chemistry-part-1-book

148 CHEMISTRY Maxwell and Boltzmann have shown that the molecular speed distribution curve of actual distribution of molecular speeds chlorine and nitrogen given in Fig. 5.9. depends on temperature and molecular mass Though at a particular temperature the of a gas. Maxwell derived a formula for individual speed of molecules keeps calculating the number of molecules changing, the distribution of speeds remains possessing a particular speed. Fig. 5.8 shows same. schematic plot of number of molecules vs. molecular speed at two different temperatures T1 and T2 (T2 is higher than T1). The distribution of speeds shown in the plot is called Maxwell-Boltzmann distribution of speeds. Fig. 5.8: Maxwell-Boltzmann distribution of speeds Fig. 5.9: Distribution of molecular speeds for chlorine and nitrogen at 300 K The graph shows that number of molecules possessing very high and very low speed is very We know that kinetic energy of a particle is small. The maximum in the curve represents given by the expression: speed possessed by maximum number of molecules. This speed is called most Kinetic Energy = 1 mu2 probable speed, ump. This is very close to the 2 average speed of the molecules. On increasing the temperature most probable Therefore, if we want to know average speed increases. Also, speed distribution curve broadens at higher temperature. translational kinetic energy, 1 mu2 , for the Broadening of the curve shows that number 2 of molecules moving at higher speed increases. Speed distribution also depends movement of a gas particle in a straight line, upon mass of molecules. At the same we require the value of mean of square of temperature, gas molecules with heavier mass have slower speed than lighter gas speeds, u2 , of all molecules. This is molecules. For example, at the same represented as follows: temperature lighter nitrogen molecules move faster than heavier chlorine molecules. u2 = u12 +u22 +..........un2 Hence, at any given temperature, nitrogen n molecules have higher value of most probable speed than the chlorine molecules. Look at The mean square speed is the direct measure of the average kinetic energy of gas molecules. If we take the square root of the mean of the square of speeds then we get a value of speed which is different from most probable speed and average speed. This speed is called root mean square speed and is given by the expression as follows: urms = u2 Root mean square speed, average speed and the most probable speed have following relationship: urms > uav > ump 2020-21

STATES OF MATTER 149 The ratio between the three speeds is given • Particles of a gas move in all possible below : directions in straight lines. During their random motion, they collide with each ump: uav : urms : : 1 : 1.128 : 1.224 other and with the walls of the container. Pressure is exerted by the gas as a result 5.8 KINETIC MOLECULAR THEORY OF of collision of the particles with the walls of GASES the container. So far we have learnt the laws (e.g., Boyle’s law, • Collisions of gas molecules are perfectly Charles’ law etc.) which are concise statements elastic. This means that total energy of of experimental facts observed in the laboratory molecules before and after the collision by the scientists. Conducting careful remains same. There may be exchange of experiments is an important aspect of scientific energy between colliding molecules, their method and it tells us how the particular individual energies may change, but the system is behaving under different conditions. sum of their energies remains constant. If However, once the experimental facts are there were loss of kinetic energy, the motion established, a scientist is curious to know why of molecules will stop and gases will settle the system is behaving in that way. For down. This is contrary to what is actually example, gas laws help us to predict that observed. pressure increases when we compress gases but we would like to know what happens at • At any particular time, different particles molecular level when a gas is compressed ? A in the gas have different speeds and hence theory is constructed to answer such different kinetic energies. This assumption questions. A theory is a model (i.e., a mental is reasonable because as the particles picture) that enables us to better understand collide, we expect their speed to change. our observations. The theory that attempts to Even if initial speed of all the particles was elucidate the behaviour of gases is known as same, the molecular collisions will disrupt kinetic molecular theory. this uniformity. Consequently, the particles must have different speeds, which go on Assumptions or postulates of the kinetic- changing constantly. It is possible to show molecular theory of gases are given below. that though the individual speeds are These postulates are related to atoms and changing, the distribution of speeds molecules which cannot be seen, hence it is remains constant at a particular said to provide a microscopic model of gases. temperature. • Gases consist of large number of identical • If a molecule has variable speed, then it particles (atoms or molecules) that are so must have a variable kinetic energy. Under small and so far apart on the average that these circumstances, we can talk only the actual volume of the molecules is about average kinetic energy. In kinetic negligible in comparison to the empty space theory, it is assumed that average kinetic between them. They are considered as point energy of the gas molecules is directly masses. This assumption explains the proportional to the absolute temperature. great compressibility of gases. It is seen that on heating a gas at constant volume, the pressure increases. On heating • There is no force of attraction between the the gas, kinetic energy of the particles particles of a gas at ordinary temperature and increases and these strike the walls of the pressure. The support for this assumption container more frequently, thus, exerting comes from the fact that gases expand and more pressure. occupy all the space available to them. Kinetic theory of gases allows us to derive • Particles of a gas are always in constant and theoretically, all the gas laws studied in the random motion. If the particles were at rest previous sections. Calculations and predictions and occupied fixed positions, then a gas would have had a fixed shape which is not observed. 2020-21

150 CHEMISTRY based on kinetic theory of gases agree very well negative deviation from ideal behaviour, the pV with the experimental observations and thus value decreases with increase in pressure and establish the correctness of this model. reaches to a minimum value characteristic of a gas. After that pV value starts increasing. The 5.9 BEHAVIOUR OF REAL GASES: curve then crosses the line for ideal gas and DEVIATION FROM IDEAL GAS after that shows positive deviation BEHAVIOUR continuously. It is thus, found that real gases do not follow ideal gas equation perfectly under Our theoritical model of gases corresponds all conditions. very well with the experimental observations. Difficulty arises when we try to test how far Deviation from ideal behaviour also the relation pV = nRT reproduce actual becomes apparent when pressure vs volume pressure-volume-temperature relationship plot is drawn. The pressure vs volume plot of of gases. To test this point we plot pV vs p plot experimental data (real gas) and that of gases because at constant temperature, pV theoretically calculated from Boyle’s law (ideal will be constant (Boyle’s law) and pV vs p gas) should coincide. Fig 5.11 shows these graph at all pressures will be a straight line plots. It is apparent that at very high pressure parallel to x-axis. Fig. 5.10 shows such a plot the measured volume is more than the constructed from actual data for several gases calculated volume. At low pressures, measured at 273 K. and calculated volumes approach each other. Fig. 5.10 Plot of pV vs p for real gas and Fig. 5.11 Plot of pressure vs volume for real gas ideal gas and ideal gas It can be seen easily that at constant It is found that real gases do not follow, temperature pV vs p plot for real gases is not a Boyle’s law, Charles law and Avogadro law straight line. There is a significant deviation perfectly under all conditions. Now two from ideal behaviour. Two types of curves are questions arise. seen. In the curves for dihydrogen and helium, as the pressure increases the value of pV also (i) Why do gases deviate from the ideal increases. The second type of plot is seen in behaviour? the case of other gases like carbon monoxide and methane. In these plots first there is a (ii) What are the conditions under which gases deviate from ideality? 2020-21

STATES OF MATTER 151 We get the answer of the first question if we because instead of moving in volume V, these look into postulates of kinetic theory once are now restricted to volume (V–nb) where nb again. We find that two assumptions of the is approximately the total volume occupied by kinetic theory do not hold good. These are the molecules themselves. Here, b is a constant. Having taken into account the corrections for (a) There is no force of attraction between the pressure and volume, we can rewrite equation molecules of a gas. (5.17) as (b) Volume of the molecules of a gas is  + an 2  (V − nb) = nRT (5.31) negligibly small in comparison to the space  p V 2  occupied by the gas. Equation (5.31) is known as van der Waals If assumption (a) is correct, the gas will equation. In this equation n is number of moles never liquify. However, we know that gases do of the gas. Constants a and b are called van liquify when cooled and compressed. Also, der Waals constants and their value depends liquids formed are very difficult to compress. on the characteristic of a gas. Value of ‘a’ is This means that forces of repulsion are measure of magnitude of intermolecular powerful enough and prevent squashing of attractive forces within the gas and is molecules in tiny volume. If assumption (b) is independent of temperature and pressure. correct, the pressure vs volume graph of experimental data (real gas) and that Also, at very low temperature, theoritically calculated from Boyles law (ideal intermolecular forces become significant. As gas) should coincide. the molecules travel with low average speed, these can be captured by one another due to Real gases show deviations from ideal gas attractive forces. Real gases show ideal law because molecules interact with each other. behaviour when conditions of temperature and At high pressures molecules of gases are very pressure are such that the intermolecular close to each other. Molecular interactions start forces are practically negligible. The real gases operating. At high pressure, molecules do not show ideal behaviour when pressure strike the walls of the container with full impact approaches zero. because these are dragged back by other molecules due to molecular attractive forces. The deviation from ideal behaviour can be This affects the pressure exerted by the measured in terms of compressibility factor molecules on the walls of the container. Thus, Z, which is the ratio of product pV and nRT. the pressure exerted by the gas is lower than Mathematically the pressure exerted by the ideal gas. pideal = preal + an 2 (5.30) Z = pV (5.32) n RT observed V2 pressure For ideal gas Z = 1 at all temperatures and correction pressures because pV = n RT. The graph of Z term vs p will be a straight line parallel to pressure axis (Fig. 5.12, page 152). For gases which Here, a is a constant. deviate from ideality, value of Z deviates from unity. At very low pressures all gases shown Repulsive forces also become significant. Repulsive interactions are short-range have Z ≈1 and behave as ideal gas. At high interactions and are significant when molecules are almost in contact. This is the pressure all the gases have Z > 1. These are situation at high pressure. The repulsive forces more difficult to compress. At intermediate cause the molecules to behave as small but pressures, most gases have Z < 1. Thus gases impenetrable spheres. The volume occupied show ideal behaviour when the volume by the molecules also becomes significant 2020-21

152 CHEMISTRY Videal = n RT . On putting this value of nRT p p in equation (5.33) we have Z = Vreal (5.34) Videal From equation (5.34) we can see that compressibility factor is the ratio of actual molar volume of a gas to the molar volume of it, if it were an ideal gas at that temperature and pressure. In the following sections we will see that it is not possible to distinguish between gaseous state and liquid state and that liquids may be considered as continuation of gas phase into a region of small volumes and very high molecular attraction. We will also see how we Fig. 5.12 Variation of compressibility factor for can use isotherms of gases for predicting the some gases conditions for liquifaction of gases. occupied is large so that the volume of the 5.10 LIQUIFACTION OF GASES molecules can be neglected in comparison to it. In other words, the behaviour of the gas First complete data on pressure-volume- becomes more ideal when pressure is very low. temperature relations of a substance in both Upto what pressure a gas will follow the ideal gaseous and liquid state was obtained by gas law, depends upon nature of the gas and Thomas Andrews on Carbon dioxide. He plotted its temperature. The temperature at which a real isotherms of carbon dioxide at various gas obeys ideal gas law over an appreciable temperatures (Fig. 5.13). Later on it was found range of pressure is called Boyle temperature that real gases behave in the same manner as or Boyle point. Boyle point of a gas depends carbon dioxide. Andrews noticed that at high upon its nature. Above their Boyle point, real temperatures isotherms look like that of an gases show positive deviations from ideality and ideal gas and the gas cannot be liquified even at Z values are greater than one. The forces of very high pressure. As the temperature is attraction between the molecules are very feeble. lowered, shape of the curve changes and data Below Boyle temperature real gases first show show considerable deviation from ideal decrease in Z value with increasing pressure, behaviour. At 30.98°C carbon dioxide remains which reaches a minimum value. On further gas upto 73 atmospheric pressure. (Point E in increase in pressure, the value of Z increases Fig. 5.13). At 73 atmospheric pressure, liquid continuously. Above explanation shows that at carbon dioxide appears for the first time. The low pressure and high temperature gases show temperature 30.98°C is called critical ideal behaviour. These conditions are different temperature (TC) of carbon dioxide. This is the for different gases. highest temperature at which liquid carbon dioxide is observed. Above this temperature it More insight is obtained in the significance is gas. Volume of one mole of the gas at critical of Z if we note the following derivation temperature is called critical volume (VC) and pressure at this temperature is called critical Z = pVreal (5.33) pressure (pC). The critical temperature, pressure n RT and volume are called critical constants. Further increase in pressure simply compresses the If the gas shows ideal behaviour then 2020-21

