142 Chemistry Term 1 STPM CHAPTER 4 (a) On mixing, the total volume is 3 dm3 . Partial pressure of H2 : P(H2 ) 3 = (1.25 106 ) 1 P(H2 ) = 4.17 105 Pa Partial pressure of CO2 : P(CO2 ) 3 = (2.50 105 ) 2 P(CO2 ) = 1.67 105 Pa (b) Total pressure = (4.17 + 1.67) 105 Pa = 5.84 105 Pa Example 4.8 A 50 dm3 vessel contains 0.65 mol CO2 , 0.80 mol O2 and 1.02 mol N2 at 298 K. (a) Calculate partial pressure of each gas in the mixture. (b) What is the total pressure of the mixture? (c) If carbon dioxide is removed through absorption by aqueous sodium hydroxide, (i) what are the partial pressures of the remaining gas? (ii) what is the total pressure of the new mixture? Solution (a) For CO2 : P (50 10–3) = 0.65 8.31 298 P = 32.2 103 Pa = 32.2 kPa For O2 : P = 0.80 0.65 32.2 kPa = 39.6 kPa For N2 : P = 1.02 0.65 32.2 kPa = 50.5 kPa (b) Total pressure = (32.2 + 39.6 + 50.5) kPa = 122.3 kPa (c) (i) The partial pressures of O2 and N2 still remain the same. P(O2 ) = 39.6 kPa P(N2 ) = 50.5 kPa (ii) Total pressure = (39.6 + 50.5) kPa = 90.1 kPa Quick Check 4.5 1 Nitrogen reacts with hydrogen according to the equation: N2 (g) + 3H2 (g) → 2NH3 (g) A 1 m3 vessel contains 0.5 mol of N2 and 1.0 mol of H2 at 450 °C. (a) Calculate the total pressure in the vessel before the reaction. (b) Calculate total pressure after the reaction. Exam Tips Exam Tips The partial pressure of a gas does not change even if other gases are added or removed from the container so long as the volume and temperature remain constant.
143 Chemistry Term 1 STPM CHAPTER 4 2 A 5 dm3 bulb contains oxygen at a pressure of x atm is connected to a 10 dm3 bulb containing carbon dioxide at 3.2 atm. When the valve between them is opened, the final equilibrium pressure is 5 atm. What is the value of x? 3 A mixture containing 50% CO, 30% N2 and 20% sulphur dioxide by volume has a total pressure of 240 kPa. (a) Calculate the partial pressure of each gas in the mixture. (b) If SO2 is removed from the mixture, calculate (i) the partial pressure of the remaining gases, and (ii) the total pressure of the new mixture. 4 A 2.5 dm3 vessel contains 16.0 g of oxygen at 300 K. (a) What is the pressure exerted by the sample of oxygen gas? (b) 4.0 g of hydrogen are then forced into the vessel. Calculate the total pressure of the mixture at 300 K if the volume of the system remains the same. (c) The mixture is continuously sparked and then allowed to cool to 300 K. What is the final pressure of the system. State any assumptions you made in your calculations. Maxwell-Boltzmann Distribution Curve 1 According to the kinetic theory, the molecules in a sample of gas is always in a state of random motion. 2 However, not all the molecules are moving at the same speed. Some move faster than others. 3 The molecular speed (as well as kinetic energy) of the gas particles is spread over a wide range. 4 The graph below shows the distribution of molecular speed (or molecular energy) of a sample of gas at a constant temperature. Molecular speed/energy n1 E1 Emean V1 Vmean No. of molecules The graph is called the Maxwell-Boltzmann distribution curve. 5 The graph shows that very few particles have very high or very low molecular speed (or energy). Most of the particles have the mean molecular speed or the mean energy (Emean). 6 The number of molecules having molecular speed of V1 or energy of E1 is given by n1 . 7 The total number of particles having energy between E1 and E2 is given by the area of the shaded part of the curve, as shown in the diagram. 2011/P1/Q10 Distribution of molecular energy or molecular speed E Molecular speed/energy 1 E2 No. of molecules Exam Tips Exam Tips The curve is not symmetrical.
144 Chemistry Term 1 STPM CHAPTER 4 8 The following graph shows the distribution curves of a sample of gas at three different temperatures. 300 K 1000 K 2000 K Molecular speed/energy Relative number of molecules 9 As the temperature increases: (a) there is a greater spread of energies, (b) the mean energy increases, (c) the curve gets flatter, (d) less molecules have low energies, (e) more molecules have higher energies. 10 In all cases, the total area under each curve (which corresponds to the total number of molecules in the system) remains the same. Example 4.9 Sketch the Maxwell-Boltzmann distribution curves for the molecular speed of a sample of hydrogen and carbon dioxide at 25 °C. Explain the shapes of your curve. Solution Hydrogen Carbon dioxide Molecular speed Relative number of molecules The carbon dioxide molecule is heavier than the hydrogen molecules. However, since the average kinetic energy of both gases is the same, more hydrogen molecules (which are lighter) will have higher molecular speed compared to the heavier carbon dioxide molecules.
145 Chemistry Term 1 STPM CHAPTER 4 Deviation of Real Gases from Ideal Behaviour 1 A gas that obeys the ideal gas equation, PV = nRT, perfectly under any conditions is called an ideal gas. 2 From the ideal gas equation: PV nRT = 1 [The ratio PV nRT is called the compressibility factor of a gas.] 3 A plot of PV nRT against P gives a straight line parallel to the P axis: P PV nRT 1 4 However, when PV nRT is plotted against P for real gases, this graph is obtained. P PV nRT 1 H2 CO2 N2 Ideal gas 5 In other words, real gases show deviation from ideal behaviour. The deviation is more profound when the gases are under very high pressures. P/atm PV nRT 1 Ideal gas 0 100 200 300 400 500 600 6 Deviation also increases with decreasing temperature. P/atm PV nRT 1 Ideal gas 0 100 200 300 400 500 600 100 K 500 K 1200 K 7 These deviations arise because for real gases, (a) there are intermolecular forces between the gas molecules, (b) the gas molecules has finite volume. Deviation of real gases from ideal behaviour. The higher the pressure, the greater the deviation. The lower the temperature, the greater the deviation. Presence of intermolecular forces, gas particles have finite volume. Exam Tips Exam Tips Exam Tips Exam Tips The compressibility factor of an ideal gas is equal to 1. The compressibility of a real gas is either greater or smaller than 1 depending on the conditions. 2009/P2/Q1(a) 2011/P1/Q1 2013/P1/Q7 2014/P1/Q16(a)(b) 2017/P1/Q7 INFO Deviation from Ideal Behaviours
146 Chemistry Term 1 STPM CHAPTER 4 8 However, real gases show near ideal behaviour when the pressure on the gas is low (especially when P approaches zero) and the temperature is high. 9 At low pressures, the volume occupied by the gas is very large compared to the volume of the gas molecules. Thus, the volume of the gas molecules can be ignored. Furthermore, the gas molecules are so far apart that the intermolecular forces between them are so weak that they too can be ignored. 10 At high temperature, the kinetic energy of the gas molecules is very high. The intermolecular forces are so weak by comparison that they can be ignored. 11 At room conditions (298 K and 101 kPa), the deviation of real gases from ideal behaviour is only about 3%. Thus, for ‘normal calculations’, the ideal gas equation can still be used. 12 Small non-polar gases such as helium and hydrogen show the least deviation while large polar molecules such as ammonia and hydrogen fluoride show larger deviation. Example 4.10 The plot shows the variation of PV nRT against P for a sample of nitrogen gas at 25 °C. Explain the shape of the graph. Solution When the pressure on the sample of nitrogen is zero, it shows ideal behaviour, and PV nRT = 1. When the pressure on the sample increases, the molecules are pushed closer together and will attract one another. This attraction causes a larger decrease in the volume than expected, and PV nRT 1. Negative deviation occurs. At very high pressures, the molecules are so close that their electron clouds repel one another making the gas more difficult to compress causing the PV nRT ratio to increase, and PV nRT 1, this resulted in positive deviation. P PV nRT 1 0 P PV nRT 1 0 A B H2 and He show the least deviation. Exam Tips Exam Tips Exam Tips Low pressures: Intermolecular forces can be ignored as particles are very far apart. High temperature: K.E. of particles is high. Intermolecular forces can be ignored. Exam Tips Exam Tips Negative deviation is due to intermolecular attraction. Positive deviation is due to intermolecular repulsion. Info Chem PV nRT 1 P H2 H2 does not show negative deviation. Hence, the intermolecular attraction is insignificant.
