibrium mixture was found to contain 15% ammonia. (a) Calculate K p for the following reaction: N2 (g) + 3H2 (g)  2NH3 (g) (b) What is the value of K p for the reverse reaction: 2NH3 (g)  N2 (g) + 3H2 (g) at the same temperature? Solution (a) The partial pressure of NH3 in the equilibrium mixture = 20 000  15 100 = 3000 kPa The total partial pressure of N2 and H2 in the mixture = 20 000 – 3000 = 17 000 kPa ∴ Partial pressure of N2 = 1 4  17 000 = 4250 kPa Partial pressure of H2 = 3 4  17 000 = 12 750 kPa K p = P2 (NH3 ) P(N2 ) P3 (H2 ) = (17 000)2 (12 750)3 (4250) = 3.28  10–8 kPa–2 (b) 2NH3 (g)  N2 (g) + 3H2 (g) K p ′ = P(N2 )P3 (H2 ) P2 (NH3 ) = 1 KP = 1 3.28 × 10–8 = 3.05  107 kPa2 Example 6.5 The equilibrium constant for the following reaction at 300 K is 4.0. CH3 COOH(l) + C2 H5 OH(l)  CH3 COOC2 H5 (l) + H2 O(l) What mass of ethanoic acid needs to be added to 59.8 g of ethanol to produce 66.0 g of ester at equilibrium? Solution No. of moles of ethanol = 59.8 46 = 1.30 mol No. of moles of ester needed = 66 88 = 0.75 mol Let the number of moles of ethanoic acid required = x mol CH3 COOH C2 H5 OH CH3 COOC2 H5 H2 O Initial/mol x 1.30 0 0 Final/mol x – 0.75 1.30 – 0.75 = 0.55 0.75 0.75 At equilibrium: (0.75)(0.75) (0.55)(x – 0.75) = 4   ∴ x = 1.01 mol Hence, the mass of ethanoic acid required = 1.01  60 = 60.6 g

230 Chemistry Term 1 STPM CHAPTER 6 Quick Check 6.1 1 (a) Steam reacts with carbon to produce hydrogen and carbon monoxide. The reaction is reversible. (i) Write a balanced equation for the reaction. (ii) Write an expression for K p , the equilibrium constant and state its unit. (b) In an experiment, steam was passed over carbon at 800 °C. The following equilibrium pressures were measured: P(CO) = 180 kPa P(H2 O) = 115 kPa (i) What is the equilibrium pressure of hydrogen? (ii) Calculate the equilibrium constant, K p , at 800 °C. 2 Sulphur dioxide reacts with oxygen to produce sulphur trioxide which is then used to manufacture sulphuric acid. 2SO2 (g) + O2 (g)  2SO3 (g) A mixture containing sulphur dioxide and oxygen with a mole ratio of 2 : 1, with a total initial pressure of 3 atm is passed over vanadium pentoxide at 450 °C. The equilibrium partial pressure of sulphur trioxide was found to be 1.9 atm. (a) What is the function of vanadium pentoxide? (b) Calculate the equilibrium partial pressure of sulphur dioxide and oxygen. (c) Sketch a graph of the variation of the partial pressures of sulphur dioxide, oxygen and sulphur trioxide with time. (d) Determine the equilibrium constant, K p at 450 °C. 3 Phosphorus(V) chloride dissociates according to the equation: PCl5 (g)  PCl3 (g) + Cl2 (g) A 1.0 mol sample of PCl5 is allowed to dissociate in a 1.0 dm3 flask at 750 K. At equilibrium, 25.0% of PCl5 has dissociated. (a) What is the total pressure in the flask when the system has achieved equilibrium? (b) Calculate the equilibrium constant, Kc at 750 K. (c) At what pressure would the percentage of dissociation be increased to 40% at 750 K? Reaction Quotient (Q) and Equilibrium 1 Consider the following reversible reaction with Kc = 4.0 at 298 K. A + B  C + D Suppose we are given a mixture containing 1.0 mol dm–3 of C and D, 0.7 mol dm–3 of A and 1.2 mol dm–3 of B at 298 K. Is the system in equilibrium? If not, then what is the net direction of reaction? 2 In order to determine whether the system is at equilibrium or not, we calculate the reaction quotient, Q, of the mixture. 3 For a general reaction: aA + bB  cC + dD The reaction quotient, Q = [C]c [D]d [A]a [B]b The reaction quotient