STATES OF MATTER 153 Fig. 5.13 Isotherms of carbon dioxide at various Thus we see that a point like A in the Fig. 5.13 temperatures represents gaseous state. A point like D represents liquid state and a point under the liquid carbon dioxide and the curve represents dome shaped area represents existence of liquid the compressibility of the liquid. The steep line and gaseous carbon dioxide in equilibrium. All represents the isotherm of liquid. Even a slight the gases upon compression at constant compression results in steep rise in pressure temperature (isothermal compression) show the indicating very low compressibility of the liquid. same behaviour as shown by carbon dioxide. Below 30.98 °C, the behaviour of the gas on Also above discussion shows that gases should compression is quite different. At 21.5 °C, be cooled below their critical temperature for carbon dioxide remains as a gas only upto liquification. Critical temperature of a gas is point B. At point B, liquid of a particular volume highest temperature at which liquifaction of the appears. Further compression does not change gas first occurs. Liquifaction of so called the pressure. Liquid and gaseous carbon permanent gases (i.e., gases which show dioxide coexist and further application of continuous positive deviation in Z value) pressure results in the condensation of more requires cooling as well as considerable gas until the point C is reached. At point C, all compression. Compression brings the the gas has been condensed and further molecules in close vicinity and cooling slows application of pressure merely compresses the down the movement of molecules therefore, liquid as shown by steep line. A slight intermolecular interactions may hold the closely compression from volume V2 to V3 results in and slowly moving molecules together and the steep rise in pressure from p2 to p3 (Fig. 5.13). gas liquifies. Below 30.98 °C (critical temperature) each curve shows the similar trend. Only length of It is possible to change a gas into liquid or the horizontal line increases at lower a liquid into gas by a process in which always temperatures. At critical point horizontal a single phase is present. For example in portion of the isotherm merges into one point. Fig. 5.13 we can move from point A to F vertically by increasing the temperature, then we can reach the point G by compressing the gas at the constant temperature along this isotherm (isotherm at 31.1°C). The pressure will increase. Now we can move vertically down towards D by lowering the temperature. As soon as we cross the point H on the critical isotherm we get liquid. We end up with liquid but in this series of changes we do not pass through two-phase region. If process is carried out at the critical temperature, substance always remains in one phase. Thus there is continuity between the gaseous and liquid state. The term fluid is used for either a liquid or a gas to recognise this continuity. Thus a liquid can be viewed as a very dense gas. Liquid and gas can be distinguished only when the fluid is below its critical temperature and its pressure and volume lie under the dome, since in that situation liquid and gas are in equilibrium and a surface separating the two phases is visible. In the absence of this surface there is no 2020-21

154 CHEMISTRY fundamental way of distinguishing between between them and under normal conditions two states. At critical temperature, liquid liquids are denser than gases. passes into gaseous state imperceptibly and continuously; the surface separating two Molecules of liquids are held together by phases disappears (Section 5.11.1). A gas attractive intermolecular forces. Liquids have below the critical temperature can be liquified definite volume because molecules do not by applying pressure, and is called vapour of separate from each other. However, molecules the substance. Carbon dioxide gas below its of liquids can move past one another freely, critical temperature is called carbon dioxide therefore, liquids can flow, can be poured and vapour. Critical constants for some common can assume the shape of the container in which substances are given in Table 5.4. these are stored. In the following sections we Table 5.4 Critical Constants for Some will look into some of the physical properties of the liquids such as vapour pressure, surface Substances tension and viscosity. Problem 5.5 5.11.1 Vapour Pressure Gases possess characteristic critical temperature which depends upon the If an evacuated container is partially filled with magnitude of intermolecular forces a liquid, a portion of liquid evaporates to fill between the gas particles. Critical the remaining volume of the container with temperatures of ammonia and carbon vapour. Initially the liquid evaporates and dioxide are 405.5 K and 304.10 K pressure exerted by vapours on the walls of respectively. Which of these gases will the container (vapour pressure) increases. After liquify first when you start cooling from some time it becomes constant, an equilibrium 500 K to their critical temperature ? is established between liquid phase and Solution vapour phase. Vapour pressure at this stage Ammonia will liquify first because its is known as equilibrium vapour pressure or critical temperature will be reached first. saturated vapour pressure.. Since process of Liquifaction of CO2 will require more vapourisation is temperature dependent; the cooling. temperature must be mentioned while reporting the vapour pressure of a liquid. 5.11 LIQUID STATE Intermolecular forces are stronger in liquid When a liquid is heated in an open vessel, state than in gaseous state. Molecules in liquids the liquid vapourises from the surface. At the are so close that there is very little empty space temperature at which vapour pressure of the liquid becomes equal to the external pressure, vapourisation can occur throughout the bulk of the liquid and vapours expand freely into the surroundings. The condition of free vapourisation throughout the liquid is called boiling. The temperature at which vapour pressure of liquid is equal to the external pressure is called boiling temperature at that pressure. Vapour pressure of some common liquids at various temperatures is given in (Fig. 5.14, page 155). At 1 atm pressure boiling temperature is called normal boiling point. If pressure is 1 bar then the boiling point is called standard boiling point of the liquid. Standard boiling point of the liquid is slightly lower than the normal boiling point because 2020-21

STATES OF MATTER 155 Fig. 5.14 Vapour pressure vs temperature curve more and more molecules go to vapour phase of some common liquids. and density of vapours rises. At the same time liquid becomes less dense. It expands because 1 bar pressure is slightly less than 1 atm molecules move apart. When density of liquid pressure. The normal boiling point of water is and vapours becomes the same; the clear 100 °C (373 K), its standard boiling point is boundary between liquid and vapours 99.6 °C (372.6 K). disappears. This temperature is called critical temperature about which we have already At high altitudes atmospheric pressure is discussed in section 5.10. low. Therefore liquids at high altitudes boil at lower temperatures in comparison to that at 5.11.2 Surface Tension sea level. Since water boils at low temperature It is well known fact that liquids assume the on hills, the pressure cooker is used for shape of the container. Why is it then small cooking food. In hospitals surgical instruments drops of mercury form spherical bead instead are sterilized in autoclaves in which boiling of spreading on the surface. Why do particles point of water is increased by increasing the of soil at the bottom of river remain separated pressure above the atmospheric pressure by but they stick together when taken out ? Why using a weight covering the vent. does a liquid rise (or fall) in a thin capillary as soon as the capillary touches the surface of Boiling does not occur when liquid is the liquid ? All these phenomena are caused heated in a closed vessel. On heating due to the characteristic property of liquids, continuously vapour pressure increases. At called surface tension. A molecule in the bulk first a clear boundary is visible between liquid of liquid experiences equal intermolecular and vapour phase because liquid is more dense forces from all sides. The molecule, therefore than vapour. As the temperature increases does not experience any net force. But for the molecule on the surface of liquid, net attractive force is towards the interior of the liquid (Fig. 5.15), due to the molecules below it. Since there are no molecules above it. Liquids tend to minimize their surface area. The molecules on the surface experience a net downward force and have more energy than Fig. 5.15 Forces acting on a molecule on liquid surface and on a molecule inside the liquid 2020-21

156 CHEMISTRY the molecules in the bulk, which do not between layers of fluid as they slip past one experience any net force. Therefore, liquids tend another while liquid flows. Strong to have minimum number of molecules at their intermolecular forces between molecules hold surface. If surface of the liquid is increased by them together and resist movement of layers pulling a molecule from the bulk, attractive past one another. forces will have to be overcome. This will require expenditure of energy. The energy When a liquid flows over a fixed surface, required to increase the surface area of the the layer of molecules in the immediate contact liquid by one unit is defined as surface energy. of surface is stationary. The velocity of upper Its dimensions are J m–2. Surface tension is layers increases as the distance of layers from defined as the force acting per unit length the fixed layer increases. This type of flow in perpendicular to the line drawn on the surface which there is a regular gradation of velocity of liquid. It is denoted by Greek letter γ in passing from one layer to the next is called (Gamma). It has dimensions of kg s–2 and in SI laminar flow. If we choose any layer in the unit it is expressed as N m–1. The lowest energy flowing liquid (Fig.5.16), the layer above it state of the liquid will be when surface area is accelerates its flow and the layer below this minimum. Spherical shape satisfies this condition, that is why mercury drops are Fig. 5.16 Gradation of velocity in the laminar spherical in shape. This is the reason that sharp flow glass edges are heated for making them smooth. On heating, the glass melts and the retards its flow. surface of the liquid tends to take the rounded shape at the edges, which makes the edges If the velocity of the layer at a distance dz smooth. This is called fire polishing of glass. is changed by a value du then velocity gradient Liquid tends to rise (or fall) in the capillary du because of surface tension. Liquids wet the is given by the amount . dz A force is required things because they spread across their surfaces to maintain the flow of layers. This force is as thin film. Moist soil grains are pulled together proportional to the area of contact of layers because surface area of thin film of water is and velocity gradient i.e. reduced. It is surface tension which gives stretching property to the surface of a liquid. F ∝ A (A is the area of contact) On flat surface, droplets are slightly flattened by the effect of gravity; but in the gravity free F ∝ A. du (where, du is velocity gradient; environments drops are perfectly spherical. dz dz The magnitude of surface tension of a liquid the change in velocity with distance) depends on the attractive forces between the molecules. When the attractive forces are large, F ∝ A. du the surface tension is large. Increase in dz temperature increases the kinetic energy of the molecules and effectiveness of intermolecular ⇒ F = ηA du attraction decreases, so surface tension dz decreases as the temperature is raised. 5.11.3 Viscosity It is one of the characteristic properties of liquids. Viscosity is a measure of resistance to flow which arises due to the internal friction 2020-21

STATES OF MATTER 157 ‘ η ’ is proportionality constant and is called Greater the viscosity, the more slowly the liquid flows. Hydrogen bonding and van der coefficient of viscosity. Viscosity coefficient Waals forces are strong enough to cause high is the force when velocity gradient is unity and viscosity. Glass is an extremely viscous liquid. the area of contact is unit area. Thus ‘ η ’ is It is so viscous that many of its properties resemble solids. measure of viscosity. SI unit of viscosity coefficient is 1 newton second per square metre Viscosity of liquids decreases as the (N s m–2) = pascal second (Pa s = 1kg m–1s–1). In temperature rises because at high temperature cgs system the unit of coefficient of viscosity is molecules have high kinetic energy and can poise (named after great scientist Jean Louise overcome the intermolecular forces to slip past Poiseuille). one another between the layers. 1 poise = 1 g cm–1s–1 = 10–1kg m–1s–1 SUMMARY Intermolecular forces operate between the particles of matter. These forces differ from pure electrostatic forces that exist between two oppositely charged ions. Also, these do not include forces that hold atoms of a covalent molecule together through covalent bond. Competition between thermal energy and intermolecular interactions determines the state of matter. “Bulk” properties of matter such as behaviour of gases, characteristics of solids and liquids and change of state depend upon energy of constituent particles and the type of interaction between them. Chemical properties of a substance do not change with change of state, but the reactivity depends upon the physical state. Forces of interaction between gas molecules are negligible and are almost independent of their chemical nature. Interdependence of some observable properties namely pressure, volume, temperature and mass leads to different gas laws obtained from experimental studies on gases. Boyle’s law states that under isothermal condition, pressure of a fixed amount of a gas is inversely proportional to its volume. Charles’ law is a relationship between volume and absolute temperature under isobaric condition. It states that volume of a fixed amount of gas is directly proportional to its absolute temperature (V ∝ T). If state of a gas is represented by p1, V and T and it changes to 1 1 state at p2, V2 and T2, then relationship between these two states is given by combined gas law according to p1V1 = p2V2 . Any one of the variables of this gas can be which T1 T2 found out if other five variables are known. Avogadro law states that equal volumes of all gases under same conditions of temperature and pressure contain equal number of molecules. Dalton’s law of partial pressure states that total pressure exerted by a mixture of non-reacting gases is equal to the sum of partial pressures exerted by them. Thus p = p1+p2+p3+ ... . Relationship between pressure, volume, temperature and number of moles of a gas describes its state and is called equation of state of the gas. Equation of state for ideal gas is pV=nRT, where R is a gas constant and its value depends upon units chosen for pressure, volume and temperature. At high pressure and low temperature intermolecular forces start operating strongly between the molecules of gases because they come close to each other. Under suitable temperature and pressure conditions gases can be liquified. Liquids may be considered as continuation of gas phase into a region of small volume and very strong molecular attractions. Some properties of liquids e.g., surface tension and viscosity are due to strong intermolecular attractive forces. 2020-21

158 CHEMISTRY 5.1 EXERCISES 5.2 What will be the minimum pressure required to compress 500 dm3 of air at 1 bar to 5.3 200 dm3 at 30°C? 5.4 5.5 A vessel of 120 mL capacity contains a certain amount of gas at 35 °C and 1.2 bar pressure. The gas is transferred to another vessel of volume 180 mL at 35 °C. What 5.6 would be its pressure? 5.7 Using the equation of state pV=nRT; show that at a given temperature density of a 5.8 gas is proportional to gas pressure p. 5.9 5.10 At 0°C, the density of a certain oxide of a gas at 2 bar is same as that of dinitrogen 5.11 at 5 bar. What is the molecular mass of the oxide? 5.12 Pressure of 1 g of an ideal gas A at 27 °C is found to be 2 bar. When 2 g of another 5.13 ideal gas B is introduced in the same flask at same temperature the pressure 5.14 becomes 3 bar. Find a relationship between their molecular masses. 5.15 5.16 The drain cleaner, Drainex contains small bits of aluminum which react with caustic soda to produce dihydrogen. What volume of dihydrogen at 20 °C and one bar will 5.17 be released when 0.15g of aluminum reacts? 5.18 5.19 What will be the pressure exerted by a mixture of 3.2 g of methane and 4.4 g of carbon dioxide contained in a 9 dm3 flask at 27 °C ? What will be the pressure of the gaseous mixture when 0.5 L of H2 at 0.8 bar and 2.0 L of dioxygen at 0.7 bar are introduced in a 1L vessel at 27°C? Density of a gas is found to be 5.46 g/dm3 at 27 °C at 2 bar pressure. What will be its density at STP? 34.05 mL of phosphorus vapour weighs 0.0625 g at 546 °C and 0.1 bar pressure. What is the molar mass of phosphorus? A student forgot to add the reaction mixture to the round bottomed flask at 27 °C but instead he/she placed the flask on the flame. After a lapse of time, he realized his mistake, and using a pyrometer he found the temperature of the flask was 477 °C. What fraction of air would have been expelled out? Calculate the temperature of 4.0 mol of a gas occupying 5 dm3 at 3.32 bar. (R = 0.083 bar dm3 K–1 mol–1). Calculate the total number of electrons present in 1.4 g of dinitrogen gas. How much time would it take to distribute one Avogadro number of wheat grains, if 1010 grains are distributed each second ? Calculate the total pressure in a mixture of 8 g of dioxygen and 4 g of dihydrogen confined in a vessel of 1 dm3 at 27°C. R = 0.083 bar dm3 K–1 mol–1. Pay load is defined as the difference between the mass of displaced air and the mass of the balloon. Calculate the pay load when a balloon of radius 10 m, mass 100 kg is filled with helium at 1.66 bar at 27°C. (Density of air = 1.2 kg m–3 and R = 0.083 bar dm3 K–1 mol–1). Calculate the volume occupied by 8.8 g of CO2 at 31.1°C and 1 bar pressure. R = 0.083 bar L K–1 mol–1. 2.9 g of a gas at 95 °C occupied the same volume as 0.184 g of dihydrogen at 17 °C, at the same pressure. What is the molar mass of the gas? A mixture of dihydrogen and dioxygen at one bar pressure contains 20% by weight of dihydrogen. Calculate the partial pressure of dihydrogen. 2020-21