147 Chemistry Term 1 STPM CHAPTER 4 Example 4.11 1 mol of carbon dioxide in a volume of 2 dm3 exerts a pressure of 1.20 106 Pa at 298 K. (a) Calculate the pressure exerted by 1 mol of an ideal gas in a volume of 2 dm3 and 298 K. (b) Explain the difference between the values you obtained in (a) and that of carbon dioxide. Solution (a) Using: PV = nRT P (2 10–3) = 1 8.31 298 P = 1.24 106 Pa (b) The pressure exerted by carbon dioxide is less than expected. This shows that under the conditions of the experiment, carbon dioxide does not show ideal bahaviour. The lower pressure is due to the presence of intermolecular attractive forces between carbon dioxide molecules. This leads to the collision between the molecules and the wall of the container to be less energetic. 4.2 Liquids Kinetic Theory of the Liquid State 1 The particles in a liquid are held together by intermolecular forces of intermediate strength into clusters with little empty space between them. The arrangement is random. Liquid cluster ‘Empty space’ 2 The particles can move randomly throughout the body of the liquid but not out of it. As a result, a liquid has a fixed volume but has no fix shapes. 3 The energy of the particles is less than that of a gas, but higher than that of a solid. 4 A liquid is not easily compressed. Intermolecular attractive forces decrease the energy of collision between the particles and the walls of the container. Collision is less energetic Attractive force Walls of the container 2016/P1/Q6
148 Chemistry Term 1 STPM CHAPTER 4 Vapour Pressure 1 When an open container of water is left standing for sometimes, the level of water drops. Some of the water has escaped from the bulk of liquid. 2 This process is called evaporation. 3 Evaporation occurs when some of the liquid molecules at the surface of the liquid have enough energy to overcome the intermolecular forces that hold them together, and leave the surface as vapour. 4 In an open container, evaporation will occur until there is no liquid left [Figure (a)]. 5 However, if the liquid is in a closed container, the vapour cannot escape and is trapped in the container and will collide with the walls of the container. The collisions exert a pressure called the vapour pressure of the liquid [Figure (b)]. O O O O O O O O O O Vapour Liquid (a) (b) 6 Some of the vapour molecules after colliding with the walls of the container or with one another might re-enter the liquid. This is called condensation. O O O O O O Vapour Liquid 7 In the beginning, as there are very little vapour molecules present, the rate of evaporation is higher than the rate of condensation. This causes the vapour pressure to increase with time. 8 Eventually the rate of evaporation and condensation becomes the same. The number of vapour molecules remains constant. The vapour pressure attains a maximum value which remains constant as long as the temperature remains unchanged. 9 The maximum vapour pressure exerted by a vapour in equilibrium with its liquid at a fixed temperature is called the saturated vapour pressure. Exam Tips Exam Tips Exam Tips Exam Tips The strength of the intermolecular forces determines the physical state of a substance. The vapour molecules of a liquid behave just like a gas would. Evaporation occurs at all temperatures below the boiling point of the liquid. Definition of saturated vapour pressure
149 Chemistry Term 1 STPM CHAPTER 4 Evaporation = Condensation Evaporation Condensation Vapour pressure Time Saturated vapour pressure 10 Vapour pressure, like gas pressure, increases with increasing temperature. At higher temperature, more liquid can undergo evaporation. This not only increases the number of vapour molecules but also the rate and intensity of the collisions. Vapour pressure Temperature 11 The magnitude of the vapour pressure at a constant temperature is a measure of the strength of the forces that holds the liquid particles together. 12 A liquid with weak intermolecular forces will have a higher tendency to evaporate compared to one with stronger intermolecular forces, and will exert a higher vapour pressure. 13 The vapour pressures of water and ethanol are given in the graph shown. Ethanol is more volatile than water. Vapour pressure Temperature Ethanol Water 14 Solid also exerts a vapour pressure when put in an enclosed evacuated container. The particles on the surface of the solid can escape as vapour if they have sufficient energy. This process is called sublimation. 15 Due to the very strong attractive forces in the solid, the rate of sublimation is very low. THe vapour pressure exerted by solids is most of the time negligible. Vapour pressure increases with temperature. The particles have higher kinetic energy. Vapour pressure exerted by a solid is very low by comparison to liquid. Exam Tips Exam Tips Exam Tips Liquids with weaker intermolecular forces are more volatile. Evaporation occurs at the surface of the liquid 2010/P1/Q2
150 Chemistry Term 1 STPM CHAPTER 4 Melting 1 The particles in the solid state are held rigidly in fixed positions by strong intermolecular forces. The particles can only vibrate or rotate about their mean positions. 2 When heat is supplied to the solid, the degree of vibration and rotation increases as the energy of the particles increases. 3 A point is reached where the energy of the particles is high enough to overcome the attractive forces and are ‘released’ from their fixed positions and can now move more freely. The solid has turned into a liquid. This process is called melting. 4 The temperature/time graph is shown. Temperature Time Liquid heats up Solid heats up Melting completes Melting begins 5 Solids with strong intermolecular forces need more energy to overcome the attractive forces than solids with weak intermolecular forces. 6 The melting points of some solids are given in the table below. Solid Ice Iron Sodium chloride Iodine Melting point/°C 0 1535 801 114 Nature of intermolecular forces Hydrogen bond Metallic bond Ionic bond van der Waals force Boiling 1 When a liquid is heated in an open container, the vapour pressure of the liquid increases. 2 When the vapour pressure is equal to the atmospheric pressure, bubbles of the vapour will form in the body of the liquid. 3 The bubbles of the vapour rise to the surface of the liquid, burst open and escape into the atmosphere. The liquid boils. 4 Boiling point of a liquid is the temperature at which the vapour pressure of the liquid is equal to the external pressure. Exam Tips Exam Tips Exam Tips Exam Tips The stronger the intermolecular forces, the higher the melting point. Boiling point is also the temperature at which the liquid is in equilibrium with its vapour at an external pressure of 1 atm. Liquid vapour The melting point is the temperature at which a solid is in equilibrium with its liquid under an external pressure of 1 atm. Solid Liquid
151 Chemistry Term 1 STPM CHAPTER 4 5 For water, this occurs at 100 °C under 1 atm pressure. Pressure/atm 100 Temperature/°C 1 6 At 100 °C, the vapour pressure of water is high enough for bubbles of water vapour to be formed in the body of water. This means that water molecules can escape as vapour from anywhere in the body of the liquid and not just from the surface. 7 Volatile liquids have lower boiling points (measured at 101 kPa) than those that are less volatile. The graph below shows the boiling point of three liquids: ether, ethanol and water. Pressure/kPa 35 100 Temperature/°C 101 78 Ether Ethanol Water 8 From the graph, it can be seen that the boiling point of a liquid also depends on the external pressure. For example, water boils at 100 °C under 1 atm, but boils at 15 °C under 0.02 atm. Pressure/atm 100 Temperature/°C 1 15 0.02 Freezing 1 Freezing is the process where a liquid turns into solid. 2 When the temperature of a liquid is lowered, the kinetic energy of the particles decreases. 3 The particles start to come closer to one another as the intermolecular forces get stronger. Exam Tips Exam Tips Exam Tips Exam Tips Volatility: Ether . Ethanol . Water Intermolecular forces: Ether , Ethanol , Water The bubbles formed at temperature lower than 100 °C are bubbles of dissolved gases in water. The solubility of gases in water decreases with increasing temperature. Boiling point of a liquid depends on the external pressure. Temperature Boiling starts Liquid heats up Vapour heats up Boiling ends Time
152 Chemistry Term 1 STPM CHAPTER 4 4 A point is reached where the intermolecular forces are strong enough to hold the particles together in a fixed orderly arrangement. The liquid freezes. 5 The freezing point is the temperature where a solid is in equilibrium with its liquid at 1 atm pressure. 4.3 Solids 1 In a solid, very strong forces of attraction hold the particles together in a fixed position. 2 Solids can be crystalline or amorphous. 3 In crystalline solids, the particles (atoms, molecules or ions) are arranged in a fixed orderly pattern called the lattice structure or crystalline structure. 4 In amorphous solids, the particles do not have fixed orderly arrangement. Two examples of amorphous solid are rubber and plastic. 5 The fixed orderly arrangement of a crystalline solid is due to the repetition of a basic arrangement extent in three dimension. 6 This basic building block is called the unit cell of the solid concerned. 7 The diagram on the left shows a unit cell for a block of cubes. 8 In this section, we shall look at the structure of four types of solid lattice: ionic, simple molecular, giant molecular and metallic. Ionic Solids 1 An example of ionic solid is sodium chloride, NaCl, where strong ionic bonds hold the Na+ and Cl– ions rigidly in place in the solid lattice. 2 Sodium chloride has a face-centred cubic structure as shown below: Na+ or Cl– Na+ ion Cl– ion The definition of freezing point is the same as that of melting point. Crystalline solid Amorphous solid 2009/P1/Q5 2017/P1/Q19(c) A unit cell
153 Chemistry Term 1 STPM CHAPTER 4 3 A slice through a layer of the lattice shows the following arrangement: Cl– Na+ + – + – – + – + – + + – + – + – – + – + – + 4 In the lattice, each Na+ ion is surrounded by six Cl– ions and each Cl– ion is in turn surrounded by six Na+ ions. 5 The coordination number of Na+ = 6 The coordination number of Cl– = 6 6 Another example of ionic solid is caesium chloride, CsCl, which has a body-centred cubic structure as below: Cs+ Cl– 7 The coordination number of Cs+ = 8 The coordination number of Cl– = 8 8 Due to the strong ionic bonds, ionic solids are hard but brittle, with high melting and boliling points. For example: Compound Melting point/oC Boiling point/oC NaCl 801.0 1413.0 CsCl 645.2 1297.0 Simple Molecular Solids 1 An example of simple molecular solid is iodine. 2 At room conditions, iodine is a dark shinny solid that sublimes when heated. It exists in the form of diatomic molecules with strong covalent bond holding the individual atoms together in the I2 molecule. I I xx •• •• x • xx xx••
154 Chemistry Term 1 STPM CHAPTER 4 3 Solid iodine has a face-centred cubic structure: 4 The forces that hold the individual molecules in the solid lattice are the weak van der Waals forces. As a result, iodine has relatively low melting point (114 °C) and low boiling point (184 °C) compared to ionic solids. Giant Molecular Solids (a.k.a. Giant Covalent Network Structures) 1 Giant molecular solids are covalent compounds but with exceptionally high melting and boiling points. 2 The particles in giant covalent solids are atoms and not simple molecules. 3 Two examples of giant molecular solids are graphite and diamond. 4 Graphite and diamond are two allotropes of carbon. Their structures consist entirely of carbon atoms joined by strong covalent bonds in a giant covalent network. 5 The structures of graphite and diamond are shown below: 6 Due to the presence of strong covalent bonds in the solid structure, graphite and diamond have very high melting and boiling points. Allotrope Melting point/oC Boiling point/oC Graphite 3652 4200 Diamond 3550 4827 7 Another example of giant molecular solid is silicon dioxide, SiO2 . 8 Sand is an impure form of silicon dioxide. 2016/P1/Q16(a)(i)
155 Chemistry Term 1 STPM CHAPTER 4 9 The structure of silicon dioxide is shown below: Silicon Oxygen 10 The strong covalent bonds between the atoms is responsible for the high melting point of silicon dioxide (1600 °C). [NOTE: Melting of giant molecular structure involves the breaking of all the covalent bonds in the solid lattice.] Metallic Solids 1 In metals, the atoms are visualised as spheres of identical size. In the solid stage, the atoms are packed as close to one another as possible. 2 The two most common lattice structures of metals are the facecentred cubic (FCC) and hexagonal close-packed (HCP) structures. 3 An example of the face-centred cubic structure (which is also called the cubic close-packed structure) is copper. Exploded view Compact view Each copper atom is in contact with 12 other copper atoms. It has a coordination number of 12. 4 Due to the strong metallic bonds in the structure, copper is hard, with a high melting point and boiling point and is a good electrical conductor. Number of Particles in a Cubic Unit Cell 1 There are essentially four types of particles in a unit cell, each occupying a different position. They are: (a) Particle at the corner of the cube (b) Particle along the edge of the cube (c) Particle at the centre of a face (d) Particle at the centre of the cube
156 Chemistry Term 1 STPM CHAPTER 4 2 In each of these positions, they contribute only a fraction of their volume to the whole unit cell. 3 A particle at the corner of a cube is shared by eight other cubes. Its contribution to the unit cell is 1 8. 4 A particle along the edge of a cube is shared by four other cubes. Its contribution to the unit cell is 1 4 . 5 A particle in the centre of a face is shared by two other cubes. Its contribution to the unit cell is 1 2 . 6 A particle at the centre of a cube belongs to the cube only. Its contribution to the unit cell is 1. 7 These are summarised in the table below: Type of particles Fraction per unit cell Corner 1 8 Along the edge 1 4 Centre of a face 1 2 Centre of the cube 1
157 Chemistry Term 1 STPM CHAPTER 4 Example 4.12 Determine the number of lattice particles in (a) a body-centred cube unit cell, (b) a face-centred unit cell. Solution (a) (8 corners) 1 8 = 1 (1 centre) 1 = 2 Number of lattice particles = 1 + 1 = 2 (b) (8 corners) 1 8 = 1 (6 faces) 1 2 = 3 Number of lattice particles = 1 + 3 = 4 Example 4.13 M O The diagram shows a unit cell of the oxide of an element M. (a) Calculate the number of atoms of M and O in the unit cell. (b) Suggest the empirical formula for the oxide. Solution (a) For element M: [(8 corners) 1 8 ] + [(1 centre) 1] = 2 For oxygen: (6 faces) 1 2 = 3 (b) The empirical formula is M2 O3 .