231 Chemistry Term 1 STPM CHAPTER 6 4 If Q = Kc , the system is in equilibrium. If Q  Kc , the system is not in equilibrium. There is too little product. A net forward reaction will occur. If Q  Kc , the system is not in equilibrium. There is too much product. A net reverse reaction will occur. 5 Looking at the example above, the reaction quotient, Q = [C][D] [A][B] = (1)(1) (0.7)(1.2) = 1.19 Since Q  Kc (which is 4.0), the system is not in equilibrium. There is a net forward reaction to produce more C and D so as to increase the value of Q until it is equal to 4.0. Example 6.6 H2 (g) + I2 (g)  2HI(g) The Kc for the above reaction at 200 °C is 25.0. A mixture containing 1.0 mol dm–3 of H2 and I2 and 1.2 mol dm–3 of HI is left standing at 200 °C. Determine the net direction of the reaction. Solution The reaction quotient for the reaction, Q = [HI]2 [H2 ][I2 ] = (1.2)2 (1)(1) = 1.44 Since Q  Kc , there is a net forward reaction to produce more HI. Example 6.7 N2 (g) + 3H2 (g)  2NH3 (g) Kc = 0.80 mol–2 dm6 A 1 dm3 flask contains 0.60 mol of NH3 , 0.020 mol N2 and 0.15 mol H2 . Predict the direction of reaction at the same temperature. Solution The reaction quotient, Q = [NH3 ]2 [N2 ][H2 ]3 = (0.60)2 (0.020)(0.15)3 = 800 Since Q  Kc , there is a net reverse reaction to produce more N2 and H2 . Info Chem Q has the same form as Kc . But the concentrations of the substances involved are not necessarily equilibrium concentrations.

232 Chemistry Term 1 STPM CHAPTER 6 Quick Check 6.2 1 The equilibrium constant for the following reaction is 0.25 atm–1 at 500 °C. 2SO2 (g) + O2 (g)  2SO3 (g) Sulphur dioxide, oxygen and sulphur trioxide were introduced into an evacuated flask until the partial pressures of SO2 , O2 and SO3 are 0.75 atm, 0.45 atm and 1.1 atm respectively at 500 °C. Determine (a) if the system is in equilibrium, (b) the direction of net reaction, if any. 2 CH3 COOH(l) + C2 H5 OH(l)  CH3 COOC2 H5 (l) + H2 O(l) The Kc for the above reaction at 298 K is 4.0. A 1 dm3 flask contains 0.50 mol of CH3 COOH, 0.50 mol of C2 H5 OH, 0.85 mol of CH3 COOC2 H5 and 1.8 mol of water at 298 K. (a) Show that the system is not in equilibrium. (b) State the direction of net reaction. 6.5 Nitrogen Dioxide and Stratospheric Ozone 1 The ozone layer in the stratosphere (the second layer of our atmosphere, about 15 – 50 km from the Earth’s surface) helps to block out most of the ultraviolet radiation (from the Sun) from reaching the Earth. 2 Nitrogen dioxide in the exhaust fumes of motor vehicles or formed during lightning strikes is very important in the formation and loss of tropospheric ozone.    3 In the stratosphere, nitrogen dioxide in the presence of ultraviolet radiation, dissociates to produce oxygen free radicals. NO2 (g) → NO(g) + O• (g) The oxygen free radicals react with other oxygen molecules to produce ozone. O2 (g) + O• (g) → O3 (g) The overall reaction is: NO2 (g) + O2 (g) → NO(g) + O3 (g) 4 On the other hand, nitrogen monoxide reacts with ozone to produce nitrogen dioxide and oxygen. NO(g) + O3 (g) → NO2 (g) + O2 (g) 5 A dynamic equilibrium thus exists between nitrogen dioxide and ozone: NO2 (g) + O2 (g)  NO(g) + O3 (g) This equilibrium helps to maintain the concentration of the ozone in the stratosphere.