STATES OF MATTER 159 5.20 What would be the SI unit for the quantity pV 2T 2/n ? 5.21 In terms of Charles’ law explain why –273 °C is the lowest possible temperature. 5.22 Critical temperature for carbon dioxide and methane are 31.1 °C and –81.9 °C respectively. Which of these has stronger intermolecular forces and why? 5.23 Explain the physical significance of van der Waals parameters. 2020-21

160 CHEMISTRY UNIT 6 THERMODYNAMICS After studying this Unit, you will It is the only physical theory of universal content concerning be able to which I am convinced that, within the framework of the applicability of its basic concepts, it will never be overthrown. • explain the terms : system and surroundings; Albert Einstein • discriminate between close, Chemical energy stored by molecules can be released as heat open and isolated systems; during chemical reactions when a fuel like methane, cooking gas or coal burns in air. The chemical energy may also be • explain internal energy, work used to do mechanical work when a fuel burns in an engine and heat; or to provide electrical energy through a galvanic cell like dry cell. Thus, various forms of energy are interrelated and • state first law of under certain conditions, these may be transformed from thermodynamics and express one form into another. The study of these energy it mathematically; transformations forms the subject matter of thermodynamics. The laws of thermodynamics deal with energy changes of • calculate energy changes as macroscopic systems involving a large number of molecules work and heat contributions rather than microscopic systems containing a few molecules. in chemical systems; Thermodynamics is not concerned about how and at what rate these energy transformations are carried out, but is • explain state functions: U, H. based on initial and final states of a system undergoing the • correlate ∆U and ∆H; change. Laws of thermodynamics apply only when a system • measure experimentally ∆U is in equilibrium or moves from one equilibrium state to another equilibrium state. Macroscopic properties like and ∆H; pressure and temperature do not change with time for a • define standard states for ∆H; system in equilibrium state. In this unit, we would like to • calculate enthalpy changes for answer some of the important questions through thermodynamics, like: various types of reactions; How do we determine the energy changes involved in a • state and apply Hess’s law of chemical reaction/process? Will it occur or not? constant heat summation; What drives a chemical reaction/process? To what extent do the chemical reactions proceed? • differentiate between extensive and intensive properties; • define spontaneous and non- spontaneous processes; • explain entropy as a thermodynamic state function and apply it for spontaneity; • explain Gibbs energy change (∆G); and • establish relationship between ∆G and spontaneity, ∆G and equilibrium constant. 2020-21

THERMODYNAMICS 161 6.1 THERMODYNAMIC TERMS be real or imaginary. The wall that separates the system from the surroundings is called We are interested in chemical reactions and the boundary. This is designed to allow us to energy changes accompanying them. For this control and keep track of all movements of we need to know certain thermodynamic matter and energy in or out of the system. terms. These are discussed below. 6.1.2 Types of the System 6.1.1 The System and the Surroundings We, further classify the systems according to A system in thermodynamics refers to that the movements of matter and energy in or out part of universe in which observations are of the system. made and remaining universe constitutes the surroundings. The surroundings include 1. Open System everything other than the system. System and the surroundings together constitute the In an open system, there is exchange of energy universe . and matter between system and surroundings [Fig. 6.2 (a)]. The presence of reactants in an The universe = The system + The surroundings However, the entire universe other than open beaker is an example of an open system*. the system is not affected by the changes Here the boundary is an imaginary surface taking place in the system. Therefore, for enclosing the beaker and reactants. all practical purposes, the surroundings are that portion of the remaining universe 2. Closed System which can interact with the system. Usually, the region of space in the In a closed system, there is no exchange of neighbourhood of the system constitutes matter, but exchange of energy is possible its surroundings. between system and the surroundings [Fig. 6.2 (b)]. The presence of reactants in a For example, if we are studying the closed vessel made of conducting material e.g., reaction between two substances A and B copper or steel is an example of a closed kept in a beaker, the beaker containing the system. reaction mixture is the system and the room where the beaker is kept is the surroundings (Fig. 6.1). Fig. 6.1 System and the surroundings Note that the system may be defined by Fig. 6.2 Open, closed and isolated systems. physical boundaries, like beaker or test tube, or the system may simply be defined by a set of Cartesian coordinates specifying a particular volume in space. It is necessary to think of the system as separated from the surroundings by some sort of wall which may * We could have chosen only the reactants as system then walls of the beakers will act as boundary. 2020-21

162 CHEMISTRY 3. Isolated System of the system. It may be chemical, electrical, mechanical or any other type of energy you In an isolated system, there is no exchange of may think of, the sum of all these is the energy energy or matter between the system and the of the system. In thermodynamics, we call it surroundings [Fig. 6.2 (c)]. The presence of the internal energy, U of the system, which may reactants in a thermos flask or any other closed change, when insulated vessel is an example of an isolated • heat passes into or out of the system, system. • work is done on or by the system, • matter enters or leaves the system. 6.1.3 The State of the System These systems are classified accordingly as The system must be described in order to make you have already studied in section 6.1.2. any useful calculations by specifying quantitatively each of the properties such as (a) Work its pressure (p), volume (V), and temperature Let us first examine a change in internal (T ) as well as the composition of the system. energy by doing work. We take a system We need to describe the system by specifying containing some quantity of water in a thermos it before and after the change. You would recall flask or in an insulated beaker. This would not from your Physics course that the state of a allow exchange of heat between the system system in mechanics is completely specified at and surroundings through its boundary and a given instant of time, by the position and we call this type of system as adiabatic. The velocity of each mass point of the system. In manner in which the state of such a system thermodynamics, a different and much simpler may be changed will be called adiabatic concept of the state of a system is introduced. process. Adiabatic process is a process in It does not need detailed knowledge of motion which there is no transfer of heat between the of each particle because, we deal with average system and surroundings. Here, the wall measurable properties of the system. We specify separating the system and the surroundings the state of the system by state functions or is called the adiabatic wall (Fig 6.3). state variables. Let us bring the change in the internal The state of a thermodynamic system is energy of the system by doing some work on described by its measurable or macroscopic (bulk) properties. We can describe the state of Fig. 6.3 An adiabatic system which does not a gas by quoting its pressure (p), volume (V), permit the transfer of heat through its temperature (T ), amount (n) etc. Variables like boundary. p, V, T are called state variables or state functions because their values depend only it. Let us call the initial state of the system as on the state of the system and not on how it is state A and its temperature as TA. Let the reached. In order to completely define the state internal energy of the system in state A be of a system it is not necessary to define all the called UA. We can change the state of the system properties of the system; as only a certain in two different ways. number of properties can be varied independently. This number depends on the nature of the system. Once these minimum number of macroscopic properties are fixed, others automatically have definite values. The state of the surroundings can never be completely specified; fortunately it is not necessary to do so. 6.1.4 The Internal Energy as a State Function When we talk about our chemical system losing or gaining energy, we need to introduce a quantity which represents the total energy 2020-21

THERMODYNAMICS 163 One way: We do some mechanical work, say the route taken. Volume of water in a pond, for example, is a state function, because 1 kJ, by rotating a set of small paddles and change in volume of its water is independent of the route by which water is filled in the thereby churning water. Let the new state be pond, either by rain or by tubewell or by both. called B state and its temperature, as TB. It is (b) Heat found that T∆BT> TA and the change in temperature, = TB–TA. Let the internal We can also change the internal energy of a system by transfer of heat from the cehnaernggyeoinf tihnetesrynsatel menienrgsyt,a∆teUB=UbeB–UUBAa. nd the surroundings to the system or vice-versa without expenditure of work. This exchange Second way: We now do an equal amount (i.e., of energy, which is a result of temperature 1kJ) electrical work with the help of an difference is called heat, q. Let us consider immersion rod and note down the temperature bringing about the same change in temperature change. We find that the change in temperature (the same initial and final states as before in is same as in the earlier case, say, TB – TA. section 6.1.4 (a) by transfer of heat through thermally conducting walls instead of In fact, the experiments in the above adiabatic walls (Fig. 6.4). manner were done by J. P. Joule between 1840–50 and he was able to show that a given amount of work done on the system, no matter how it was done (irrespective of path) produced the same change of state, as measured by the change in the temperature of the system. So, it seems appropriate to define a Fig. 6.4 A system which allows heat transfer quantity, the internal energy U, whose value through its boundary. is characteristic of the state of a system, whereby the adiabatic work, wad required to bring about a change of state is equal to the difference between the value of U in one state and that in another state, ∆U i.e., ∆U = U 2 − U1 = w ad Therefore, internal energy, U, of the system is a state function. By conventions of IUPAC in chemical We take water at temperature, TA in a thermodynamics. The positive sign expresses container having thermally conducting walls, that wad is positive when work is done on the system and the internal energy of system say made up of copper and enclose it in a huge increases. Similarly, if the work is done by the system,wad will be negative because internal heat reservoir at temperature, TB. The heat energy of the system decreases. absorbed by the system (water), q can be measured in terms of temperature difference , Can you name some other familiar state ∆TUB –= TA. In this case change in internal energy, functions? Some of other familiar state q, when no work is done at constant functions are V, p, and T. For example, if we bring a change in temperature of the system volume. from 25°C to 35°C, the change in temperature is 35°C–25°C = +10°C, whether we go straight By conventions of IUPAC in chemical up to 35°C or we cool the system for a few thermodynamics. The q is positive, when degrees, then take the system to the final heat is transferred from the surroundings to temperature. Thus, T is a state function and the system and the internal energy of the the change in temperature is independent of system increases and q is negative when heat is transferred from system to the surroundings resulting in decrease of the internal energy of the system.. * Earlier negative sign was assigned when the work is done on the system and positive sign when the work is done by the system. This is still followed in physics books, although IUPAC has recommended the use of new sign convention. 2020-21

164 CHEMISTRY (c) The general case Solution (i) ∆ U = w ad, wall is adiabatic Let us consider the general case in which a (ii) ∆ U = – q, thermally conducting walls change of state is brought about both by (iii) ∆ U = q – w, closed system. doing work and by transfer of heat. We write 6.2 APPLICATIONS change in internal energy for this case as: Many chemical reactions involve the generation of gases capable of doing mechanical work or ∆U = q + w (6.1) the generation of heat. It is important for us to quantify these changes and relate them to the For a given change in state, q and w can changes in the internal energy. Let us see how! vary depending on how the change is carried 6.2.1 Work out. However, q +w = ∆U will depend only on First of all, let us concentrate on the nature of initial and final state. It will be independent of work a system can do. We will consider only the way the change is carried out. If there is mechanical work i.e., pressure-volume work. no transfer of energy as heat or as work (isolated system) i.e., if w = 0 and q = 0, then For understanding pressure-volume ∆ U = 0. work, let us consider a cylinder which contains one mole of an ideal gas fitted with a The equation 6.1 i.e., ∆U = q + w is frictionless piston. Total volume of the gas is mathematical statement of the first law of Vi and pressure of the gas inside is p. If thermodynamics, which states that external pressure is pex which is greater than p, piston is moved inward till the pressure The energy of an isolated system is inside becomes equal to pex. Let this change constant. Fig. 6.5(a) Work done on an ideal gas in a It is commonly stated as the law of cylinder when it is compressed by a conservation of energy i.e., energy can neither constant external pressure, pex be created nor be destroyed. (in single step) is equal to the shaded area. Note: There is considerable difference between the character of the thermodynamic property energy and that of a mechanical property such as volume. We can specify an unambiguous (absolute) value for volume of a system in a particular state, but not the absolute value of the internal energy. However, we can measure only the changes in the internal energy, ∆U of the system. Problem 6.1 Express the change in internal energy of a system when (i) No heat is absorbed by the system from the surroundings, but work (w) is done on the system. What type of wall does the system have ? (ii) No work is done on the system, but q amount of heat is taken out from the system and given to the surroundings. What type of wall does the system have? (iii) w amount of work is done by the system and q amount of heat is supplied to the system. What type of system would it be? 2020-21