158 Chemistry Term 1 STPM CHAPTER 4 Quick Check 4.6 1 Given that the interionic distance between Na+ and Cl– is 2.81 × 10–10 m, and the molar volume of sodium chloride is 26.98 cm3 . Calculate the Avogadro’s Constant, which is the number of Na+Cl– ion pairs in one mole of sodium chloride. 2 The above diagram refers to the unit cell for iodine. Calculate the number of iodine atoms in one unit cell of iodine. 3 The unit cell of a metal sulphide is shown below: Sulphur M Determine the empirical formula of the sulphide. Allotropy 1 Some elements can exist in more than one form due to the different arrangements of the atoms. 2 These elements are said to exhibit allotropy. The different forms of the same substances are called allotropes. 3 One example is carbon. The three major allotropes of carbon are graphite, diamond and fullerene. Graphite 1 In graphite, each carbon atom undergoes sp2 hybridisation and is covalently bonded to three other carbon atoms to form flat sheets of hexagonal rings. The carbon-carbon bond angles in the layers are 120°. 2 The hexagonal rings are arranged in layers with weak van der Waals forces between the layers. Other elements that exhibit allotropy are oxygen, tin and phosphorus.
159 Chemistry Term 1 STPM CHAPTER 4 Strong covalent bonds hold the atoms together in the individual layers. Weak van der Waals forces hold the layers together. Graphite is soft and slippery. High melting point Relatively low density due to the ‘open’ structure Good conductor of electricity High melting point High density 3 The weak van der Waals forces allow the layers to slide over one another. As a result, graphite is soft and slippery and is used as a lubricant. 4 The carbon-carbon bond length in graphite is 0.142 nm, which is intermediate between that of a single bond (0.154 nm) and double bond (0.134 nm). This accounts for the high melting point of graphite (3650 °C). 5 Graphite has a relatively low density (2.25 g cm–3) due to the open structure. The distance between adjacent layers is 0.341 nm which is about 2.2 times the carbon-carbon bond length. 6 As each carbon atom in graphite uses only three of its valence shell electrons to form covalent bond, there is a ‘free electron’ per carbon atom which delocalises over the whole structure. Hence, graphite is a conductor and is frequently used as electrodes in electrolysis. 7 Another use of graphite is in the making of ‘pencil lead’, which is a mixture of graphite and clay. Diamond 1 In diamond, each carbon atom undergoes sp 3 hybridisation and is bonded covalently to four other carbon atoms located at the apex of a tetrahedron to form a giant network of a six-member ring. The carbon-carbon bond angles are 109.5°. 2 The strong covalent bonds (bond length = 0.154 nm) in the network accounts for the high melting point of diamond (3550 °C), and the close-packed structure gives diamond a higher density (3.50 g cm–3) than graphite.
160 Chemistry Term 1 STPM CHAPTER 4 3 The melting point of diamond (3550 °C) is lower than that of graphite (3650 °C) because its carbon-carbon bonds are weaker. 4 All the valence electrons in diamond are involved in the formation of covalent bonds. There are no free electrons available. Thus, diamond is a non-conductor. 5 Diamond is the hardest element on Earth. It is used to make drills for high velocity cutting machines. The high refractive index of diamond also makes it a suitable material for ornaments. Diamond is bright and shiny. 6 Due to its high reflective index, diamond is used extensively as jewellery. 7 Graphite and diamond can co-exist at room conditions. However, graphite is the stable allotrope. (This type of allotropy is called monotropy.) C(diamond) → C(graphite) ∆H = –2.0 kJ mol–1 8 Due to the very high activation energy of the reaction, the rate of conversion of diamond into graphite is negligible, especially at room conditions. In other words, diamond is energetically less stable than graphite, but is kinetically stable. 9 The table below lists the differences between graphite and diamond. Property Graphite Diamond Type of hybridisation sp2 sp3 Carbon-carbon bond length/nm 0.142 0.154 Bond angle 120° 109.5° Melting point/°C 3650 3550 Density/g cm–3 2.25 3.50 Electrical conductivity Conductor Non-conductor Texture Black, shiny and slippery Bright and sparkling Uses Electrode, lubricants, pencil-lead High velocity cutting tools, ornaments Fullerene 1 Fullerene, another allotrope of carbon, was discovered in 1985. It has a molecular formula of C60 and a molar mass of 720 g mol–1. 2 Unlike graphite and diamond, fullerene is spherical. The carbon atoms are joined by covalent bonds as shown below: Non-conductor of electricity The hardest element on Earth Monotropy – All the allotropes can co-exist at room conditions. However only one of them is stable. Energy Diamond Graphite Fullerene resembles a hollow sphere. 2013/P1/Q8 2018/P1/Q8
161 Chemistry Term 1 STPM CHAPTER 4 3 All the carbon atoms in fullerene undergo sp2 hybridisation. 4 Fullerene is used to make superconductors, micro-ball bearings, as catalysts in some chemical reactions and also in drug delivery systems. 5 To date, other fullerene molecules with molecular formulae of C70, C82 and C90 are also known. 4.4 Phase Diagrams 1 A phase is that part of a system which is of uniform composition and texture that is separated from the other parts by a distinct boundary. 2 For example, a closed container of water consists of two phases: The liquid phase (water) and the gaseous phase (water vapour). Water vapour Boundary Water 3 However, an aqueous solution of sodium chloride consists of only one phase, the liquid phase. 4 A mixture of gases is a one phase system. 5 Most matter can exist in more than one physical state (or phase): Solid, liquid and gas/vapour. 6 At 1 atm and 110 °C, water exists in gaseous phase. At 1 atm and 25 °C, water exists in liquid phase. Hence, a substance can exist in different phases, depending on the temperature and pressure. 7 The introversion between the various phases of a substance is summarised in a phase diagram. 8 A phase diagram is a plot of experimental results. Phase Diagram of Water 1 The phase diagram of water (not to scale) is shown below. Temperature/°C Pressure/atm A B C T 200 0.006 0.01 374 Solid Liquid Gas Cylindrical fullerenes known as nanotubes have also been discovered. Info Chem Due to its hollow structure, fullerence has lower density than graphite and diamond. Definition of ‘phase’ Exam Tips Exam Tips Gas is the name given to a substance that exists in the gaseous state at room conditions. Vapour refers to the gaseous state of a substance that exists either in the solid or liquid state at room conditions. E.g. Oxygen gas and water vapour. 2009/P1/Q3 2013/P1/Q20 2014/P1/Q7 2015/P1/Q11 2017/P1/Q8 2016/P1/Q8 2018/P1/Q17 INFO Phase Diagram of Pure Substances
162 Chemistry Term 1 STPM CHAPTER 4 2 The graph is divided by three lines to form three areas. Within each area, only one phase can exist. They are summarised as follows: Area ATB BTC ATC Phase Solid Liquid Gas/Vapour 3 A line forms a boundary between two phases. Line AT BT TC Phase Solid/Gas Solid/Liquid Liquid/Vapour 4 A point on any line shows the conditions where the two phases can exist in equilibrium. 5 The line AT (the solid/vapour line) Temperature/°C Pressure/atm A T 0.006 –5 0.01 25 X Y (a) The line AT shows the variation of the vapour pressure of ice with temperature. (b) It also shows the variation of the sublimation temperature of ice with pressure. (c) For example, a sample of ice (at –5 °C) will sublime when placed at room temperature (25 °C) but under a pressure of less than 0.006 atm. (The line X → Y). (d) Theoretically, the line TA cuts the temperature axis at absolute zero or –273 °C. 6 The TC line (the liquid/vapour line) Temperature/°C Pressure/atm T 1 100 t 1 P1 C (a) The line TC shows the variation of the vapour pressure of water with temperature. Solid B C T A Liquid Liquid Gas Solid Gas Pressure Temperature fiff fiff fiff
163 Chemistry Term 1 STPM CHAPTER 4 (b) It also shows the variation of the boiling point of water with pressure. (c) The boiling point of water at 1 atm is 100 °C. However, the boiling point of water at pressure higher than 1 atm is more than 100 °C. (d) For example, the boiling point of water at pressure P2 is t 2 . 7 The TB line (the solid/liquid line) Temperature/°C Pressure/atm T 1 t 0 2 P2 B (a) The line TB shows the variation of the melting point of ice with pressure. (b) It also shows the variation of the freezing point of water with pressure. (c) The line has a negative slope. This means that the melting point of ice decreases with increasing pressure. (d) For example, at 1 atm, ice melts at 0 °C but at a higher pressure, P2 , it melts at a lower temperature (t 2 ). (e) This is abnormal. For most substances, the melting point increases with increasing pressure because melting is accompanied by an increase in volume. Solid Liquid Volume increases Increasing pressure would shift the equilibrium to the left, thus making the solid more difficult to melt. (f) However, water is abnormal because its melting is accompanied by a decrease in volume. H2 O(s) H2 O(l) Volume decreases Increasing pressure would shift the equilibrium to the right, making ice easier to melt. Liquid benzene Solid benzene Increasing pressure More benzene freezes Info Chem This abnormality is due to the presence of hydrogen bonding. This is the principle of the pressure cooker.