233 Chemistry Term 1 STPM CHAPTER 6 6.6 Factors Affecting Equilibrium System and Le Chatelier’s Principle 1 How would changes in concentration, pressure, temperature and present of a catalyst affect an equilibrium system? 2 Qualitatively, the effect can be predicted using the Le Chatelier’s Principle which states that when a system in dynamic equilibrium is disturbed, the position of equilibrium will change to remove the effect of the disturbance so that equilibrium is reestablished. 3 As most important industrial processes (such as the Haber process and the Contact process) are reversible, industrial engineers have to find the optimum conditions so as to get maximum yield at the shortest time possible and at the same time keep the operational cost low. Effect of Concentration 1 Consider the following reversible reaction at equilibrium: A B Kc = [B] [A] 2 If more A were added to the equilibrium mixture, there is a momentary decrease in the reaction quotient, and Q , Kc . 3 In order for equilibrium to reestablish, the system must convert some of the added A into B. The equilibrium will shift to the right-hand side. At the new equilibrium, more B will be present. 4 The scenario is the same if some B was removed from the original equilibrium mixture. The equilibrium will shift to the right-hand side to replace some of the B that was removed. 5 Adding more B or removing some A will cause the equilibrium to shift to the left to remove some of the B added, or to replace some of the A removed. 6 For example, an aqueous solution of bismuth(II) chloride is cloudy due to the hydrolysis of the salt to produce insoluble bismuth oxychloride according to the equilibrium: BiCl3 (aq) + H2 O(l)  BiOCl(s) + 2HCl(aq) 7 When a little concentrated hydrochloric acid is added to the mixture, the solution turns clear. Addition of HCl displaces the equilibrium to the left-hand side. The BiOCl solid dissolves to form BiCl3 (aq). 8 However, when water is added to the new clear solution, the equilibrium will shift to the right, and BiOCl is precipitated. The solution turns cloudy again. Exam Tips Exam Tips • Increasing the concentration of the reactants (or decreasing the concentration of the products) favours the forward reaction. • Increasing the concentration of the products (or decreasing the concentration of the reactants) favours the reverse reaction INFO Le Chatelierʼs Principle and Applications 2008/P1/Q11 2009/P1/Q43, Q44 2009/P2/Q6(b) 2010/P1/Q12 2011/P1/Q12 2014/P1/Q20(b)(iii) 2017/P1/Q12 2013/P1/Q12, Q17(d)(i) 2015/P1/Q17(2) 2018/P1/Q11, Q20(b)

234 Chemistry Term 1 STPM CHAPTER 6 9 Change in concentration has no effect on the equilibrium constant. Example 6.8 Consider the following equilibrium system: Ag+(aq) + Fe2+(aq)  Ag(s) + Fe3+(aq) (a) Write the expression for the equilibrium constant, Kc , of the reaction. (b) State what happens when the following changes were made after the system has achieved equilibrium. (i) Adding more aqueous silver nitrate (ii) Adding more iron(III) nitrate (iii) Adding more silver Solution (a) Kc = [Fe3+] [Ag+][Fe2+] (b) (i) The equilibrium shifts towards the right-hand side and more Ag and Fe3+ will be produced. (ii) The equilibrium shifts to the left-hand side and more Ag+ and Fe2+ will be produced. (iii) There is no change in the equilibrium position, because Ag does not appear in the expression for Kc . Example 6.9 Concentration Time [N2 O4 ] [NO2 ] The graph above refers to the following equilibrium system: N2 O4 (g)  2NO2 (g) Kc = 1.30 mol dm–3 (a) Sketch the change in the concentration of N2 O4 and NO2 when more N2 O4 (g) were added, at constant temperature, to the above equilibrium mixture until equilibrium is reestablished. (b) What is the Kc value for the new equilibrium mixture? Exam Tips Concentration has no effect on the value of the equilibrium constant.