THERMODYNAMICS 165 be achieved in a single step and the final If the pressure is not constant but changes during the process such that it is always volume be Vf . During this compression, infinitesimally greater than the pressure of the suppose piston moves a distance, l and is gas, then, at each stage of compression, the volume decreases by an infinitesimal amount, cross-sectional area of the piston is A dV. In such a case we can calculate the work done on the gas by the relation [Fig. 6.5(a)]. then, volume change = l × A = ∆V = (Vf – Vi ) We also know, pressure = force Vf area ∫w = − pex dV ( 6.3) Therefore, force on the piston = pex . A Vi If w is the work done on the system by movement of the piston then Here, pex at each stage is equal to (pin + dp) in case of compression [Fig. 6.5(c)]. In an w = force × distance = pex . A .l = pex . (–∆V) = – pex ∆V = – pex (Vf – Vi ) (6.2) expansion process under similar conditions, The negative sign of this expression is the external pressure is always less than the required to obtain conventional sign for w, which will be positive. It indicates that in case pressure of the system i.e., pex = (pin– dp). In of compression work is done on the system. general case we can write, pex = (pin + dp). Such Here (Vf – Vi ) will be negative and negative processes are called reversible processes. multiplied by negative will be positive. Hence the sign obtained for the work will be positive. A process or change is said to be If the pressure is not constant at every reversible, if a change is brought out in stage of compression, but changes in number of finite steps, work done on the gas will be such a way that the process could, at any summed over all the steps and will be equal moment, be reversed by an infinitesimal to − ∑ p∆V [Fig. 6.5 (b)] change. A reversible process proceeds infinitely slowly by a series of equilibrium states such that system and the surroundings are always in near equilibrium with each other. Processes Fig. 6.5 (b) pV-plot when pressure is not constant Fig. 6.5 (c) pV-plot when pressure is not constant and changes in finite steps during and changes in infinite steps compression from initial volume, Vi to (reversible conditions) during final volume, Vf . Work done on the gas is represented by the shaded area. compression from initial volume, Vi to final volume, Vf . Work done on the gas is represented by the shaded area. 2020-21

166 CHEMISTRY other than reversible processes are known Isothermal and free expansion of an as irreversible processes. ideal gas In chemistry, we face problems that can For isothermal (T = constant) expansion of an be solved if we relate the work term to the internal pressure of the system. We can ideal gas into vacuum ; w = 0 since p = 0. relate work to internal pressure of the system ex under reversible conditions by writing equation 6.3 as follows: Also, Joule determined experimentally that q = 0; therefore, ∆U = 0 Equation 6.1, ∆U = q + w can be expressed for isothermal irreversible and Vf Vf reversible changes as follows: ∫ ∫wrev = − pexdV = − (pin ± dp)dV 1. For isothermal irreversible change Vi Vi q = – w = pex (Vf – Vi ) Since dp × dV is very small we can write 2. For isothermal reversible change Vf (6.4) Vf q = – w = nRT ln Vi ∫wrev = − pindV Vi Now, the pressure of the gas (pin which we Vf can write as p now) can be expressed in terms = 2.303 nRT log Vi of its volume through gas equation. For n mol 3. For adiabatic change, q = 0, of an ideal gas i.e., pV =nRT ∆U = wad ⇒ p = nRT Problem 6.2 V Two litres of an ideal gas at a pressure of Therefore, at constant temperature (isothermal 10 atm expands isothermally at 25 °C into process), a vacuum until its total volume is 10 litres. How much heat is absorbed and how much ∫w rev= Vf nRT dV = −nRT ln V f work is done in the expansion ? Vi − Solution VVi We have q = – w = pex (10 – 2) = 0(8) = 0 Vf (6.5) No work is done; no heat is absorbed. = – 2.303 nRT log Vi Problem 6.3 Free expansion: Expansion of a gas in Consider the same expansion, but this vacuum (pex = 0) is called free expansion. No time against a constant external pressure work is done during free expansion of an ideal of 1 atm. gas whether the process is reversible or Solution irreversible (equation 6.2 and 6.3). We have q = – w = pex (8) = 8 litre-atm Problem 6.4 Now, we can write equation 6.1 in number Consider the expansion given in problem of ways depending on the type of processes. 6.2, for 1 mol of an ideal gas conducted reversibly. Let us substitute w = – pex∆V (eq. 6.2) in equation 6.1, and we get ∆U = q − pex ∆V If a process is carried out at constant volume Solution (∆V = 0), then We have q = – w = 2.303 nRT log Vf ∆U = qV Vs 10 = 2.303 × 1 × 0.8206 × 298 × log 2 the subscript V in q denotes that heat is V supplied at constant volume. 2020-21

THERMODYNAMICS 167 = 2.303 x 0.8206 x 298 x log 5 Remember ∆H = qp, heat absorbed by the = 2.303 x 0.8206 x 298 x 0.6990 system at constant pressure. = 393.66 L atm ∆H is negative for exothermic reactions 6.2.2 Enthalpy, H which evolve heat during the reaction and ∆H is positive for endothermic reactions (a) A Useful New State Function which absorb heat from the surroundings. We know that the heat absorbed at constant At constant volume (∆V = 0), ∆U = qV, volume is equal to change in the internal therefore equation 6.8 becomes energy i.e., ∆U = q . But most of chemical reactions are carriVed out not at constant ∆H = ∆U = q V volume, but in flasks or test tubes under constant atmospheric pressure. We need to The difference between ∆H and ∆U is not usually significant for systems consisting of define another state function which may be suitable under these conditions. only solids and / or liquids. Solids and liquids do not suffer any significant volume changes We may write equation (6.1) as upon heating. The difference, however, ∆U = qp – p∆V at constant psryesstseumrea, nwdhe–rpe∆qVp becomes significant when gases are involved. is heat absorbed by the Let us consider a reaction involving gases. If represent expansion work done by the system. VA is the total volume of the gaseous reactants, VB is the total volume of the gaseous products, Let us represent the initial state by nA is the number of moles of gaseous reactants subscript 1 and final state by 2 and nB is the number of moles of gaseous products, all at constant pressure and We can rewrite the above equation as temperature, then using the ideal gas law, we U2–U1 = qp – p (V2 – V1) On rearranging, we get write, pVA = nART qp = (U2 + pV2) – (U1 + pV1) (6.6) and pVB = nBRT Now we can define another thermodynamic function, the enthalpy H [Greek word Thus, pVB – pVA = nBRT – nART = (nB–nA)RT enthalpien, to warm or heat content] as : or p (VB – VA) = (nB – nA) RT H = U + pV (6.7) or p ∆V = ∆ngRT so, equation (6.6) becomes (6.9) qp= H2 – H1 = ∆H Here, ∆ng refers to the number of moles of Although q is a path dependent function, gaseous products minus the number of moles H is a state function because it depends on U, p and V, all of which are state functions. of gaseous reactants. Therefore, ∆H is independent of path. Hence, qp is also independent of path. Substituting the value of p∆V from For finite changes at constant pressure, we equation 6.9 in equation 6.8, we get can write equation 6.7 as ∆H = ∆U + ∆ngRT (6.10) ∆H = ∆U + ∆pV The equation 6.10 is useful for calculating ∆H from ∆U and vice versa. Since p is constant, we can write (6.8) Problem 6.5 ∆H = ∆U + p∆V If water vapour is assumed to be a perfect It is important to note that when heat is gas, molar enthalpy change for absorbed by the system at constant pressure, vapourisation of 1 mol of water at 1bar we are actually measuring changes in the and 100°C is 41kJ mol–1. Calculate the enthalpy. internal energy change, when 2020-21

168 CHEMISTRY 1 mol of water is vapourised at 1 bar Fig. 6.6(a) A gas at volume V and temperature T pressure and 100°C. Fig. 6.6 (b) Partition, each part having half the Solution volume of the gas (i) The change H2O (l ) → H2O (g) ∆H = ∆U + ∆ngRT or ∆U = ∆H – ∆ngRT , substituting the values, we get ∆U = 41.00 kJ mol−1 − 1 × 8.3 J mol−1K −1 × 373 K = 41.00 kJ mol-1 – 3.096 kJ mol-1 = 37.904 kJ mol–1 (b) Extensive and Intensive Properties (c) Heat Capacity In thermodynamics, a distinction is made In this sub-section, let us see how to measure between extensive properties and intensive heat transferred to a system. This heat appears properties. An extensive property is a as a rise in temperature of the system in case property whose value depends on the of heat absorbed by the system. quantity or size of matter present in the system. For example, mass, volume, The increase of temperature is proportional internal energy, enthalpy, heat capacity, etc. to the heat transferred are extensive properties. q = coeff × ∆T Those properties which do not depend The magnitude of the coefficient depends on the quantity or size of matter present are on the size, composition and nature of the system. We can also write it as q = C ∆T known as intensive properties. For The coefficient, C is called the heat capacity. example temperature, density, pressure etc. Thus, we can measure the heat supplied are intensive properties. A molar property, by monitoring the temperature rise, provided we know the heat capacity. χm, is the value of an extensive property χ of When C is large, a given amount of heat the system for 1 mol of the substance. If n results in only a small temperature rise. Water χ has a large heat capacity i.e., a lot of energy is is the amount of matter, χm = n is needed to raise its temperature. independent of the amount of matter. Other C is directly proportional to amount of substance. The molar heat capacity of a examples are molar volume, Vm and molar heat capacity, Cm. Let us understand the distinction between extensive and intensive properties by considering a gas enclosed in a container of volume V and at temperature T [Fig. 6.6(a)]. Let us make a partition such substance, Cm =  C  , is the heat capacity for n that volume is halved, each part [Fig. 6.6 (b)] now has one half of the original volume, one mole of the substance and is the quantity V of heat needed to raise the temperature of 2 , but the temperature will still remain the same i.e., T. It is clear that volume is an one mole by one degree celsius (or one extensive property and temperature is an kelvin). Specific heat, also called specific heat intensive property. capacity is the quantity of heat required to raise the temperature of one unit mass of a 2020-21

THERMODYNAMICS 169 substance by one degree celsius (or one i) at constant volume, q V kelvin). For finding out the heat, q, required to raise the temperatures of a sample, we ii) at constant pressure, qp multiply the specific heat of the substance, (a) ∆U Measurements For chemical reactions, heat absorbed at c, by the mass m, and temperatures change, constant volume, is measured in a bomb ∆T as calorimeter (Fig. 6.7). Here, a steel vessel (the bomb) is immersed in a water bath. The whole q = c × m × ∆T = C ∆T (6.11) device is called calorimeter. The steel vessel is immersed in water bath to ensure that no heat (d) The Relationship between Cp and CV is lost to the surroundings. A combustible for an Ideal Gas substance is burnt in pure dioxygen supplied At constant volume, the heat capacity, C is Fig. 6.7 Bomb calorimeter in the steel bomb. Heat evolved during the denoted by CV and at constant pressure, this reaction is transferred to the water around the is denoted by Cp . Let us find the relationship bomb and its temperature is monitored. Since between the two. the bomb calorimeter is sealed, its volume does not change i.e., the energy changes associated We can write equation for heat, q with reactions are measured at constant volume. Under these conditions, no work is at constant volume as qV = CV ∆T = ∆U done as the reaction is carried out at constant volume in the bomb calorimeter. Even for at constant pressure as qp = Cp ∆T = ∆H reactions involving gases, there is no work done as ∆V = 0. Temperature change of the The difference between Cp and CV can be calorimeter produced by the completed derived for an ideal gas as: reaction is then converted to q , by using the For a mole of an ideal gas, ∆H = ∆U + ∆(pV ) V = ∆U + ∆(RT ) known heat capacity of the calorimeter with the help of equation 6.11. = ∆U + R∆T ∴ ∆H = ∆U + R ∆T (6.12) On putting the values of ∆H and ∆U, we have Cp ∆T = CV ∆T + R∆T Cp = CV + R Cp – CV = R (6.13) 6.3 MEASUREMENT OF ∆U AND ∆H: CALORIMETRY We can measure energy changes associated with chemical or physical processes by an experimental technique called calorimetry. In calorimetry, the process is carried out in a vessel called calorimeter, which is immersed in a known volume of a liquid. Knowing the heat capacity of the liquid in which calorimeter is immersed and the heat capacity of calorimeter, it is possible to determine the heat evolved in the process by measuring temperature changes. Measurements are made under two different conditions: 2020-21

170 CHEMISTRY (b) ∆H Measurements of the bomb calorimeter is 20.7kJ/K, what is the enthalpy change for the above Measurement of heat change at constant pressure reaction at 298 K and 1 atm? Solution (generally under atmospheric pressure) can be Suppose q is the quantity of heat from the reaction mixture and CV is the heat capacity done in a calorimeter shown in Fig. 6.8. We know of the calorimeter, then the quantity of heat that ∆Η = qp (at constant p) and, therefore, heat absorbed by the calorimeter. absorbed or evolved, qp at constant pressure is q = CV × ∆T also called the heat of reaction or enthalpy of Quantity of heat from the reaction will have the same magnitude but opposite reaction, ∆rH. sign because the heat lost by the system (reaction mixture) is equal to the heat In an exothermic reaction, heat is evolved, gained by the calorimeter. and system loses heat to the surroundings. q = − CV × ∆T = − 20.7 kJ/K × (299 − 298) K Therefore, qp will be negative and ∆rH will also = − 20.7 kJ be negative. Similarly in an endothermic (Here, negative sign indicates the reaction, heat is absorbed, qp is positive and exothermic nature of the reaction) ∆rH will be positive. Thus, ∆U for the combustion of the 1g of graphite = – 20.7 kJK–1 For combustion of 1 mol of graphite, 12.0 g mol−1 × (−20.7 kJ) = 1g = – 2.48 ×102 kJ mol–1 , Since ∆ ng = 0, ∆ H = ∆ U = – 2.48 ×102 kJ mol–1 Fig. 6.8 Calorimeter for measuring heat changes 6.4 ENTHALPY CHANGE, ∆rH OF A at constant pressure (atmospheric REACTION – REACTION ENTHALPY pressure). In a chemical reaction, reactants are converted Problem 6.6 into products and is represented by, 1g of graphite is burnt in a bomb calorimeter in excess of oxygen at 298 K Reactants → Products and 1 atmospheric pressure according to the equation The enthalpy change accompanying a C (graphite) + O2 (g) → CO2 (g) reaction is called the reaction enthalpy. The During the reaction, temperature rises enthalpy change of a chemical reaction, is given from 298 K to 299 K. If the heat capacity by the symbol ∆rH ∆rH = (sum of enthalpies of products) – (sum of enthalpies of reactants) ∑ ∑= a Hi products − b Hi reactants (6.14) ii ∑Here symbol (sigma) is used for summation and ai and bi are the stoichiometric 2020-21