164 Chemistry Term 1 STPM CHAPTER 4 Increasing pressure Water Ice Ice melts 8 The point C Temperature/°C Pressure/atm T 200 374 C P Q M L The line TC ends at point C, which is called the critical point. (a) The critical point refers to the maximum temperature above which a sample of vapour cannot be condensed to liquid by increasing pressure alone. For water, this happens at 374 °C. (b) At temperature above 374 °C, water vapour cannot be condensed to form water by increasing pressure only (Line L → M). (c) In order for water vapour to condense, its temperature must be lowered to less than 374 °C before applying pressure (Line L → P → Q). (d) The pressure required to condense a vapour at its critical temperature is called the critical pressure. 9 The point T The point T, where the three lines meet, is called the triple point. Temperature/°C Pressure/atm T 0.006 0.01 Solid Liquid Gas (a) It is the conditions of temperature and pressure where the three phases of a substance can exist in equilibrium. Critical point also refers to the temperature and pressure, above which the vapour phase and the liquid phase is indistinguishable. That is, no phase boundaries exit. Info Chem It is easier to melt ice at high pressures compared to at low pressures. Exam Tips Exam Tips Substances with critical temperatures lower than room temperature cannot be condensed at room conditions by applying pressure alone.
165 Chemistry Term 1 STPM CHAPTER 4 (b) For water, this happens at 0.006 atm and 0.01 °C. (c) The triple point is not the same as the freezing/melting point. The freezing/melting point refers to the temperature where a solid is in equilibrium with its liquid at 1 atm pressure. Example 4.14 Explain why increasing pressure can sometimes cause a vapour to condense. Solution Increasing pressure pushes the gas particles closer to one another. This causes the intermolecular forces to increase. At sufficiently high pressure, the particles are so close that they ‘stick’ to one another, causing it to condense. Example 4.15 Explain why a sample of vapour at temperature above its critical temperature cannot be condensed no matter how high the pressure is. Solution At temperature higher than the critical temperature, the kinetic energies of the particles are so high that no matter how close they approach one another, the intermolecular forces are not strong enough to bind them together to form liquid. Condensation by pressure Why vapour cannot be condensed at temperature . critical temperature? Quick Check 4.7 1 (a) What do you understand by sublimation? (b) Explain how you would cause ice at –10 °C to sublime at room temperature. Illustrate your answer with a suitable diagram. 2 The phase diagram of water is shown. (a) What is the physical state of water at point A? (b) Sketch a temperature/time graph to illustrate what happens when a sample of water at point A is cooled gradually to point B. (c) Suggest how you would change the physical state of water at point A at constant temperature. State what phase changes occur. Temperature Pressure B A
166 Chemistry Term 1 STPM CHAPTER 4 Phase Diagram of Carbon Dioxide 1 The phase diagram of carbon dioxide (not to scale) is shown below: Temperature/°C Pressure/atm B A C T Solid Liquid Gas –57 31 5 73 2 The phase diagram of carbon dioxide has two important features that are different from the phase diagram of water. 3 The triple point pressure of carbon dioxide is higher than 1 atm. (a) This means that at pressures less than 5 atm, liquid carbon dioxide cannot exist. (b) If solid carbon dioxide is warmed gently at 1 atm, it will sublime (Line P → Q). Temperature/°C Pressure/atm B A C T Solid Liquid Gas –57 31 5 73 1 P Q (c) This property of solid carbon dioxide (also called dry ice) is made used of to cool food stuffs such as ice cream. Dry ice will not melt and thus it will not contaminate or dissolve the ice cream. (d) Dry ice is also used in cloud-seeding. When it is sprayed into the clouds, it absorbs heat in the cloud and undergoes sublimation. This lowers the temperature of the cloud and causes the water vapour to condense and fall as rain. (e) Dry ice is also used in certain fuctions or concerts to form ‘fog’ or mist. When dry ice sublimes, it absorbs heat from the surrounding. The temperature of the surrounding air drops, causing water vapour to condense forming the ‘fog’ or the mist. 4 The solid-liquid line has a positive slope. (a) The melting point of carbon dioxide increases with increasing temperature. Info Chem In a fire extinguisher, partial liquefaction of CO2 occurs. CO2 (g) CO2 (l) When pressure is released, the liquid CO2 absorbs heat from the surroundings and vaporised causing the water vapour in the air to condense forming a mist. Info Chem Any substance at temperature and pressure above its critical point is known as a supercritical fluid. P s l g T Supercritical fluid region It has the properties of a gas and liquid. It can effuse like a gas, and dissolves substances like a liquid. 2016/P1/Q16(a)(ii)
167 Chemistry Term 1 STPM CHAPTER 4 (b) This is because melting of carbon dioxide is accompanied by an increase in volume. CO2 (s) CO2 (l) Volume increases Thus, according to Le Chatelier’s Principle, increasing pressure will shift the equilibrium to the left. This makes melting more difficult at high pressures. 5 Note that the critical temperature of carbon dioxide is 31 °C. A sample of carbon dioxide gas at temperature above 31 °C cannot be condensed by increasing pressure alone. Quick Check 4.8 1 Explain why solid carbon dioxide is denser than liquid carbon dioxide. 2 Solid iodine sublimes when it is heated under room conditions. Sketch the phase diagram of iodine. Mark the pressure at 1 atm clearly on your sketch. 3 The density of solid iron is less than that of molten iron. (a) Sketch the phase diagram of iron. (b) Explain the slope of the solid/liquid line. 4 The phase diagram of a substance, W, is shown below. Temperature/°C Pressure/atm 1 X (a) Label the three areas in the phase diagram. (b) The critical temperature of W is at 132 °C. Mark this point on the diagram clearly. (c) Can a vapour sample of W be condensed at room temperature by increasing the pressure alone? Explain your answer. (d) Mark on the graph the normal melting point and boiling point of W. (e) State what happens when a sample of W at point X is subjected to the following changes: (i) The pressure is increased at constant temperature. (ii) The temperature is increased at constant pressure. (f) Comment on the density of solid W and that of liquid W. 5 The critical temperatures of some substances are given in the table below: Substance H2 CO2 NH3 C2 H5 OH H2 O Critical temperature/°C –268 31 131 243 374 Which of the above substances can be condensed at room temperature by increasing pressure alone?
168 Chemistry Term 1 STPM CHAPTER 4 Exam Tips Effect of Non-volatile Solutes on the Vapour Pressure of a Solvent The vapour pressure of pure water at 298 K is 3.17 kPa. What would be the vapour pressure of a solution of glucose in water at the same temperature? Lowering of Vapour Pressure 1 The vapour pressure of a pure liquid (e.g. water) is independent of the amount of liquid present (as long as some liquid is present to establish equilibrium with the vapour) and is only affected by temperature. 2 However, when a non-volatile solute (e.g. glucose) is added to water, the vapour pressure exerted by the solution is now less than the vapour pressure exerted by the pure water alone at the same temperature. 3 For example, the vapour pressure of a glucose solution containing 43.2 g of glucose in 100 g of water is 3.04 kPa instead of 3.17 kPa. There is a lowering of the vapour pressure of water by glucose. Water vapour Water molecules • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •• • • • • • • • •• • • • • • • • 4 In pure water, water has the whole surface for evaporation to occur. 5 In a glucose solution, part of the surface is occupied by glucose molecules which are non-volatile. Hence, less water molecule can undergo evaporation. However, the water vapour molecules still have the whole surface of the solution to return to liquid. 6 At equilibrium, there would be less water vapour molecules present. This causes a decrease in the vapour pressure. 7 The more concentrated the solution, the lower is the vapour pressure. This is illustrated in the phase diagram for water: Temperature Pressure Dilute solution Concentrated solution Pure H2 O Exam Tips Vapour pressure of a pure liquid depends only on temperature. Info Chem A non-volatile solute is one with negligible vapour pressure. Water vapour Water molecules • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •• • • • • • • • •• • • • • • • • Glucose molecules Exam Tips Exam Tips It is the water that undergoes evaporation and not the glucose.
169 Chemistry Term 1 STPM CHAPTER 4 8 The vapour pressure of a solution containing a non-volatile solute is given by: P = XAPA O where, P = vapour pressure of the solution XA = mole fraction of the solvent in the solution PA O = vapour pressure of pure solvent at the same temperature 9 The lowering of vapour pressure is a colligative property. The magnitude of the lowering of the vapour pressure depends only on the amount of solute particles (molecules or ions) present and not the nature of the solute. Example 4.16 The vapour pressure of water at 298 K is 3.17 kPa. (a) Calculate the vapour pressure of a solution containing 50 g of glucose (C6 H12O6 ) in 100 g of water at 298 K. (b) Calculate the vapour pressure of a solution containing 95 g of sucrose (C12H22O11) in 100 g of water at 298 K. Solution (a) No. of moles of glucose = 50 180 = 0.278 mol No. of moles of water = 100 18 = 5.56 mol Mole fraction of water = 5.56 (5.56 + 0.278) = 0.95 ∴ Vapour pressure = 0.95 3.17 kPa = 3.01 kPa (b) No. of moles of sucrose = 95 342 = 0.278 mol Since the number of moles of solute used is the same in both cases (adding to the same amount of solvent), the vapour pressure = 3.01 kPa. Colligative property Mole fraction = No. of moles of solvent Total no. of moles Quick Check 4.9 1 The vapour pressure of benzene is 35.3 kPa at 55 °C. Calculate the mass of naphthalene (C10H8 ) needed to add to one mole of benzene to produce a solution with a vapour pressure of 22.0 kPa at 55 °C. 2 The vapour pressure of a liquid X is 34.5 kPa at 298 K. What is the mole fraction of X in a solution with vapour pressure of 28.9 kPa?