235 Chemistry Term 1 STPM CHAPTER 6 Solution (a) Concentration Time [N2 O4 ] [NO2 ] Addition of more [N2 O4 ] Addition of more N2 O4 shifts the equilibrium to the right to produce more NO2 . (b) The Kc value is still 1.30 mol dm–3 because there is no change in temperature. Effect of Pressure 1 Pressure only affects reactions involving gases. This is because pressure has no effect on the volume occupied by a solid or liquid. 2 When the pressure on a gaseous system is increased, the volume occupied by the gas decreases (Boyle’s Law). The gas molecules are pushed closer to one another. Hence, to relieve the increased internal pressure, the equilibrium will shift in direction as to decrease the total number of gaseous particles present. 3 Conversely, when the pressure of a gaseous system is decreased, the equilibrium will shift in direction so as to increase the number of gaseous particles present. 4 Consider the following system in equilibrium: 2SO2 (g) + O2 (g) 2SO3 (g) (a) When the pressure of the system is increased, the equilibrium will shift to the right-hand side as the forward reaction is accompanied by a decrease in the total number of gaseous particles. (b) When the pressure of the system is decreased, the equilibrium will shift to the left-hand side as the reverse reaction is accompanied by an increase in the total number of gaseous particles. 5 However, a change in pressure does not affect the following system: H2 (g) + I2 (g) 2HI(g) This is because there is no change in the total number of gaseous particles either in the forward or in the reverse reaction. Effect of increasing pressure: At the new equilibrium, there will be more N2 O4 than before. Gas particle 2008/P1/Q6(c)

236 Chemistry Term 1 STPM CHAPTER 6 6 Summary, Increasing pressure favours the reaction that decreases the total number of gaseous particles. Decreasing pressure favours the reaction that increases the total number of gaseous particles. Addition of a Noble Gas at Constant Volume 1 Consider the following system in equilibrium: PCl5 (g) PCl3 (g) + Cl2 (g) Kp = [P(PCl3 )][P(Cl2 )] [P(PCl5 )] 2 When a noble gas (such as helium) is added to a gaseous system in equilibrium at constant volume, the total pressure of the system increases, because the number of gaseous particles present increases. 3 However, the partial pressures of PCl5 , PCl3 and Cl2 do not change. This is because the partial pressure of a gas in a mixture of gases does not depend on the presence or absence of other gases as long as the volume remains the same [Refer to Section 4.1, Dalton’s law of partial pressure]. 4 Since there is no change in the partial pressures of all the gases present, there is no change in the position of equilibrium. Addition of a Noble Gas at Constant Pressure 1 Consider the following system in equilibrium: PCl5 (g) PCl3 (g) + Cl2 (g) 2 When a noble gas is added to the system under constant pressure, the volume occupied by the gas will increase so as to maintain the same pressure. 3 This causes the partial pressures of PCl5 , PCl3 and Cl2 to decrease. The net effect is equivalent to decreasing the total pressure of the original equilibrium system. 4 As a result, the equilibrium will shift to the right to increase the total number of gaseous particles. Example 6.10 Explain the effect of pressure on the following equilibrium system: 2NO2 (g)  N2 O4 (g) Solution Increasing pressure will shift the equilibrium to the right-hand side because the forward reaction is accompanied by a decrease in the total number of moles of gaseous particles. Conversely, decreasing pressure will shift the equilibrium to the left-hand side, because the reverse reaction is accompanied with an increase in the total number of gaseous particles. Exam Tips Changes in pressure do not affect the value of the equilibrium constant. No change in the partial pressure of the gases present. Partial pressure of the gases decreases. Info Chem NO2 is brown. N2 O4 is colourless. 2013/P1/Q17(d)(ii)