THERMODYNAMICS 171 coefficients of the products and reactants ethanol at 298 K is pure liquid ethanol at 1 bar; standard state of solid iron at 500 K is respectively in the balanced chemical pure iron at 1 bar. Usually data are taken at 298 K. equation. For example, for the reaction Standard conditions are denoted by adding (g) + 2O2 (g) → CO2 (g) + 2H2O (l) the superscript 0 to the symbol ∆H, e.g., ∆H0 = ai HPr − b Hi reac tan ts ∑ ∑CH4 oducts (b) Enthalpy Changes during Phase Transformations ∆r H ii Phase transformations also involve energy changes. Ice, for example, requires heat for = [Hm (CO2 ,g) + 2Hm (H2O, l)]– [Hm (CH4 , g) melting. Normally this melting takes place at constant pressure (atmospheric pressure) and + 2Hm (O2, g)] during phase change, temperature remains where Hm is the molar enthalpy. constant (at 273 K). H2O(s) → H2O(l); ∆fusH 0 = 6.00 kJ moI–1 Enthalpy change is a very useful quantity. Here ∆fusH 0 is enthalpy of fusion in standard state. If water freezes, then process is reversed Knowledge of this quantity is required when and equal amount of heat is given off to the surroundings. one needs to plan the heating or cooling The enthalpy change that accompanies required to maintain an industrial chemical melting of one mole of a solid substance in standard state is called standard reaction at constant temperature. It is also enthalpy of fusion or molar enthalpy of fusion, ∆fusH 0. required to calculate temperature dependence Melting of a solid is endothermic, so all of equilibrium constant. enthalpies of fusion are positive. Water requires (a) Standard Enthalpy of Reactions Enthalpy of a reaction depends on the conditions under which a reaction is carried out. It is, therefore, necessary that we must specify some standard conditions. The standard enthalpy of reaction is the enthalpy change for a reaction when all the participating substances are in their standard states. The standard state of a substance at a specified temperature is its pure form at 1 bar. For example, the standard state of liquid Table 6.1 Standard Enthalpy Changes of Fusion and Vaporisation (Tf and Tb are melting and boiling points, respectively) 2020-21

172 CHEMISTRY heat for evaporation. At constant temperature Solution of its boiling point Tb and at constant pressure: We can represent the process of evaporation as H2O(l) → H2O(g); ∆ H0 = + 40.79 kJ moI–1 ∆ H0 vap vap is the standard enthalpy of vaporisation. Amount of heat required to vaporize H2O(1) vaporisation → H2O(g) 1mol 1mol one mole of a liquid at constant No. of moles in 18 g H2O(l) is temperature and under standard pressure = 18 18g −1 = 1mol (1bar) is called its standard enthalpy of g mol vaporization or molar enthalpy of Heat supplied to evaporate18g water at vaporization, ∆vapH 0. 298 K = n × ∆ H 0 Sublimation is direct conversion of a solid vap into its vapour. Solid CO2 or ‘dry ice’ sublimes = (1 mol) × (44.01 kJ mol–1) at 195K with ∆subH 0=25.2 kJ mol–1; naphthalene sublimes slowly and for this = 44.01 kJ ∆sub H0 = 73.0 kJ mol–1 . (assuming steam behaving as an ideal Standard enthalpy of sublimation, gas). ∆subH 0 is the change in enthalpy when one ∆vapU = ∆vap H V − p∆V = ∆vap H V − ∆n g RT mole of a solid substance sublimes at a ∆vap H V − ∆ng RT = 44.01 kJ constant temperature and under standard − (1)(8.314 JK −1mol−1)(298K )(10−3 kJ J−1 ) pressure (1bar). ∆vapU V = 44.01kJ − 2.48 kJ = 41.53 kJ The magnitude of the enthalpy change depends on the strength of the intermolecular Problem 6.8 interactions in the substance undergoing the phase transfomations. For example, the strong Assuming the water vapour to be a perfect hydrogen bonds between water molecules hold gas, calculate the internal energy change them tightly in liquid phase. For an organic when 1 mol of water at 100°C and 1 bar liquid, such as acetone, the intermolecular pressure is converted to ice at 0°C. Given dipole-dipole interactions are significantly the enthalpy of fusion of ice is 6.00 kJ mol- weaker. Thus, it requires less heat to vaporise 1 heat capacity of water is 4.2 J/g°C 1 mol of acetone than it does to vaporize 1 mol of water. Table 6.1 gives values of standard The change take place as follows: enthalpy changes of fusion and vaporisation for some substances. Step - 1 1 mol H2O (l, 100°C) 1 mol (l, 0°C) Enthalpy change ∆H1 Problem 6.7 Step - 2 1 mol H2O (l, 0°C) 1 mol A swimmer coming out from a pool is H2O( S, 0°C) Enthalpy change ∆H2 covered with a film of water weighing about 18g. How much heat must be Total enthalpy change will be - supplied to evaporate this water at ∆H = ∆H1 + ∆H2 ∆H1 = - (18 x 4.2 x 100) J mol-1 298 K ? Calculate the internal energy of = - 7560 J mol-1 = - 7.56 k J mol-1 vaporisation at 298K. ∆ H 0 for water vap ∆H2 = - 6.00 kJ mol-1 at 298K= 44.01kJ mol–1 2020-21

THERMODYNAMICS 173 Table 6.2 Standard Molar Enthalpies of Formation (∆f H ) at 298K of a Few Selected Substances Therefore, aggregation (also known as reference ∆H = - 7.56 kJ mol-1 + (-6.00 kJ mol-1) = -13.56 kJ mol-1 states) is called Standard Molar Enthalpy of Formation. Its symbol is ∆f H 0, where There is negligible change in the volume the subscript ‘ f ’ indicates that one mole of during the change form liquid to solid state. the compound in question has been formed Therefore, p∆v = ∆ng RT = 0 in its standard state from its elements in their ∆H = ∆U = - 13.56kJ mol-1 most stable states of aggregation. The reference (c) Standard Enthalpy of Formation state of an element is its most stable state of The standard enthalpy change for the formation of one mole of a compound from aggregation at 25°C and 1 bar pressure. its elements in their most stable states of For example, the reference state of dihydrogen is H2 gas and those of dioxygen, carbon and sulphur are O2 gas, Cgraphite and Srhombic respectively. Some reactions with standard molar enthalpies of formation are as follows. 2020-21

174 CHEMISTRY H2(g) + ½O2 (g) → H2O(1); Here, we can make use of standard enthalpy of formation and calculate the enthalpy ∆f H0 = –285.8 kJ mol–1 change for the reaction. The following general equation can be used for the enthalpy change C (graphite, s) + 2H2(g) → Ch4 (g); calculation. ∆f H0 = –74.81 kJ mol–1 ∆rH0= ∑ ai∆fH0(products) – ∑ bi∆fH0(reactants) ii 2C (graphite, s)+3H2 (g)+ ½O2(g) → C2H5OH(1); (6.15) ∆ H0 = – 277.7kJ mol–1 where a and b represent the coefficients of the f products and reactants in the balanced equation. Let us apply the above equation for It is important to understand that a decomposition of calcium carbonate. Here, standard molar enthalpy of formation, ∆fH 0, coefficients ‘a’ and ‘b’ are 1 each. Therefore, is just a special case of ∆rH 0, where one mole ∆rH0 = ∆f H0 = [CaO(s)]+ ∆f H0 [CO2(g)] of a compound is formed from its constituent – ∆f H0 = [CaCO3(s)] elements, as in the above three equations, =1 (–635.1 kJ mol–1) + 1(–393.5 kJ mol–1) where 1 mol of each, water, methane and –1(–1206.9 kJ mol–1) ethanol is formed. In contrast, the enthalpy = 178.3 kJ mol–1 change for an exothermic reaction: Thus, the decomposition of CaCO3 (s) is an endothermic process and you have to heat it CaO(s) + CO2(g) → CaCo3(s); for getting the desired products. ∆rH0 = – 178.3kJ mol–1 (d) Thermochemical Equations is not an enthalpy of formation of calcium A balanced chemical equation together with the value of its ∆rH is called a thermochemical carbonate, since calcium carbonate has been equation. We specify the physical state (alongwith allotropic state) of the substance in formed from other compounds, and not from an equation. For example: C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l); its constituent elements. Also, for the reaction ∆rH0 = – 1367 kJ mol–1 given below, enthalpy change is not standard The above equation describes the enthalpy of formation, ∆fH0 for HBr(g). combustion of liquid ethanol at constant temperature and pressure. The negative sign H2(g) + Br2(l) → 2HBr(g); of enthalpy change indicates that this is an ∆ exothermic reaction. r H0 = – 178.3kJ mol–1 It would be necessary to remember the Here two moles, instead of one mole of the following conventions regarding thermo- chemical equations. product is formed from the elements, i.e., . 1. The coefficients in a balanced thermo- ∆ H0 = 2∆f H0 chemical equation refer to the number of r moles (never molecules) of reactants and products involved in the reaction. Therefore, by dividing all coefficients in the 2. The numerical value of ∆rH 0 refers to the balanced equation by 2, expression for number of moles of substances specified by an equation. Standard enthalpy change enthalpy of formation of HBr (g) is written as ∆rH0 will have units as kJ mol–1. ½H2(g) + ½Br2(1) → HBr(g); ∆ H0 = – 36.4 kJ mol–1 f Standard enthalpies of formation of some common substances are given in Table 6.2. By convention, standard enthalpy for formation, ∆ H 0, of an element in reference f state, i.e., its most stable state of aggregation is taken as zero. Suppose, you are a chemical engineer and want to know how much heat is required to decompose calcium carbonate to lime and carbon dioxide, with all the substances in their standard state. CaCO3(s) → CaO(s) + CO2(g); ∆r H0 = ? 2020-21

THERMODYNAMICS 175 To illustrate the concept, let us consider (e) Hess’s Law of Constant Heat the calculation of heat of reaction for the Summation following reaction : We know that enthalpy is a state function, Fe2O3 (s) + 3H2 (g) → 2Fe (s) + 3H2O (l), therefore the change in enthalpy is independent of the path between initial state From the Table (6.2) of standard enthalpy of (reactants) and final state (products). In other formation (∆f H 0), we find : words, enthalpy change for a reaction is the ∆f H 0 (H2O,l) = –285.83 kJ mol–1; same whether it occurs in one step or in a ∆f H 0 (Fe2O3,s) = – 824.2 kJ mol–1; series of steps. This may be stated as follows Also ∆f H 0 (Fe, s) = 0 and in the form of Hess’s Law. ∆f H 0 (H2, g) = 0 as per convention If a reaction takes place in several steps then its standard reaction enthalpy is the Then, sum of the standard enthalpies of the intermediate reactions into which the ∆ H 0 = 3(–285.83 kJ mol–1) overall reaction may be divided at the same f 1 temperature. – 1(– 824.2 kJ mol–1) = (–857.5 + 824.2) kJ mol–1 Let us understand the importance of this law with the help of an example. = –33.3 kJ mol–1 Note that the coefficients used in these Consider the enthalpy change for the calculations are pure numbers, which are reaction equal to the respective stoichiometric C (graphite,s) + 1 O2 (g) → CO (g); ∆ H 0 = ? coefficients. The unit for ∆rH 0 is 2 r kJ mol–1, which means per mole of reaction. Once we balance the chemical equation in a Although CO(g) is the major product, some particular way, as above, this defines the mole CO2 gas is always produced in this reaction. of reaction. If we had balanced the equation Therefore, we cannot measure enthalpy change differently, for example, for the above reaction directly. However, if we 1 Fe2O3 (s) + 3 H2 (g) → Fe (s) + 3 H2O (l) can find some other reactions involving related 2 2 2 species, it is possible to calculate the enthalpy then this amount of reaction would be one change for the above reaction. mole of reaction and ∆H 0 would be Let us consider the following reactions: r C (graphite,s) + O2 (g) → CO2 (g); ∆ 0 = 3 f H 2 2 (–285.83 kJ mol–1) ∆ H 0 = – 393.5 kJ mol–1 (i) r 1 CO (g) + 1 (g) → CO2 (g) – 2 (–824.2 kJ mol–1) 2 O2 = (– 428.7 + 412.1) kJ mol–1 ∆ H 0 = – 283.0 kJ mol–1 (ii) r = –16.6 kJ mol–1 = ½ ∆ H 0 We can combine the above two reactions r 1 in such a way so as to obtain the desired It shows that enthalpy is an extensive quantity. reaction. To get one mole of CO(g) on the right, 3. When a chemical equation is reversed, the value of ∆rH 0 is reversed in sign. For we reverse equation (ii). In this, heat is example absorbed instead of being released, so we N2(g) + 3H2 (g) → 2NH3 (g); change sign of ∆rH 0 value ∆ H 0 = – 91.8 kJ. mol–1 CO2 (g) → CO (g) + 1 r 2 O2 (g); 2NH3(g) → N2(g) + 3H2 (g); ∆ 0 r ∆ H 0 = + 91.8 kJ mol–1 H = + 283.0 kJ mol–1 (iii) r 2020-21