170 Chemistry Term 1 STPM CHAPTER 4 3 Calculate the vapour pressure of an aqueous solution containing 38.0 g of sodium chloride in 150 g of water at 300 K. [Vapour pressure of water at 300 K is 3.18 kPa.] 4 When 0.10 mol of sodium chloride is added to a fixed amount of water at 300 K, the vapour pressure of water is decreased by x kPa. Determine, in terms of x, the lowering of the vapour pressure of water when the following substances are added separately to the same amount of water at 300 K: (a) 0.10 mol of glucose (b) 0.10 mol of aluminium chloride (c) 0.30 mol of sucrose (d) 0.15 mol of chromium(III) sulphate Elevation of Boiling Point 1 A liquid boils when its vapour pressure is equal to the external pressure (usually 1 atm). 2 For water, this happens at 100 °C. 3 If some glucose is dissolved in a certain mass of water, the boiling point of the solution will be higher. 4 Consider the phase diagram of water shown: Temperature/°C Pressure/atm Glucose solution Pure H2 O Δt 100 t 1 5 Due to the lowering of the vapour pressure of water by glucose, it requires a higher temperature for the vapour pressure of the solution to be equal to 1 atm. 6 The water in the glucose solution will boil at t °C which is higher than 100 °C. ∆t is known as the elevation of boiling point. 7 The more concentrated the solution, the higher is the elevation of the boiling point. The Ebullioscopic Constant 1 The ebullioscopic constant is the elevation of boiling point of a solvent when 1 mole of a non-volatile solute is dissolved in 1 kg of the solvent. 2 Elevation of boiling point, like the lowering of vapour pressure, is a colligative property. Its magnitude depends only on the amount of dissolved particles. It does not depend on the nature of the solute. Exam Tips Exam Tips It is the water in the glucose solution that boils and not the glucose. Definition of ebullioscopic constant The solute must not dissociate or associate in the solvent.
171 Chemistry Term 1 STPM CHAPTER 4 Dissociation of sodium chloride 3 The unit for ebullioscopic constant is °C mol–1 kg or K mol–1 kg. 4 Sometimes the unit of °C mol–1 100 g is also used. This is the elevation when 1 mol of a non-volatile solute is dissolved in 100 g of the solvent. 5 The ebullioscopic constant for water is 0.52 °C mol–1 kg. This means that the boiling point of water will be elevated by 0.52 °C when 1 mol of any non-volatile solute (which does not dissociate or associate) is dissolved in 1 kg of water. 6 However, solutes that undergo dissociation in water will have a higher than expected elevation in the boiling point of water. 7 For example, when 1 mol of sodium chloride is dissolved in 1 kg of water, the boiling point of water is elevated by (2 0.52 °C). This is because of the dissociation of sodium chloride into ions: NaCl(s) + aq → Na+(aq) + Cl– (aq) 8 The table below lists the ebullioscopic constant of some common liquids: Liquid Ebullioscopic constant/°C mol–1 kg Water 0.52 Ethanoic acid 3.07 Benzene 2.53 Propanone 1.71 Carbon tetrachloride 4.95 Ether 1.82 Cyclohexane 2.79 Trichloromethane 3.90 Example 4.17 Calculate the boiling point of an aqueous solution containing 72.0 g of glucose in 750 g of water. Solution No. of moles of glucose = 72 180 = 0.40 mol Concentration of solution = 0.40 1000 750 = 0.533 mol kg–1 Elevation of boiling point = 0.533 0.52 = 0.28 °C Boiling point of the solution = 100 + 0.28 = 100.28 °C
172 Chemistry Term 1 STPM CHAPTER 4 Presence of impurities lowers the freezing point of a pure liquid. The solute must not dissociate or associate in the solvent. Depression of Freezing Point 1 Consider the phase diagram of water below: Temperature/°C Pressure/atm Solution Pure H2 O Δt t 0 1 2 At 1 atm, the freezing point of water is 0 °C. 3 However, the freezing point of water in a glucose solution is lower than 0 °C. ∆t is called the depression of freezing point. 4 On cooling, the kinetic energy of the water molecules decreases. This causes the intermolecular attraction to get stronger and the water molecules are pulled closer to one another. 5 At 0 °C, the attractive forces are strong enough to hold the molecules together in a fixed orderly arrangement. 6 However, in the presence of foreign particles such as glucose, the fixed orderly arrangement is more difficult to achieve even as the temperature drops to 0 °C. The temperature has to be further lowered before freezing occurs. 7 Depression of freezing point of a solvent is also a colligative property. 8 The more concentrated the solution, the larger is the depression of freezing point. The Cryoscopic Constant 1 The cryoscopic constant is the depression of freezing point of a solvent when 1 mol of a non-volatile solute is dissolved in 1 kg of the solvent. 2 The cryoscopic constant of some common liquids is listed below: Liquid Cryoscopic constant/°C mol–1 kg Water 1.86 Ethanoic acid 3.90 Benzene 5.12 Carbon tetrachloride 30.0 Ether 1.79 Naphthalene 6.94
173 Chemistry Term 1 STPM CHAPTER 4 Example 4.18 Calculate the freezing point of a solution containing 12.6 g of glucose dissolved in 150 g of water. Solution No. of moles of glucose = 12.6 180 = 0.07 mol Concentration of solution = 0.07 1000 150 = 0.467 mol kg–1 Depression of freezing point = 0.467 1.86 = 0.87 °C Freezing point of solution = – 0.87 °C Quick Check 4.10 1 Calculate the boiling point of the following solution: 34.5 g of aluminium sulphate in 1.5 kg of water 2 Calculate the freezing point of the following solution: 5.60 g of 1,2-ethanediol in 2.0 kg of water 3 Which will have a lower freezing temperature: a solution containing 120 g of sodium chloride in 1 kg of water or 120 g of calcium chloride in 120 g of water? 4 Explain the following: (a) 0.25 mol of ethanoic acid depresses the freezing point of 1 kg of benzene half as much as 0.25 mol of naphthalene. (b) 0.30 mol of aluminium chloride, AlCl3 , depresses the freezing point of 1 kg of cyclohexane half as much as 0.30 mol of naphthalene. 5 When 0.136 g of ethanoic acid is dissolved in 100 g of water, the freezing point of water is depressed by 0.0421 °C. However, when 0.136 g of ethanoic acid is dissolved in 100 g of benzene, the freezing point of benzene is lowered by 0.0211 °C. (a) Calculate the molar mass of ethanoic acid in water. (b) Calculate the molar mass of ethanoic acid in benzene. (c) Comment on your results. [Cryoscopic constant of: water = 1.86; benzene = 5.12 K mol–1 kg]
174 Chemistry Term 1 STPM CHAPTER 4 SUMMARY SUMMARY 1 The two important assumptions of the kinetic theory of gases are: (a) There are no intermolecular forces between the gaseous particles. (b) The volume of the gas particles are negligible compared to the volume occupied by the gas. 2 The pressure exerted by a gas is a consequent of the collisions between the gas particles and the walls of the container. 3 The average kinetic energy of the gas particles is directly proportional to the absolute temperature of the gas. 4 Boyle’s law states that the volume occupied by a fixed mass of gas, at constant temperature, is inversely proportional to its pressure. 5 Charles’ law states the volume occupied by a fixed mass of gas, at constant pressure, is directly proportional to its absolute temperature. 6 Avogadro’s law states that equal volume of all gases, under the same temperature and pressure, contains the same number of moles of gas. 7 Dalton’s law of partial pressure states that the total pressure of a mixture of gases, which do not react with one another, is the sum of the partial pressures of the constituent gases. 8 The ideal gas equation: PV = nRT 9 Real gases show deviation from ideal behaviours because (a) there are intermolecular forces between the gas particles, (b) the gas particles have finite volume. 10 Real gases show deviation from ideal behaviour under very high pressures and very low temperatures. 11 Real gases show ideal behaviours under very low pressures and very high temperatures. 12 The vapour pressure is the pressure exerted by a vapour in equilibrium with its own liquid at a fix temperature. 13 The melting point is the temperature at which a solid is in equilibrium with its liquid at 1 atm. 14 The boiling point is the temperature at which a liquid is in equilibrium with its vapour at 1 atm. 15 The four main classes of solids are ionic solids, simple molecular solids, giant molecular solids and metallic solids. 16 The fractional contribution of a lattice particle to a unit cell is: Type of lattice point Corner Face Edge Centre Fractional contribution 1 8 1 2 1 4 1 17 Allotropes are different forms of the same element. 18 The triple point is the temperature and pressure where the three phases of a substance can exist in equilibrium. 19 The critical point is the maximum temperature above which a sample of gas cannot be condensed by increasing pressure alone. 20 The melting point of ice decreases with pressure because the melting process is accompanied by a decrease in volume. 21 The melting point of carbon dioxide increases with pressure because the melting process is accompanied by an increase in volume. 22 Colligative properties include: (a) Lowering of vapour pressure (b) Elevation of boiling point (c) Depression of melting point