237 Chemistry Term 1 STPM CHAPTER 6 Effect of Temperature Temperature is the only factor that affects the magnitude of the equilibrium constant K p and Kc . For Exothermic Reaction 1 Consider the following reaction: N2 (g) + 3H2 (g)  2NH3 (g) ∆H = negative 2 When the temperature of the above equilibrium system is decreased (that is heat is removed from the system), the equilibrium must shift so as to release heat to replace the heat being removed. 3 The equilibrium would shift towards the right-hand side as the forward reaction is exothermic. More ammonia will be formed, Kc increases. 4 The Kc for the following reaction at various temperature is given below: H2 (g) + I2 (g)  2HI(g) ∆H = negative Temperature/°C Kc 25 795 227 160 430 25 5 The equilibrium constant increases with decreasing temperature. Example 6.11 Discuss how the rate constants for the following reaction are affected by changes in temperature. A + B k1 k–1 C + D ∆H = Negative Solution Using Kc = k1 k–1 Increasing temperature increases the values of k1 and k–1. However, the value of Kc decreases. This means that the increase in k–1 is more than the increase in k1 . Decreasing temperature decreases the values of k1 and k–1. However, the value of Kc increases. This means that the decrease in k–1 is more than the decrease in k1 . Kc decreases with increasing temperature. Exam Tips Exam Tips The equation can be written by incorporating heat into the equation: N2 + 3H2  2NH3 + Heat Decreasing temperature (removing heat) would cause the equilibrium to shift to the right-hand side. 2016/P1/Q20(a)(ii)

238 Chemistry Term 1 STPM CHAPTER 6 For Endothermic Reaction 1 Consider the following reaction in equilibrium: PCl5 (g)  PCl3 (g) + Cl2 (g) ∆H = positive 2 When the temperature of the above equilibrium system is increased (that is heat is supplied to the system), the equilibrium will shift so as to absorb the heat supplied. Hence, the equilibrium will shift to the right, and more PCl5 would decompose. 3 The equilibrium constant will increase with increasing temperature. 4 The Kc for the following reaction at various temperature is given below: N2 O4 (g)  2NO2 (g) ∆H = positive Temperature/°C Kc 77 3.9 127 48 177 347 227 1700 5 For a reaction where ∆H = 0, temperature has no effect on such systems. 6 The sketches below show the variation of the equilibrium constant for an endothermic reaction and an exothermic reaction with temperature. Temperature Endothermic reaction Kc or Kp Temperature Exothermic reaction Kc or Kp Example 6.12 Consider the following reaction: N2 O4 (g)  2NO2 (g) The degree of dissociation of N2 O4 at 100 kPa and 150 °C is 15% while the degree of dissociation at 350 °C is 23% at 100 kPa. (a) State if the forward reaction is endothermic or exothermic. Explain your answer. (b) Calculate the Kp for the reaction at (i) 150 °C and (ii) 350 °C Kc increases with increasing temperature. The relationship between Kc (or Kp   ) with temperature is an exponential one. Increasing temperature ΔH = –ve ΔH = +ve Decreasing temperature A + B C + D Exam Tips PCI5 + Heat  PCI3 + CI2 Increasing temperature (adding heat) would shift the equilibrium to the right-hand side. Exam Tips Exam Tips • Endothermic reactions are favoured by high temperatures. • Exothermic reactions are favoured by low temperatures.