176 CHEMISTRY Adding equation (i) and (iii), we get the Similarly, combustion of glucose gives out desired equation, 2802.0 kJ/mol of heat, for which the overall 1 equation is : 2 C (graphite, s) + O2 (g) → CO (g); C6H12O6 (g) + 6O2(g) → 6CO2(g) + 6H2O(1); ∆ H 0 = – 2802.0 kJ mol–1 C for which ∆ H 0 = (– 393.5 + 283.0) r Our body also generates energy from food = – 110.5 kJ mol–1 by the same overall process as combustion, In general, if enthalpy of an overall reaction although the final products are produced after A→B along one route is ∆rH and ∆rH1, ∆rH2, ∆rH3..... representing enthalpies of reactions a series of complex bio-chemical reactions leading to same product, B along another involving enzymes. route,then we have Problem 6.9 ∆rH = ∆rH1 + ∆rH2 + ∆rH3 ... (6.16) The combustion of one mole of benzene It can be represented as: takes place at 298 K and 1 atm. After A ∆rH combustion, CO2(g) and H2O (1) are ∆H1 produced and 3267.0 kJ of heat is C B liberated. Calculate the standard ∆rH2 ∆rH3 enthalpy of formation, ∆ H 0 of benzene. D f Standard enthalpies of formation of CO2(g) and H2O(l) are –393.5 kJ mol–1 and – 285.83 kJ mol–1 respectively. 6.5 ENTHALPIES FOR DIFFERENT TYPES Solution OF REACTIONS The formation reaction of benezene is It is convenient to give name to enthalpies given by : specifying the types of reactions. 6C (graphite) + 3H2 (g) → C6H6 (l); (a) Standard Enthalpy of Combustion (symbol : ∆cH 0) ∆ H 0 = ? ... (i) f Combustion reactions are exothermic in nature. These are important in industry, The enthalpy of combustion of 1 mol of rocketry, and other walks of life. Standard enthalpy of combustion is defined as the benzene is : enthalpy change per mole (or per unit amount) of a substance, when it undergoes combustion C6H6 ( l ) + 15 O2 → 6CO2 (g) + 3H2O ( l ); and all the reactants and products being in 2 their standard states at the specified temperature. ∆ H 0 = – 3267 kJ mol–1... (ii) C The enthalpy of formation of 1 mol of CO2(g) : C (graphite) + O2 (g) → CO2 (g); Cooking gas in cylinders contains mostly ∆ H 0 = – 393.5 kJ mol–1... (iii) butane (C4H10). During complete combustion f of one mole of butane, 2658 kJ of heat is released. We can write the thermochemical The enthalpy of formation of 1 mol of reactions for this as: H2O(l) is : H2 (g) + 1 O2 (g) → H2O (l); 2 13 C4 H10 ( g ) + 2 O2 ( g ) → 4CO2 ( g ) + 5H2O(1); ∆ H 0 = – 285.83 kJ mol–1... (iv) C ∆C H 0 = – 2658.0 kJ mol–1 multiplying eqn. (iii) by 6 and eqn. (iv) by 3 we get: 2020-21

THERMODYNAMICS 177 6C (graphite) + 6O2 (g) → 6CO2 (g); In this case, the enthalpy of atomization is same as the enthalpy of sublimation. ∆ H 0 = – 2361 kJ mol–1 f (c) Bond Enthalpy (symbol: ∆bondH 0) 3H2 (g) + 3 O2 (g) → 3H2O (1); 2 Chemical reactions involve the breaking and making of chemical bonds. Energy is required ∆ H 0 = – 857.49 kJ mol–1 to break a bond and energy is released when f a bond is formed. It is possible to relate heat of reaction to changes in energy associated Summing up the above two equations : with breaking and making of chemical bonds. With reference to the enthalpy changes 6C ( graphite) + 3H2 ( g ) + 15 O2 ( g ) → 6CO2 (g) associated with chemical bonds, two different 2 terms are used in thermodynamics. + 3H2O (l); ∆f H 0 = – 3218.49 kJ mol–1... (v) (i) Bond dissociation enthalpy Reversing equation (ii); (ii) Mean bond enthalpy 6CO2 (g) + 3H2O (l) → C6H6 (l) + 15 O2 ; Let us discuss these terms with reference 2 to diatomic and polyatomic molecules. ∆ H 0 = – 3267.0 kJ mol–1... (vi) Diatomic Molecules: Consider the following f process in which the bonds in one mole of dihydrogen gas (H2) are broken: Adding equations (v) and (vi), we get H2(g) → 2H(g) ; ∆H–HH 0 = 435.0 kJ mol–1 6C (graphite) + 3H2 (g) → C6H6 (l); The enthalpy change involved in this process is the bond dissociation enthalpy of H–H bond. ∆f H 0 = – 48.51 kJ mol–1... (iv) The bond dissociation enthalpy is the change in enthalpy when one mole of covalent bonds (b) Enthalpy of Atomization of a gaseous covalent compound is broken to (symbol: ∆aH 0) form products in the gas phase. Consider the following example of atomization Note that it is the same as the enthalpy of of dihydrogen atomization of dihydrogen. This is true for all diatomic molecules. For example: H2(g) → 2H(g); ∆aH0 = 435.0 kJ mol–1 You can see that H atoms are formed by Cl2(g) → 2Cl(g) ; ∆Cl–ClH 0 = 242 kJ mol–1 breaking H–H bonds in dihydrogen. The enthalpy change in this process is known as O2(g) → 2O(g) ; ∆O=OH 0 = 428 kJ mol–1 enthalpy of atomization, ∆aH 0. It is the In the case of polyatomic molecules, bond enthalpy change on breaking one mole of bonds completely to obtain atoms in the gas dissociation enthalpy is different for different phase. bonds within the same molecule. In case of diatomic molecules, like Polyatomic Molecules: Let us now consider dihydrogen (given above), the enthalpy of a polyatomic molecule like methane, CH4. The atomization is also the bond dissociation overall thermochemical equation for its enthalpy. The other examples of enthalpy of atomization reaction is given below: atomization can be CH4(g) → C(g) + 4H(g); CH4(g) → C(g) + 4H(g); ∆aH0 = 1665 kJ mol–1 ∆ H 0 = 1665 kJ mol–1 Note that the products are only atoms of C a and H in gaseous phase. Now see the following reaction: In methane, all the four C – H bonds are Na(s) → Na(g) ; ∆aH0 = 108.4 kJ mol–1 identical in bond length and energy. However, the energies required to break the individual C – H bonds in each successive step differ : 2020-21

178 CHEMISTRY CH4(g) → CH3(g)+H(g);∆bond H0 = +427 kJ mol–1 given in Table 6.3. The reaction enthalpies are CH3(g) → CH2(g)+H(g);∆bond H0 = +439 kJ mol–1 CH2(g) → CH(g)+H(g);∆bond H0 = +452 kJ mol–1 very important quantities as these arise from CH(g) → C(g)+H(g);∆bond H0 = +347 kJ mol–1 the changes that accompany the breaking of Therefore, old bonds and formation of the new bonds. CH4(g) → C(g)+4H(g);∆a H0 = 1665 kJ mol–1 In such cases we use mean bond enthalpy We can predict enthalpy of a reaction in gas of C – H bond. For example in CH4, ∆C–HH 0 is calculated as: phase, if we know different bond enthalpies. ∆C–HH 0 = ¼ (∆a H0) = ¼ (1665 kJ mol–1) The standard enthalpy of reaction, ∆rH0 is related to bond enthalpies of the reactants and = 416 kJ mol–1 products in gas phase reactions as: We find that mean C–H bond enthalpy in methane is 416 kJ/mol. It has been found that ∑∆ H0 = bond enthalpiesreactants mean C–H bond enthalpies differ slightly from r compound to compound, as in ∑− bond enthalpiesproducts CH3CH2Cl,CH3NO2, etc, but it does not differ in a great deal*. Using Hess’s law, bond (6.17)** enthalpies can be calculated. Bond enthalpy values of some single and multiple bonds are This relationship is particularly more useful when the required values of ∆ H0 are f not available. The net enthalpy change of a reaction is the amount of energy required to break all the bonds in the reactant molecules minus the amount of energy required to break all the bonds in the product molecules. Remember that this relationship is Table 6.3(a) Some Mean Single Bond Enthalpies in kJ mol–1 at 298 K HC N O F Si P S Cl Br I 435.8 414 389 464 569 293 318 339 431 368 297 H 347 293 276 238 C 159 351 439 289 264 259 330 243 - N - 201 O 201 272 - 209 - 201 197 - F 289 213 Si 138 184 368 351 - 205 272 213 P 213 - S 155 540 490 327 255 218 209 Cl 192 180 Br 176 213 226 360 I 151 213 230 331 213 251 243 Table 6.3(b) Some Mean Multiple Bond Enthalpies in kJ mol–1 at 298 K N=N 418 C=C 611 O = O 498 N≡N 946 C≡ C 837 615 741 C=N 891 C=O 1070 C≡N C≡O * Note that symbol used for bond dissociation enthalpy and mean bond enthalpy is the same. ** If we use enthalpy of bond formation, (∆f H 0 d), which is the enthalpy change when one mole of a particular type of bon bond is formed from gaseous atom, then ∆ H 0 = ∑ ∆ H – ∑ ∆ H0 0 f f bonds of productsf bonds of reactants 2020-21

THERMODYNAMICS 179 approximate and is valid when all substances 2. Na(g) → Na+ (g) + e−1(g) , the ionization of (reactants and products) in the reaction are in gaseous state. sodium atoms, ionization enthalpy ∆iH0 = 496 kJ mol–1 (d) Lattice Enthalpy 3. 1 Cl2 (g ) → Cl(g) , the dissociation of The lattice enthalpy of an ionic compound is 2 the enthalpy change which occurs when one mole of an ionic compound dissociates into its chlorine, the reaction enthalpy is half the ions in gaseous state. bond dissociation enthalpy. Na+Cl− (s) → Na+(g) + Cl− (g); 1 ∆bondH0 = 121 kJ mol–1 ∆latticeH 0 = +788 kJ mol–1 2 Since it is impossible to determine lattice enthalpies directly by experiment, we use an 4. Cl(g) + e−1(g) → Cl(g) electron gained by indirect method where we construct an enthalpy diagram called a Born-Haber Cycle chlorine atoms. The electron gain enthalpy, (Fig. 6.9). ∆egH 0 = – 348.6 kJ mol–1. You have learnt about ionization enthalpy Let us now calculate the lattice enthalpy and electron gain enthalpy in Unit 3. In of Na+Cl–(s) by following steps given below : fact, these terms have been taken from 1. Na(s) → Na(g) , sublimation of sodium thermodynamics. Earlier terms, ionization energy and electron affinity were in metal, ∆subH 0 = 108.4 kJ mol–1 practice in place of the above terms (see the box for justification). Ionization Energy and Electron Affinity Ionization energy and electron affinity are defined at absolute zero. At any other temperature, heat capacities for the reactants and the products have to be taken into account. Enthalpies of reactions for M(g) → M+(g) + e– (for ionization) M(g) + e– → M–(g) (for electron gain) at temperature, T is T ∫∆ H r 0(T ) = ∆H 0(0) + ∆ C0dT r rP 0 The value of Cp for each species in the above reaction is 5/2 R (CV = 3/2R) ∆∆rrCCpp00 So, = + 5/2 R (for ionization) = – 5/2 R (for electron gain) Therefore, ∆rH0 (ionization enthalpy) = E0 (ionization energy) + 5/2 RT ∆rH0 (electron gain enthalpy) = – A( electron affinity) – 5/2 RT Fig. 6.9 Enthalpy diagram for lattice enthalpy 5. Na+ (g) + Cl− (g) → Na+Cl−(s) of NaCl The sequence of steps is shown in Fig. 6.9, and is known as a Born-Haber cycle. The 2020-21