175 Chemistry Term 1 STPM CHAPTER 4 1 Which of the following gas is likely to deviate the most from ideal behaviour? A HCl C N2 B HF D CH4 2 Which of the following is not true about fullerenes? A Exists in the form of hollow spheres, ellipsoids and tubes B Generally, fullerenes are more reactive than graphite C Fullerenes are generally better conductors than graphite D All the carbon atoms in the molecule undergo sp2 hybridisation 3 A certain amount of gas occupies 23.5 dm3 at 200 kPa and 50°C. At what temperature would it occupy 47.0 dm3 at 200 kPa? A 100°C C 373°C B 200°C D 746 °C 4 Which of the following would have the highest average kinetic energy at 300 K? A H2 C NH3 B He D All the same 5 A vessel contains nitrogen at a pressure of 120 kPa. A sample of oxygen gas at a pressure of 50 kPa is forced into the vessel at constant volume. What is the partial pressure of nitrogen in the mixture, assuming that the temperature remains unchanged? A 70 kPa C 170 kPa B 120 kPa D 7200 kPa 6 Under what conditions of temperature and pressure will a gas deviate most from ideal behaviour? Temperature Pressure A Low Low B High High C Room Standard D Low High Objective Questions 7 A gas is considered as ideal if A 1 mol of it occupies 22.4 dm3 at s.t.p. B it is compressible. C it does not occupy any volume. D it obeys the ideal gas equation. 8 According to the kinetic theory of matter, as the temperature of a substance increases, A its melting point decreases. B its kinetic energy increases. C its boiling point increases. D none of the above 9 The phase diagram of substance P is shown below. Temperature (°C) 0 20 40 60 80 100 120 140 160 180 0 0.2 0.4 0.6 0.8 1.0 Pressure (Atmosphere) What of the statements is not true about P? A The melting point of P varies slightly with increasing pressure B Solid P sublimes when heated at room conditions C P is a solid under standard temperature and pressure D P could be carbon dioxide 10 The volume of a gas at 350 K is V cm3 . What is its volume at 700 K at constant pressure? A V cm3 B 2V cm3 C V 2 cm3 D V ((700 + 273) (350 + 273)) cm3 STPM PRACTICE 4
176 Chemistry Term 1 STPM CHAPTER 4 11 The reaction occurring in the airbags for cars is: 2NaN3 (s) → 2Na(s) + 3N2 (g) What is the mass of sodium azide (in g) that is required to produce 19.0 dm3 of nitrogen at 293 K and 103 kPa? A 17.5 B 34.9 C 52.0 D 87.2 12 Which of the following solutions has the lowest freezing point? A 0.25 mol dm–3 sodium chloride B 0.25 mol dm–3 glucose C 0.25 mol dm–3 ethanol D 0.25 mol dm–3 lead(II) nitrate 13 According to the kinetic theory of matter, different gases at the same temperature have the same A pressure B volume C molecular speed D average kinetic energy 14 Which of the following best explains why solid carbon dioxide sublimes when left at room conditions? A The van der Waals forces between the molecules are weak. B The critical point temperature of carbon dioxide is below room temperature. C The triple point pressure of carbon dioxide is around 5 atm. D The triple point temperature of carbon dioxide is below room temperature. 15 Which of the following is not correct regarding the critical point of a pure substance? A It is the maximum temperature above which a gas cannot be condensed by increasing pressure alone. B It is the maximum temperature above which the liquid phase and the vapour phase are indistinguishable. C It is the maximum temperature above which no sublimation can take place. D It is the maximum temperature above which the boundary between liquid and vapour disappear. 16 Which of the following solids has a giant molecular structure? A Ice B Sodium chloride C Silicon dioxide D Vanadium 17 The direct conversion of a solid to gas is called A evaporation B sublimation C boiling D vaporisation 18 Which of the following statements about the structure of graphite is not correct? A Each carbon is bonded to three other carbon atoms. B Each carbon atom undergoes sp2 hybridisation. C The structure is made up of rings of six carbon atoms. D Each carbon-carbon bond angle is 109.5°. 19 The phase diagram of water is as follows: Temperature Pressure/atm Y What is the name given to point Y in the diagram? A Melting point B Critical point C Triple point D Transition temperature 20 When iodine is heated under room conditions, it sublimes. It can be deduced that A the triple point pressure is higher than 101 kPa B the critical point pressure is higher than 101 kPa C the covalent bond in iodine is weak D the solid/ liquid line has a negative slope
177 Chemistry Term 1 STPM CHAPTER 4 B Positive deviation is due to the presence of intermolecular repulsion. C All gases behave ideally as the pressure on the gas approaches zero. D The repulsion between CO2 molecules is weaker than that between N2 molecules at pressure greater than x. 25 Which of the following statements about fullerene, C60, is not true? A The carbon-carbon bond angle is 120°. B It has a higher density than diamond. C It is spherical. D The carbon atoms undergo sp2 hybridisation. 26 The phase diagram for substance W is given below: A B C T Pressure/atm Temperature/°C 200 0.006 0.01 374 Which statement about the above phase diagram is not true? A Substance W exists as a supercritical fluid at 380 K and 250 atm. B The melting point of W is 0.01 o C. C At pressure below 0.006 atm, substance W will undergo sublimation. D It is easier to melt W at high pressure than at low pressure. 27 Which graph does not represent Boyle’s law for a fixed mass of an ideal gas? A pV p C p 1 V B p V D V 1 p 21 The phase diagram of a substance X is shown below: Pressure / atm 1.0 Temperature / °C What can be deduced from the phase diagram? A Solid X is more dense than liquid X. B X could be water. C X will sublime at room conditions. D Liquid X cannot exist at pressures less than 1 atm. 22 When a flask is filled with gas Y at 200 °C, the pressure produced is 2.50 atm. When the temperature is increased to 400 °C, the pressure increases to 3.15 atm. Which statement(s) is/are correct of the observations? I The gas shows negative deviation. II The number of gas molecules increases by two times at 400 °C. III There are no intermolecular forces between the gas molecules. A I only C I and II B II only D II and III 23 0.30 g of a gas Y occupies 81 cm3 at 308 kPa and 27 °C. What is the relative molecular mass of Y? A 15.0 C 37.6 B 30.0 D 60.8 24 The plot of PV nRT for three gases at 298 K is shown below: 1 P Ideal gas H2 CO2 N2 x PV nRT Which of the following statements is not true? A The intermolecular attraction between H2 molecules is negligible.
178 Chemistry Term 1 STPM CHAPTER 4 Structured and Essay Questions 1 (a) Define an ideal gas. (b) Use the kinetic theory of gases to explain the following observations: (i) The pressure of a fixed mass of gas increases with temperature at constant volume. (ii) The volume of a fixed mass of gas decreases with increasing pressure at constant temperature. (iii) The volume of a fixed mass of gas increases with increasing temperature at constant pressure. 2 (a) State the ideal gas equation. (b) Give two reasons why real gases deviate from ideal behaviour. (c) Name a gas that behaves most ideally at room conditions. (d) On the grid given, sketch the graph of PV nRT against P for 1 mol of (a) an ideal gas and (b) ammonia. Explain the negative deviation of the ammonia gas. P PV nRT (e) Under what conditions of temperature and pressure would a real gas show ideal behaviour? 3 How does the kinetic theory of gases account for the fact that gases are mutually miscible, exert a pressure and are compressible? 4 (a) State the two important assumptions of the kinetic theory of gases. (b) (i) Given the atomic radius of argon atom is 0.192 nm, calculate the volume of one mole of argon atoms. (ii) Calculate the volume occupied by 1 mol of argon at 298 K and 101 kPa. (iii) What percentage of this volume is occupied by the atoms themselves? (iv) Explain how your answer to (b)(iii) justifies one of the assumptions stated in (a). 5 Explain the following phenomena: (a) Increasing pressure can sometimes cause a gas to liquefy. (b) The volume occupied by one mole of ammonia at s.t.p. is less than 22.4 dm3 . (c) Gases quickly take up the shape of their container and always fill it. (d) The noble gases become less ideal in their behaviour as one descends the group in the Periodic Table from helium to xenon. (e) A real gas is easier to compress at moderate pressure than at high pressure. 6 (a) Derive the ideal gas equation from known gas laws. (b) The volume of 32.5 g of a gas Y was measured at various pressures but at a constant temperature of 298 K. Pressure/(× 105 Pa) Volume/(× 10–3 m3 ) 4.0 5.80 8.0 2.86 15.0 1.49 20.0 1.10 Plot a suitable graph and use the ideal gas equation to calculate the relative molecular mass of Y.