239 Chemistry Term 1 STPM CHAPTER 6 Solution (a) Increasing temperature shifts the equilibrium to the right (more N2 O4 dissociates). Therefore, the forward reaction is favoured by high temperature. The forward reaction is endothermic. (b) (i) At 150 °C N2 O4 2NO2 Initial/mol 1 0 Final/mol 1 – 0.15 = 0.85 0.15  2 = 0.30 Partial pressure of NO2 = 0.3 (0.85 + 0.30)  100 = 26.1 kPa Partial pressure of N2 O4 = 100 – 26.1 = 73.9 kPa K p = (26.1)2 73.9 = 9.22 kPa (ii) At 350 °C N2 O4 2NO2 Initial/mol 1 0 Final/mol 1 – 0.23 = 0.77 0.23  2 = 0.46 Partial pressure of NO2 = 0.46 (0.77 + 0.46)  100 = 37.4 kPa Partial pressure of N2 O4 = 100 – 37.4 = 62.6 kPa K p = (37.4)2 62.6 = 22.3 kPa Effect of Catalyst 1 Catalyst is a substance that increases the rate of a chemical reaction, but itself is not consumed in the reaction. It can be recovered chemically unchanged at the end of the reaction. 2 A catalyst allows the reaction to take place by another pathway of lower activation energy as shown by the energy profile below: Reaction pathway Energy Reactants Products Ea (uncatalysed) Ea (catalysed) Exam Tips Exam Tips Catalyst does not affect the equilibrium mixture or the equilibrium constant.

240 Chemistry Term 1 STPM CHAPTER 6 3 In the presence of a catalyst, the activation energy for the forward reaction and the activation energy for the reverse reaction are reduced by the same amount. 4 This means that the rate of the forward and the reverse reactions are increased by the same factor. Hence, a catalyst does not affect the position or composition of an equilibrium system. It only shortens the time taken for the equilibrium to be achieved. 6.7 Equilibrium and Temperature 1 The relationship between equilibrium constant and temperature is given by the equation: Kc = Ae –∆H RT Where Kc = Equilibrium constant A = Constant ∆H = Enthalpy change of reaction R = Universal gas constant T = Absolute temperature 2 Taking log throughout: ln Kc = ln A – ∆H RT 3 A graph of ln Kc against 1 T gives a straight line with the slope equal to – ∆H R : For endothermic reaction In Kc Gradient = – ΔH R 1 T For exothermic reaction In Kc 1 T Gradient = – ΔH R A catalyst increases the rate for the forward and reverse reactions by the same factor. Exponential relationship Info Chem The equation can also be written as: In Kc = – ∆H RT + C Exam Tips Exam Tips The equation can also be written as: In K2 K1 = ∆H R ( 1 T1 – 1 T2 ) where K1 and K2 are the equilibrium constants at temperature T1 and T2 respectively. 2014/P1/Q20(b)(ii)

241 Chemistry Term 1 STPM CHAPTER 6 4 Summary Change Rate Rate constant Equilibrium composition Equilibrium constant Increasing concentration Increases Unchanged Changes Unchanged Decreasing concentration Decreases Unchanged Changes Unchanged Increasing pressure Increases Unchanged Changes Unchanged Decreasing pressure Decreases Unchanged Changes Unchanged Increasing temperature Increases Increases Changes Changes Decreasing temperature Decreases Decreases Changes Changes Addition of catalyst Increases Increases Unchanged Unchanged Example 6.13 The data for the system: H2 (g) + I2 (g)  2HI(g) is given below: T/°C 750 945 1089 1200 Kp 106.0 20.2 8.0 4.4 Plot a suitable graph to determine the enthalpy change for the reaction. Solution 1 T /K–1 9.8  10–4 8.2  10–4 7.3  10–4 6.8  10–4 ln Kp 4.66 3.01 2.08 1.48 1 2 3 4 5 ln KP 6 7 8 9 10 – (fi 10–4)/K 1 –1 T Gradient = –––––––––– = 1.6 fi 104 K 4.35 – 1.65 (9.5 – 6.95) fi 10–4 Gradient of the graph = 1.06  104 ∴ –∆H R = 1.06  104 ∆H = –88.09  103 J = –88.09 kJ