180 CHEMISTRY importance of the cycle is that, the sum of The enthalpy of solution of AB(s), ∆solH0, in the enthalpy changes round a cycle is water is, therefore, determined by the selective zero. Applying Hess’s law, we get, values of the lattice enthalpy,∆latticeH0 and enthalpy of hydration of ions, ∆hydH0 as ∆latticeH0 = 411.2 + 108.4 + 121 + 496 – 348.6 ∆sol H 0 = ∆latticeH 0 + ∆hydH 0 ∆latticeH0 = + 788kJ For most of the ionic compounds, ∆sol H0 is positive and the dissociation process is for NaCl(s) d Na+(g) + Cl–(g) endothermic. Therefore the solubility of most salts in water increases with rise of Internal energy is smaller by 2RT ( because temperature. If the lattice enthalpy is very ∆ng = 2) and is equal to + 783 kJ mol–1. high, the dissolution of the compound may not take place at all. Why do many fluorides tend Now we use the value of lattice enthalpy to to be less soluble than the corresponding calculate enthalpy of solution from the chlorides? Estimates of the magnitudes of expression: enthalpy changes may be made by using tables of bond energies (enthalpies) and lattice ∆solH0 = ∆latticeH0 + ∆hydH0 energies (enthalpies). For one mole of NaCl(s), (f) Enthalpy of Dilution lattice enthalpy = + 788 kJ mol–1 It is known that enthalpy of solution is the and ∆hydH 0 = – 784 kJ mol–1( from the enthalpy change associated with the addition of a specified amount of solute to the specified ∆ H0 literature) amount of solvent at a constant temperature sol and pressure. This argument can be applied = + 788 kJ mol–1 – 784 kJ mol–1 to any solvent with slight modification. Enthalpy change for dissolving one mole of = + 4 kJ mol–1 gaseous hydrogen chloride in 10 mol of water can be represented by the following equation. The dissolution of NaCl(s) is accompanied For convenience we will use the symbol aq. for water by very little heat change. HCl(g) + 10 aq. → HCl.10 aq. (e) Enthalpy of Solution (symbol : ∆solH0 ) ∆H = –69.01 kJ / mol Enthalpy of solution of a substance is the Let us consider the following set of enthalpy enthalpy change when one mole of it dissolves changes: in a specified amount of solvent. The enthalpy of solution at infinite dilution is the enthalpy (S-1) HCl(g) + 25 aq. → HCl.25 aq. change observed on dissolving the substance ∆H = –72.03 kJ / mol in an infinite amount of solvent when the interactions between the ions (or solute (S-2) HCl(g) + 40 aq. → HCl.40 aq. molecules) are negligible. ∆H = –72.79 kJ / mol When an ionic compound dissolves in a (S-3) HCl(g) + ∞ aq. → HCl. ∞ aq. solvent, the ions leave their ordered positions on ∆H = –74.85 kJ / mol the crystal lattice. These are now more free in solution. But solvation of these ions (hydration The values of ∆H show general dependence in case solvent is water) also occurs at the same of the enthalpy of solution on amount of solvent. time. This is shown diagrammatically, for an As more and more solvent is used, the enthalpy ionic compound, AB (s) of solution approaches a limiting value, i.e, the value in infinitely dilute solution. For hydrochloric acid this value of ∆H is given above in equation (S-3). 2020-21

THERMODYNAMICS 181 If we subtract the first equation (equation between them, it is at an extremely slow rate. S-1) from the second equation (equation S-2) It is still called spontaneous reaction. So in the above set of equations, we obtain– spontaneity means ‘having the potential to proceed without the assistance of external HCl.25 aq. + 15 aq. → HCl.40 aq. agency’. However, it does not tell about the ∆H = [ –72.79 – (–72.03)] kJ / mol rate of the reaction or process. Another aspect of spontaneous reaction or process, as we see = – 0.76 kJ / mol is that these cannot reverse their direction on This value (–0.76kJ/mol) of ∆H is enthalpy their own. We may summarise it as follows: of dilution. It is the heat withdrawn from the surroundings when additional solvent is A spontaneous process is an added to the solution. The enthalpy of dilution irreversible process and may only be of a solution is dependent on the original reversed by some external agency. concentration of the solution and the amount (a) Is Decrease in Enthalpy a Criterion of solvent added. for Spontaneity ? 6.6 SPONTANEITY If we examine the phenomenon like flow of water down hill or fall of a stone on to the The first law of thermodynamics tells us about ground, we find that there is a net decrease in the relationship between the heat absorbed potential energy in the direction of change. By and the work performed on or by a system. It analogy, we may be tempted to state that a puts no restrictions on the direction of heat chemical reaction is spontaneous in a given flow. However, the flow of heat is unidirectional direction, because decrease in energy has from higher temperature to lower taken place, as in the case of exothermic temperature. In fact, all naturally occurring reactions. For example: processes whether chemical or physical will tend to proceed spontaneously in one 13 direction only. For example, a gas expanding 2 N2(g) + 2 H2(g) = NH3(g) ; to fill the available volume, burning carbon in dioxygen giving carbon dioxide. ∆r H0 = – 46.1 kJ mol–1 But heat will not flow from colder body to 11 warmer body on its own, the gas in a container 2 H2(g) + 2 Cl2(g) = HCl (g) ; will not spontaneously contract into one corner or carbon dioxide will not form carbon ∆r H0 = – 92.32 kJ mol–1 and dioxygen spontaneously. These and many other spontaneously occurring changes show H2(g) + 1 O2(g) → H2O(l) ; unidirectional change. We may ask ‘what is the 2 driving force of spontaneously occurring changes ? What determines the direction of a ∆r H0 = –285.8 kJ mol–1 spontaneous change ? In this section, we shall The decrease in enthalpy in passing from establish some criterion for these processes whether these will take place or not. reactants to products may be shown for any Let us first understand what do we mean exothermic reaction on an enthalpy diagram by spontaneous reaction or change ? You may think by your common observation that as shown in Fig. 6.10(a). spontaneous reaction is one which occurs immediately when contact is made between Thus, the postulate that driving force for a the reactants. Take the case of combination of chemical reaction may be due to decrease in hydrogen and oxygen. These gases may be energy sounds ‘reasonable’ as the basis of mixed at room temperature and left for many evidence so far ! years without observing any perceptible change. Although the reaction is taking place 1 Now let us examine the following reactions: 2 N2(g) + O2(g) → NO2(g); = +33.2 kJ mol–1 ∆r H0 C(graphite, s) + 2 S(l) → CS2(l); ∆r H0 = +128.5 kJ mol–1 2020-21

182 CHEMISTRY Fig. 6.10 (a) Enthalpy diagram for exothermic reactions These reactions though endothermic, are spontaneous. The increase in enthalpy may be represented on an enthalpy diagram as shown in Fig. 6.10(b). Fig. 6.11 Diffusion of two gases Fig. 6.10 (b) Enthalpy diagram for endothermic respectively and separated by a movable reactions partition [Fig. 6.11 (a)]. When the partition is withdrawn [Fig.6.11( b)], the gases begin to Therefore, it becomes obvious that while diffuse into each other and after a period of decrease in enthalpy may be a contributory time, diffusion will be complete. factor for spontaneity, but it is not true for all cases. Let us examine the process. Before partition, if we were to pick up the gas (b) Entropy and Spontaneity molecules from left container, we would be sure that these will be molecules of gas A and Then, what drives the spontaneous process in similarly if we were to pick up the gas a given direction ? Let us examine such a case molecules from right container, we would be in which ∆H = 0 i.e., there is no change in sure that these will be molecules of gas B. But, enthalpy, but still the process is spontaneous. if we were to pick up molecules from container when partition is removed, we are not sure Let us consider diffusion of two gases into whether the molecules picked are of gas A or each other in a closed container which is gas B. We say that the system has become less isolated from the surroundings as shown in predictable or more chaotic. Fig. 6.11. We may now formulate another postulate: The two gases, say, gas A and gas B are in an isolated system, there is always a represented by black dots and white dots tendency for the systems’ energy to become more disordered or chaotic and this could be a criterion for spontaneous change ! At this point, we introduce another thermodynamic function, entropy denoted as S. The above mentioned disorder is the manifestation of entropy. To form a mental 2020-21

THERMODYNAMICS 183 picture, one can think of entropy as a measure ∆S = qrev (6.18) of the degree of randomness or disorder in the T system. The greater the disorder in an isolated system, the higher is the entropy. As far as a The total entropy change ( ∆Stotal) for the chemical reaction is concerned, this entropy system and surroundings of a spontaneous change can be attributed to rearrangement of atoms or ions from one pattern in the reactants process is given by to another (in the products). If the structure of the products is very much disordered than ∆Stotal = ∆Ssystem + ∆Ssurr > 0 (6.19) that of the reactants, there will be a resultant increase in entropy. The change in entropy When a system is in equilibrium, the accompanying a chemical reaction may be estimated qualitatively by a consideration of entropy is maximum, and the change in the structures of the species taking part in the entropy, ∆S = 0. reaction. Decrease of regularity in structure would mean increase in entropy. For a given We can say that entropy for a spontaneous substance, the crystalline solid state is the process increases till it reaches maximum and state of lowest entropy (most ordered), The at equilibrium the change in entropy is zero. gaseous state is state of highest entropy. Since entropy is a state property, we can calculate the change in entropy of a reversible Now let us try to quantify entropy. One way process by to calculate the degree of disorder or chaotic distribution of energy among molecules would ∆Ssys = qsys,rev be through statistical method which is beyond T the scope of this treatment. Other way would be to relate this process to the heat involved in We find that both for reversible and a process which would make entropy a thermodynamic concept. Entropy, like any irreversible expansion for an ideal gas, under other thermodynamic property such as isothermal conditions, ∆U = 0, but ∆Stotal i.e., internal energy U and enthalpy H is a state ∆Ssys + ∆Ssurr is not zero for irreversible function and ∆S is independent of path. process. Thus, ∆U does not discriminate Whenever heat is added to the system, it between reversible and irreversible process, increases molecular motions causing whereas ∆S does. increased randomness in the system. Thus heat (q) has randomising influence on the Problem 6.10 system. Can we then equate ∆S with q ? Wait ! Experience suggests us that the distribution Predict in which of the following, entropy of heat also depends on the temperature at increases/decreases : which heat is added to the system. A system at higher temperature has greater randomness (i) A liquid crystallizes into a solid. in it than one at lower temperature. Thus, temperature is the measure of average (ii) Temperature of a crystalline solid is chaotic motion of particles in the system. raised from 0 K to 115 K. Heat added to a system at lower temperature causes greater randomness than when the (iii) 2NaHCO3 (s) → Na2CO3 (s) + same quantity of heat is added to it at higher CO2 (g) + H2O (g) temperature. This suggests that the entropy change is inversely proportional to the (iv) H2 (g) → 2H (g) temperature. ∆S is related with q and T for a reversible reaction as : Solution (i) After freezing, the molecules attain an ordered state and therefore, entropy decreases. (ii) At 0 K, the contituent particles are static and entropy is minimum. If temperature is raised to 115 K, these 2020-21

184 CHEMISTRY begin to move and oscillate about their = 4980.6 JK–1 mol–1 equilibrium positions in the lattice and This shows that the above reaction is system becomes more disordered, spontaneous. therefore entropy increases. (iii) Reactant, NaHCO3 is a solid and it (c) Gibbs Energy and Spontaneity has low entropy. Among products there are one solid and two gases. We have seen that for a system, it is the total Therefore, the products represent a entropy change, ∆Stotal which decides the condition of higher entropy. spontaneity of the process. But most of the chemical reactions fall into the category of (iv) Here one molecule gives two atoms i.e., number of particles increases either closed systems or open systems. leading to more disordered state. Therefore, for most of the chemical reactions Two moles of H atoms have higher entropy than one mole of dihydrogen there are changes in both enthalpy and molecule. entropy. It is clear from the discussion in Problem 6.11 previous sections that neither decrease in enthalpy nor increase in entropy alone can For oxidation of iron, determine the direction of spontaneous change for these systems. 4Fe(s) + 3O2 (g) → 2Fe2O3 (s) For this purpose, we define a new thermodynamic function the Gibbs energy or entropy change is – 549.4 JK–1mol–1at Gibbs function, G, as 298 K. Inspite of negative entropy change of this reaction, why is the reaction G = H – TS (6.20) spontaneous? Gibbs function, G is an extensive property (∆rH 0for and a state function. this reaction is The change in Gibbs energy for the system, ∆Gsys can be written as –1648 × 103 J mol–1) ∆Gsys = ∆Hsys − T ∆Ssys − Ssys ∆T Solution One decides the spontaneity of a reaction At constant temperature, ∆T = 0 by considering ( )∆Stotal ∆Ssys + ∆Ssurr . For calculating ∴ ∆Gsys = ∆Hsys − T ∆Ssys ∆S ,surr we have to consider the heat Usually the subscript ‘system’ is dropped absorbed by the surroundings which is and we simply write this equation as equal to – ∆rH0. At temperature T, entropy change of the surroundings is ∆G = ∆H − T ∆S (6.21) ( )= − −1648 ×103 J mol−1 Thus, Gibbs energy change = enthalpy 298 K change – temperature × entropy change, and is referred to as the Gibbs equation, one of the = 5530 JK–1mol–1 most important equations in chemistry. Here, Thus, total entropy change for this we have considered both terms together for reaction spontaneity: energy (in terms of ∆H) and entropy (∆S, a measure of disorder) as ∆r Stotal = 5530JK –1mol–1 + indicated earlier. Dimensionally if we analyse, we find that ∆G has units of energy because, ( )−549.4JK –1mol–1 both ∆H and the T∆S are energy terms, since T∆S = (K) (J/K) = J. Now let us consider how ∆G is related to reaction spontaneity. 2020-21