179 Chemistry Term 1 STPM CHAPTER 4 7 (a) Starting with the ideal gas equation, derive an expression relating the density and pressure of an ideal gas. (b) The variation of the density of a gas Y with pressure at a constant temperature of 28.5 °C is given in the table below. Pressure/kPa Density/g dm–3 33.7 0.960 67.6 1.937 100.0 2.928 (i) Explain the variation of the density of Y with pressure. (ii) Plot a graph of density pressure against pressure, and use the graph to determine the relative molecular mass of Y. 8 (a) State Dalton’s Law of partial pressure. (b) 2.60 dm3 of argon at a pressure of 2.0 105 Pa, 12.5 dm3 of ethane at a pressure of 1.2 105 Pa and 0.80 dm3 of carbon dioxide at 2.8 104 kPa were introduced into a 4.0 dm3 vessel. The temperature remains constant throughout the process. (i) Calculate the partial pressure of each gas in the mixture. (ii) What is the total pressure of the mixture? (iii) If the carbon dioxide is removed from the mixture at constant temperature, calculate the partial pressure of the argon and ethane in the container and the total pressure of the mixture. 9 Use the kinetic theory of matter to explain the differences between solid, liquid and gas. 10 (a) Explain what is meant by the term vapour pressure of water. (b) Sketch a graph of how the vapour pressure of water changes with temperature in the range of 25 °C to 90 °C. Explain the shape of your curve. (c) Explain how the vapour pressure of water is affected by the addition of an non-volatile substance. 11 (a) What is meant by the terms melting point and boiling point of a substance? (b) Sketch the phase diagram of water, and label its melting point and boiling point. (c) In very cold weather, the rivers often freeze but the sea into which the rivers flow stays unfrozen. Suggest two reasons why it is so. You may illustrate your answer with reference to the phase diagram in (b). 12 (a) What is meant by the triple point? (b) The triple point of carbon dioxide is –217 K and 500 kPa. Solid carbon dioxide is denser than liquid carbon dioxide. Use the above information to construct the phase diagram of carbon dioxide. Label the areas and mark the triple point. (c) Explain how the freezing point of carbon dioxide is affected by the addition of another substance. (d) Explain why solid carbon dioxide sublimes when left standing at room conditions. 13 The phase diagram of a substance Q is shown below. Temperature Pressure/kPa 101 a b c d
180 Chemistry Term 1 STPM CHAPTER 4 (a) What do the following points represent? (i) b (ii) c (b) Sketch the temperature/time curve you would obtain if a sample of Q at point a is heated at constant pressure to point d. Indicate on your curve the temperature corresponding to points a to d. (c) Suggest a possible identify of Q. Explain your answer. 14 The phase diagram of water (not to scale) is as follows: Temperature Pressure a b c d (a) Label the three areas in the graph. (b) What is the significance of point b? (c) Why is point b not the freezing point of water? (d) Point d is the critical point of water. What do you understand by the term in italics? 15 The variation of the volume (V) occupied by an ideal gas with temperature (T), at constant pressure p, is shown in the graph below: X 0 Temperature / °C Volume (a) Name a gas that shows the most ideal behaviour. Explain your answer. (b) State the temperature marked X. (c) On the graph above, sketch and label a graph of the variation in volume with temperature at a higher pressure p’ while other conditions remain the same. (d) At pressure p and 0 °C, a gas Z shows positive deviation from ideal behaviours. Mark the expected volume of gas Z in the above graph. 16 Some informations about carbon dioxide is given in the table below: Triple point Critical point 5 atm and –57 °C 73 atm and 31 °C (a) Sketch and label the phase diagram for carbon dioxide. (b) Explain how carbon dioxide can be liquefied at room conditions. (c) A fire extinguisher contains carbon dioxide under high pressure at room temperature. Explain what would be observe when the extinguisher is used to put out a small fire.
CHAPTER REACTION KINETICS 5 Experimental Determination of Order of Reaction • Initial rate method • Graphical method Factors Affecting Rates of Reactions • Concentration • Pressure • Temperature • Catalyst Collision Theory Rate of Reaction Reaction Equation and Order of Reaction Rate Constant and Temperature Reaction Rate and Stoichiometry Reaction Kinetics Rate Equation and Reaction Mechanism • Alkaline hydrolysis – 2-methyl-2-iodopropane – 1-iodobutane Students should be able to: Rate of reaction • define rate of reaction, rate equation, order of reaction, rate constant, half-life of a first-order reaction, rate determining step, activation energy and catalyst; • explain qualitatively, in terms of collision theory, the effects of concentration and temperature on the rate of a reaction. Rate law • calculate the rate constant from initial rates; • predict an initial rate from rate equations and experimental data; • use titrimetric method to study the rate of a given reaction. The effect of temperature on reaction kinetics • explain the relationship between the rate constants with the activation energy and temperature using Arrhenius equation k = Ae –Ea RT ; • use the Boltzmann distribution curve to explain the distribution of molecular energy. The role of catalysts in reactions • explain the effect of catalysts on the rate of a reaction; • explain how a reaction, in the presence of a catalyst, follows an alternative path with a lower activation energy; • explain the role of atmospheric oxides of nitrogen as catalysts in the oxidation of atmospheric sulphur dioxide; • explain the role of vanadium(V) oxide as a catalyst in the Contact process; • describe enzymes as biological catalysts. Order of reactions and rate constants • deduce the order of a reaction (zero-, first- and second-) and the rate constant by the initial rates method and graphical methods; • verify that a suggested reaction mechanism is consistent with the observed kinetics; • use the half-life (t 1 — 2 ) of a first-order reaction in calculations. Learning earning Outcomes Concept Map
182 Chemistry Term 1 STPM CHAPTER 5 5.1 Rate of Reaction 1 The rate of reaction is defined as the rate of change of the concentration of a substance (be it the reactant or product) with time. The unit is mol dm–3 s–1. Rate = [Initial concentration] – [Final concentration] Time taken = ∆[Concentration] ∆t 2 Below is the concentration/time graph for the general reaction. A → B c t1 t2 d b a x y Concentration of A/mol dm–3 0 Time/s 3 The rate of reaction at a particular moment can be obtained from the tangent to the concentration/time graph at that particular point of time. 4 The rates of reaction at time zero, at time t1 and at time t 2 are given by Rate at time zero = y x mol dm–3 s–1 Rate at time t 1 = a b mol dm–3 s–1 Rate at time t 2 = c d mol dm–3 s–1 The rate at time ‘zero’ is called the initial rate. 5 It can be seen that the rate of reaction at time t 2 is less than that at time t 1 . This is because the concentration of the reactant decreases with time. 6 Generally, the rate of reaction is proportional to the concentration of the reactants, except in cases where the order of reaction (Refer to Section 5.5) with respect to that reactant is zero. Definition of ‘rate of reaction’ These are called instanteneous rates. Initial rate = maximum rate Rate of reaction decreases with time as reactants are used up. [Rate]/mol dm–3 s–1 Time/s 2016/P1/Q9 2018/P1/Q10
183 Chemistry Term 1 STPM CHAPTER 5 Example 5.1 Consider the following concentration/time graph for a particular experiment. 0.2 20 0 40 60 80 100 120 140 160 0.4 0.6 0.8 1.0 [A]/mol dm–3 t/s (a) Determine the rate of reaction at (i) time = 0 (ii) time = 60 s (iii) time = 100 s (b) Calculate the average rate for the first 50 s. Solution (a) (i) Initial rate = (1.0 – 0.0) 40 = 0.025 mol dm–3 s–1 (ii) Rate at time 60 s = (0.4 – 0.16) (88 – 30) = 0.24 58 = 4.14 10–3 mol dm–3 s–1 (iii) Rate at time 100 s= (0.24 – 0.10) (136 – 60) = 1.84 10–3 mol dm–3 s–1 (b) Average rate for the first 50 s = 1 – 0.33 50 = 1.34 10–2 mol dm–3 s–1
184 Chemistry Term 1 STPM CHAPTER 5 Quick Check 5.1 1 The relative rate of reaction X to reaction Y is 0.67. If reaction X takes 1 minute 25 seconds to complete, calculate the time taken for reaction Y to complete. 2 Ethyl ethanoate undergoes hydrolysis according to the equation: CH3 COOC2 H5 + H2 O → CH3 COOH + C2 H5 OH The concentration of the ester decreases from an initial concentration of 0.40 mol dm–3 to 0.15 mol dm–3 at a time interval of 20 minutes. (a) Calculate the average rate of reaction for the first 20 minutes. (b) Calculate the amount of ethanoic acid formed at the end of 20 minutes. 3 The change in the concentration of reactant A with time for the following reaction: A → B is given in the table. Plot a graph of [A] against time. Determine from the graph (a) the initial rate, (b) the rate of reaction at 40th minute, (c) the rate of reaction at 80th minute. Time/min [A]/mol dm–3 0 1.00 10 0.81 20 0.65 30 0.53 40 0.44 50 0.36 60 0.30 70 0.26 80 0.22 Reaction Rate and Stoichiometry 1 In a chemical reaction, the concentration of the reactants decreases with time while the concentration of the products increases with time. 2 The rate of reaction can be expressed in terms of the rate of disappearance of the reactants, –∆[Reactant] ∆t or the rate of appearance of the products, ∆[Product] ∆t . [The negative sign in front of the rate of disappearance of reactants is to make the rate positive because the concentration of the reactants decreases with time]. 3 Take a simple reaction: A → B The rate of disappearance of A = –∆[A] ∆t The rate of appearance of B = ∆[B] ∆t And, –∆[A] ∆t = ∆[B] ∆t This means, the rate of disappearance of A is the same as the rate of appearance of B. Differential form of the rate equation
185 Chemistry Term 1 STPM CHAPTER 5 4 For the following reaction: 2X → Y The rate of reaction can be expressed either as Rate of disappearance of reactant = – ∆[X] ∆t Rate of appearance of product = ∆[Y] ∆t But the rate of disappearance of X = 2 (rate of appearance of Y) That is, – ∆[X] ∆t = 2 ∆[Y] ∆t Or ∆[Y] ∆t = – 1 2 ∆[X] ∆t That is, rate = – 1 2 ∆[X] ∆t = ∆[Y] ∆t 5 For a general reaction: aA + bB → cC + dD The rate of disappearance of reactants = – 1 a ∆[A] ∆t = – 1 b ∆[B] ∆t The rate of appearance of products = 1 c ∆[C] ∆t = 1 d ∆[D] ∆t Hence, – 1 a ∆[A] ∆t = – 1 b ∆[B] ∆t = 1 c ∆[C] ∆t = 1 d ∆[D] ∆t 6 Note that a, b, c and d are the stoichiometric coefficients of the substances involved in the balanced equation. Example 5.2 Dinitrogen pentoxide decomposes according to the equation: 2N2 O5 (g) → 4NO2 (g) + O2 (g) At a particular instance, the rate of disappearance of N2 O5 is 1.65 10–3 mol dm–3 s–1. What is the rate of formation of NO2 and O2 at that instance? Solution Rate = – 1 2 ∆[N2 O5 ] ∆t = 1 4 ∆[NO2 ] ∆t = ∆[O2 ] ∆t Hence, rate of formation of NO2 , ∆[NO2 ] ∆t = 4 1 2 ∆[N2 O5 ] ∆t = 2 (1.65 10–3) = 3.30 10–3 mol dm–3 s–1 The rate of formation of O2 , ∆[O2 ] ∆t = 1 4 ∆[NO2 ] ∆t = 1 4 (3.30 10–3) = 8.25 10–4 mol dm–3 s–1 Alternative method: The rate of formation of O2 : ∆[O2 ] ——– ∆t = —1 2 ∆[N2 O5 ] ———–– ∆t = —1 2 × (1.65 × 10–3) = 8.25 ×10–4 mol dm–3 s–1
186 Chemistry Term 1 STPM CHAPTER 5 Quick Check 5.2 1 The reaction taking place in the Haber process is: N2 (g) + 3H2 (g) → 2NH3 (g) At a particular moment, the rate is given as: –∆[N2 ] ∆t = 1.5 10–3 mol dm–3 s–1 What is the rate of reaction express in terms of (a) H2 and (b) NH3 ? 2 Consider the following reaction: 2Cu2+(aq) + 6CN– (aq) → 2[Cu(CN)2 ] – (aq) + (CN)2 (aq) (a) Express the rate of reaction in terms of the four species in the equation. (b) At a particular moment, the rate of formation of [Cu(CN)2 ] – is 5.60 10–2 mol dm–3 s–1. What is the rate express in terms of (i) CN– and (ii) (CN)2 ? 3 The reaction between fluorine and chlorine dioxide is as follows: F2 (g) + 2ClO2 (g) → 2FClO2 (g) If the rate of disappearance of F2 is 0.095 g s–1, what is the rate of formation of FClO2 in mol dm–3 s–1? The Collision Theory 1 The collision theory was put forward to explain how chemical reactions take place. 2 According to the collision theory, chemical reactions occur due to the collisions between reacting particles. 3 Take the following reaction as example: H2 (g) + I2 (g) → 2HI(g) Hydrogen iodide is formed as a result of the collisions between H2 and I2 molecules. 4 However, not all collisions would lead to the formation of hydrogen iodide. The rate of collision between H2 and I2 molecules at 450 °C (from theoretical calculation using the kinetic theory) is about 1 1028 collisions per second, while the rate of formation of HI is only about 1 1015 molecules per second. 5 A large fraction of collisions do not end up in forming the product. They are called ineffective collisions. 6 This arises because as the reactant molecules approach one another, they experience repulsion between their electron clouds. Repulsion Electron cloud H2 I Electron cloud 2 Arrhenius collision theory 2013/P1/Q10 2014/P1/Q10 Only a very small fraction of the collisions gives rise to products. Repulsion between electron clouds INFO Collision Theory
187 Chemistry Term 1 STPM CHAPTER 5 7 Therefore, the H2 and I2 molecules must have sufficient energy to overcome this repulsion before their electron clouds can penetrate one another to form HI. 8 The minimum energy required before a reaction can take place is called the activation energy, Ea . 9 If the reactant particles do not have sufficient energy to overcome the activation energy, the molecules would repel one another and no product will be formed. Thus, the activation energy acts like an energy barrier for the reaction. Insufficient energy Ineffective collision Repulsion With sufficient energy Effective collision Attraction 10 From the Maxwell-Boltzmann distribution curve, at constant temperature, only a very small fraction of the molecules have energy equal to or greater than the activation energy (Ea ). This accounts for the small fraction of effective collisions. 11 The activation energy is defined as the minimum energy that must be overcome by reactant molecules before a reaction can take place. 12 The activation energies for some reactions are given in the table below: Reaction Activation energy/kJ 2N2 O(g) → 2N2 (g) + O2 (g) 482.0 2HI(g) → H2 (g) + I2 (g) 390.0 CH3 Cl(g) + H2 O(l) → CH3 OH(aq) + HCl(aq) 115.0 2N2 O5 (g) → 2N2 O4 (g) + O2 (g) 217.0 13 The second condition is that the colliding molecules must be of the correct orientation. The direction of collision should be such that it facilitates formation of new bonds. 14 For example, the reaction between the methyl free radical and chloroform to form methane: H3 C• + CHCl3 → CH4 + Cl3 C• If the methyl free radical collides with the hydrogen atom of the chloroform molecule as shown below, and with sufficient energy, a reaction will occur and methane is formed. Definition of activation energy Activation energy is the energy barrier of the reaction. Energy Ea Number of particles These particles do not have enough energy to react Only these particles have enough energy to react Collisions must have the correct orientation before a reaction can occur.
188 Chemistry Term 1 STPM CHAPTER 5 H3 C•+ H—CCl3 → CH4 + Cl3 C• H • H3C C Cl Cl Cl Products 15 However, if the methyl free radical collides with the chlorine atom of the chloroform molecule, no reaction will occur even if the collision has enough energy to overcome the activation energy. H3 C•+ Cl—CHCl2 → No products formed H C • H3C Cl Cl Cl No products formed 16 Hence, for a chemical reaction to occur: (a) the colliding particles must have sufficient energy to overcome the activation energy, (b) the particles must collide in the correct orientation. 17 The energy profile of a typical chemical reaction is shown below: [E1 is the activation energy for the forward reaction. E2 is the activation energy for the reverse reaction. ∆H is the enthalpy change of reaction.] Energy Reaction pathway Reactants Δ E1 Products H [Activated complex] E2 Effects of Concentration and Temperature on the Rate of a Reaction Concentration 1 Increasing the concentration of the reactants increases the number of particles per unit volume. This increases the rate of collision leading to an increase in the rate of reaction. 2 The graph below shows the rate of evolution of carbon dioxide gas from the reaction between the same mass of powdered calcium carbonate and dilute hydrochloric acid of different concentrations. CaCO3 (s) + 2HCl(aq) → CaCl2 (aq) + H2 O(l) + CO2 (g) Volume of carbon dioxide Time [HCI] = 1.0M [HCI] = 2.0M Info Chem The activated complex is the intermediate with the highest energy. The greater the gradient, the faster the rate of reaction. 2015/P1/Q13 2017/P1/Q10
189 Chemistry Term 1 STPM CHAPTER 5 3 The gradient of the curve using 2.0 mol dm–3 HCl is steeper than that using 1.0 mol dm–3 HCl. This shows that the rate of reaction increases when the concentration of hydrochloric acid increases. Pressure 1 Pressure only affects reaction system containing gases because the volume occupied by a gas depends on the pressure on the gas. 2 When the pressure on a gas is increased, the molecules are pushed closer to one another. This causes the concentration of the gas to increase. 3 Hence, more collisions will occur leading to an increase in the reaction rate. Temperature 1 For most reactions, the rate of reaction increases exponentially with increasing (irregardless of whether the reaction is endothermic or exothermic) temperature. Rate of reaction Temperature 2 As temperature increases, the average kinetic energy of the molecule increases, resulting in more collisions per unit time. 3 At the same time, the number of molecules having energy equal to or greater than the activation energy also increases. This is illustrated by the Maxwell-Boltzmann distribution curve below. Fraction of molecules Molecular energy T1 Ea T2 T2 > T1 Low pressure High pressure Volume accupied by a gas depends on the pressure. Increasing pressure increases the concentration of the gas. Exponential relationship between rate of reaction and temperature. Number of particles having energy Ea increases with temperature.
190 Chemistry Term 1 STPM CHAPTER 5 4 These two factors together cause a large increase in the rate of reaction when the temperature is increased. However, the latter plays a more important role. 5 An increase of 10 °C usually doubles the rate of reaction. However, calculations from the kinetic theory show that the rate of collision only increases about 0.016 times when the temperature increases by 10 °C. (Refer to Example 5.3) 6 Look at the Maxwell-Boltzmann distribution curves below at temperature T K and (T + 10) K. Fraction of molecules Energy T + 10 T D C B Ea A E The total number of particles having energy equal or greater than the activation energy, at temperature T, is given the area of triangle ABC while the total number of particles having energy equal or greater than the activation energy, at temperature (T + 10), is given the area of triangle BDE, which is about two times the area of triangle ABC. 7 However, for biological reactions involving enzymes, the rate of reaction will increase with temperature until a certain point (the optimum temperature), beyond which the rate decreases as the enzymes (which are protein molecules) get denatured. Example 5.3 According to the kinetic theory, the kinetic energy (K.E.) of a particle is directly proportional to its absolute temperature (T). K.E. ∝ T Calculate the relative speed of the particle at 310 K and 300 K. Solution Kinetic energy = 1 2 mv 2 Since 1 2 mv 2 ∝ T Therefore, v 2 ∝ T Hence, v2 2 v1 2 ∝ T2 T1 [Where T2 = 310 K and T1 = 300 K] Rate of collision increases by 1.6% only for every 10 °C change in temperature. Rate of reaction involving enzymes Optimum temperature Temperature Info Chem ∆ADE ≈ 2 × ∆ABC
191 Chemistry Term 1 STPM CHAPTER 5 By substitution: v2 v1 = T2 T1 = 310 300 = 1.016 Hence, the molecular speed at 310 K is 1.016 times the molecular speed at 300 K. In other words, the molecular speed increases only 0.016 times when the temperature increases by 10 K from 300 K to 310 K. 5.2 Rate Law 1 Earlier in the chapter we noted that the rate of reaction is proportional to the concentration of the reactants. 2 The dependence of rate of reaction on concentration is expressed in what is called a rate equation. 3 This is different from the stoichiometric equation which shows the reacting mole ratios of the reactants and products. 4 For a general equation: aA + bB → cC + dD The rate equation is expressed as: Rate = k[A] x [B] y k is known as the rate constant, x and y are integers called the order of reaction. 5 The values of x and y are obtained through experiment only. It is not necessarily the same as the stoichiometric coefficient of the reactants in the balanced stoichiometric equation. 6 For the STPM syllabus, we are concerned with three types of order of reactions: First order, second order and zero order. 7 If a reaction is first order with respect to a substance, it means when the concentration of that substance is changed x times, the rate would change by x 1 times. 8 If a reaction is second order with respect to a substance, it means when the concentration of that substance is changed x times, the rate would change by x 2 times. 9 If a reaction is zero order with respect to a substance, it means when the concentration of that substance is changed x times, the rate would change by x 0 (which is = 1) times. That is the rate of the reaction is independent of the concentration of that substance. 10 For example, the rate equation for the reaction between carbon monoxide and nitrogen dioxide: CO(g) + NO2 (g) → CO2 (g) + NO(g) is given by: Rate = k[NO2 ] 2 Exam Tips Exam Tips 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. The overall order of reaction is the sum of the powers to which the reactants is raised in an experimentally determined rate equation. Info Chem The reaction is second order with respect to NO2 and zero order with respect to CO. 2009/P1/Q8 2010/P1/Q10, Q42 2011/P1/Q44