THERMODYNAMICS 185 We know, can be ‘small’ in which case T must be large. ∆Stotal = ∆Ssys + ∆Ssurr (b) The positive entropy change of the system can be ’large’, in which case T may be small. If the system is in thermal equilibrium with The former is one of the reasons why reactions the surrounding, then the temperature of the are often carried out at high temperature. surrounding is same as that of the system. Table 6.4 summarises the effect of temperature Also, increase in enthalpy of the surrounding on spontaneity of reactions. is equal to decrease in the enthalpy of the system. (d) Entropy and Second Law of Thermodynamics Therefore, entropy change of surroundings, We know that for an isolated system the change in energy remains constant. Therefore, ∆Ssurr = ∆Hsurr = − ∆Hsys increase in entropy in such systems is the T T natural direction of a spontaneous change. This, in fact is the second law of ∆Stotal = ∆Ssys +  − ∆H sys  thermodynamics. Like first law of  T  thermodynamics, second law can also be stated in several ways. The second law of Rearranging the above equation: thermodynamics explains why spontaneous T∆Stotal = T∆Ssys – ∆Hsys exothermic reactions are so common. In For spontaneous process, ∆Stotal > 0 , so exothermic reactions heat released by the reaction increases the disorder of the T∆Ssys – ∆Hsys > Ο surroundings and overall entropy change is positive which makes the reaction ( )⇒ − ∆Hsys − T ∆Ssys > 0 spontaneous. Using equation 6.21, the above equation can (e) Absolute Entropy and Third Law of be written as Thermodynamics –∆G > O Molecules of a substance may move in a ∆G = ∆H − T ∆S < 0 (6.22) straight line in any direction, they may spin like a top and the bonds in the molecules may dT∆oS∆usyHss esiysfsutilshetwheoenrekenr.gthSyaowlph∆yiGcchhiasisntgnheoetofanaveatrieleaancbtelierogntyo, stretch and compress. These motions of the available to do useful work and is thus a molecule are called translational, rotational measure of the ‘free energy’. For this reason, it and vibrational motion respectively. When temperature of the system rises, these motions is also known as the free energy of the reaction. become more vigorous and entropy increases. On the other hand when temperature is ∆G gives a criteria for spontaneity at lowered, the entropy decreases. The entropy constant pressure and temperature. of any pure crystalline substance approaches zero as the temperature (i) If ∆G is negative (< 0), the process is approaches absolute zero. This is called spontaneous. third law of thermodynamics. This is so because there is perfect order in a crystal at (ii) If ∆G is positive (> 0), the process is non absolute zero. The statement is confined to spontaneous. pure crystalline solids because theoretical arguments and practical evidences have Note : If a reaction has a positive enthalpy shown that entropy of solutions and super change and positive entropy change, it can be cooled liquids is not zero at 0 K. The spontaneous when T∆S is large enough to importance of the third law lies in the fact that outweigh ∆H. This can happen in two ways; (a) The positive entropy change of the system 2020-21

186 CHEMISTRY it permits the calculation of absolute values proceed with a decrease in free energy, which of entropy of pure substance from thermal seems impossible. It is possible only if at data alone. For a pure substance, this can be equilibrium the free energy of the system is done by summing qrev increments from 0 K minimum. If it is not, the system would spontaneously change to configuration of T lower free energy. to 298 K. Standard entropies can be used to calculate standard entropy changes by a So, the criterion for equilibrium Hess’s law type of calculation. A + B l C + D ; is 6.7 GIBBS ENERGY CHANGE AND EQUILIBRIUM ∆rG = 0 Gibbs energy for a reaction in which all We have seen how a knowledge of the sign and magnitude of the free energy change of a reactants and products are in standard state, chemical reaction allows: ∆rG0 is related to the equilibrium constant of the reaction as follows: (i) Prediction of the spontaneity of the chemical reaction. 0 = ∆rG0 + RT ln K (6.23) or ∆rG0 = – RT ln K (ii) Prediction of the useful work that could or ∆rG0 = – 2.303 RT log K be extracted from it. We also know that So far we have considered free energy changes in irreversible reactions. Let us now (6.24) examine the free energy changes in reversible reactions. For strongly endothermic reactions, the ‘Reversible’ under strict thermodynamic value of ∆rH0 may be large and positive. In sense is a special way of carrying out a such a case, value of K will be much smaller process such that system is at all times in perfect equilibrium with its surroundings. than 1 and the reaction is unlikely to form When applied to a chemical reaction, the term ‘reversible’ indicates that a given much product. In case of exothermic reaction can proceed in either direction simultaneously, so that a dynamic reactions, ∆ H0 is large and negative, and ∆ G0 equilibrium is set up. This means that the r r reactions in both the directions should is likely to be large and negative too. In such cases, K will be much larger than 1. We may expect strongly exothermic reactions to have a large K, and hence can go to near completion. ∆rG0 also depends upon ∆rS0, if the changes in the entropy of reaction is also taken into account, the value of K or extent of chemical reaction will also be affected, Table 6.4 Effect of Temperature on Spontaneity of Reactions ∆rH0 ∆rS0 ∆rG0 Description* – +– Reaction spontaneous at all temperatures – – – (at low T ) Reaction spontaneous at low temperature – – + (at high T ) Reaction nonspontaneous at high temperature + + + (at low T ) Reaction nonspontaneous at low temperature + + – (at high T ) Reaction spontaneous at high temperature + – + (at all T ) Reaction nonspontaneous at all temperatures * The term low temperature and high temperature are relative. For a particular reaction, high temperature could even mean room temperature. 2020-21

THERMODYNAMICS 187 depending upon whether ∆rS0 is positive or ( )–13.6 × 103 J mol–1 negative. ( )= 2.303 8.314 JK –1 mol–1 (298 K ) Using equation (6.24), = 2.38 (i) It is possible to obtain an estimate of ∆G0 Hence K = antilog 2.38 = 2.4 × 102. from the measurement of ∆HV and ∆S0, Problem 6.14 and then calculate K at any temperature for economic yields of the products. At 60°C, dinitrogen tetroxide is 50 (ii) If K is measured directly in the per cent dissociated. Calculate the laboratory, value of ∆G0 at any other standard free energy change at this temperature can be calculated. temperature and at one atmosphere. Using equation (6.24), Problem 6.12 Solution 2NO2(g) N2O4(g) Calculate ∆rG 0 for conversion of oxygen If N2O4 is 50% dissociated, the mole to ozone, 3/2 O2(g) → O3(g) at 298 K. if Kp fraction of both the substances is given for this conversion is 2.47 × 10–29. by Solution x = 1 − 0.5 = 2 × 0.5 N 2 O4 1 + 0.5 : x NO2 1 + 0.5 We know ∆rG0 = – 2.303 RT log Kp and R = 8.314 JK–1 mol–1 p = 0.5 × 1 atm, p = N2O4 1.5 NO2 Therefore, ∆rG0 = – 2.303 (8.314 J K–1 mol–1) 1 × 1 atm. 1.5 × (298 K) (log 2.47 × 10–29) = 163000 J mol–1 = 163 kJ mol–1. The equilibrium constant Kp is given by Problem 6.13 ( )Kp = 2 Find out the value of equilibrium constant pNO2 = 1.5 for the following reaction at 298 K. pN2O4 (1.5)2 (0.5) Standard Gibbs energy change, ∆rG0 at = 1.33 atm. the given temperature is –13.6 kJ mol–1. Solution Since ∆rG0 = –RT ln Kp We know, log K = ∆rG0 = (– 8.314 JK–1 mol–1) × (333 K) × (2.303) × (0.1239) = – 763.8 kJ mol–1 2020-21

188 CHEMISTRY SUMMARY Thermodynamics deals with energy changes in chemical or physical processes and enables us to study these changes quantitatively and to make useful predictions. For these purposes, we divide the universe into the system and the surroundings. Chemical or physical processes lead to evolution or absorption of heat (q), part of which may be converted into work (w). These quantities are related through the first law of thermodynamics via ∆U = q + w. ∆U, change in internal energy, depends on initial and final states only and is a state function, whereas q and w depend on the path and are not the state functions. We follow sign conventions of q and w by giving the positive sign to these quantities when these are added to the system. We can measure the transfer of heat from one system to another which causes the change in temperature. The magnitude of rise in temperature depends on the heat capacity (C) of a substance. Therefore, heat absorbed or evolved is q = C∆T. Work can be measured by w = –pex∆V, in case of expansion of gases. Under reversible process, we can put p = p for infinitesimal changes in the volume ex making wrev = – p dV. In this condition, we can use gas equation, pV = nRT. At constant volume, w = 0, then ∆U = qV , heat transfer at constant volume. But in study of chemical reactions, we usually have constant pressure. We define another state function enthalpy. Enthalpy change, ∆H = ∆U + ∆ngRT, can be found directly from the heat changes at constant pressure, ∆H = qp. There are varieties of enthalpy changes. Changes of phase such as melting, vaporization and sublimation usually occur at constant temperature and can be characterized by enthalpy changes which are always positive. Enthalpy of formation, combustion and other enthalpy changes can be calculated using Hess’s law. Enthalpy change for chemical reactions can be determined by ∑( ) ∑( )∆rH = ai ∆ f H products − bi ∆ f H reactions fi and in gaseous state by ∆rH0 = Σ bond enthalpies of the reactants – Σ bond enthalpies of the products First law of thermodynamics does not guide us about the direction of chemical reactions i.e., what is the driving force of a chemical reaction. For isolated systems, ∆U = 0. We define another state function, S, entropy for this purpose. Entropy is a measure of disorder or randomness. For a spontaneous change, total entropy change is positive. Therefore, for an isolated system, ∆U = 0, ∆S > 0, so entropy change distinguishes a spontaneous change, while energy change does not. Entropy changes can be measured by the equation ∆S = qrev for a reversible process. q rev is independent of path. T T Chemical reactions are generally carried at constant pressure, so we define another state function Gibbs energy, G, which is related to entropy and enthalpy changes of the system by the equation: ∆rG = ∆rH – T ∆rS For a spontaneous change, ∆Gsys < 0 and at equilibrium, ∆Gsys = 0. Standard Gibbs energy change is related to equilibrium constant by ∆rG0 = – RT ln K. K can be calculated from this equation, if we know ∆rG0 which can be found from . Temperature is an important factor in the equation. Many reactions which are non-spontaneous at low temperature, are made spontaneous at high temperature for systems having positive entropy of reaction. 2020-21

THERMODYNAMICS 189 EXERCISES 6.1 Choose the correct answer. A thermodynamic state function is a quantity (i) used to determine heat changes (ii) whose value is independent of path (iii) used to determine pressure volume work (iv) whose value depends on temperature only. 6.2 For the process to occur under adiabatic conditions, the correct condition is: (i) ∆T = 0 (ii) ∆p = 0 (iii) q = 0 (iv) w = 0 6.3 The enthalpies of all elements in their standard states are: (i) unity (ii) zero (iii) < 0 (iv) different for each element 6.4 ∆U 0of combustion of methane is – X kJ mol–1. The value of ∆H0 is (i) = ∆U 0 (ii) > ∆U 0 (iii) < ∆U 0 (iv) = 0 6.5 The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, –890.3 kJ mol–1 –393.5 kJ mol–1, and –285.8 kJ mol–1 respectively. Enthalpy of formation of CH4(g) will be (i) –74.8 kJ mol–1 (ii) –52.27 kJ mol–1 (iii) +74.8 kJ mol–1 (iv) +52.26 kJ mol–1. 6.6 A reaction, A + B → C + D + q is found to have a positive entropy change. The reaction will be (i) possible at high temperature (ii) possible only at low temperature (iii) not possible at any temperature (v) possible at any temperature 6.7 In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy for the process? 6.8 Tcahleorriemaectteior,naonfdc∆yaUnwaamsidfoeu, nNdHt2oCbNe(s–7),4w2i.t7hkdJiomxoygl–e1 natw2a9s8cKa.rrCieadlcuoulatteinenatbhoamlpby change for the reaction at 298 K. NH2CN(g) + 3 O2(g) → N2(g) + CO2(g) + H2O(l) 2 2020-21

190 CHEMISTRY 6.9 Calculate the number of kJ of heat necessary to raise the temperature of 6.10 60.0 g of aluminium from 35°C to 55°C. Molar heat capacity of Al is 24 J mol–1 K–1. 6.11 6.12 Calculate the enthalpy change on freezing of 1.0 mol of water at10.0°C to ice at 6.13 –10.0°C. ∆fusH = 6.03 kJ mol–1 at 0°C. 6.14 Cp [H2O(l)] = 75.3 J mol–1 K–1 6.15 Cp [H2O(s)] = 36.8 J mol–1 K–1 6.16 6.17 Enthalpy of combustion of carbon to CO2 is –393.5 kJ mol–1. Calculate the heat released upon formation of 35.2 g of CO2 from carbon and dioxygen gas. 6.18 6.19 Enthalpies omf ofol–r1mraestipoenctoifveClyO.(gF)i,nCdOt2h(ge),vNal2uOe(go)fa∆nrdH N2O4(g) are –110, – 393, 81 and 9.7 kJ for the reaction: N2O4(g) + 3CO(g) → N2O(g) + 3CO2(g) Given N2(g) + 3H2(g) → 2NH3(g) ; ∆rH0 = –92.4 kJ mol–1 What is the standard enthalpy of formation of NH3 gas? Calculate the standard enthalpy of formation of CH3OH(l) from the following data: CH3OH (l) + 3 O2(g) → CO2(g) + 2H2O(l) ; ∆ H0 = –726 kJ mol–1 2 r C(graphite) + O2(g) → CO2(g) ; ∆cH0 = –393 kJ mol–1 H2(g) + 1 O2(g) → H2O(l) ; ∆ H0 = –286 kJ mol–1. 2 f Calculate the enthalpy change for the process CCl4(g) → C(g) + 4 Cl(g) and calculate bond enthalpy of C – Cl in CCl4(g). ∆vapH0(CCl4) = 30.5 kJ mol–1. ∆fH0 (CCl4) = –135.5 kJ mol–1. ∆aH0 (C) = 715.0 kJ mol–1 , where ∆aH0 is enthalpy of atomisation ∆aH0 (Cl2) = 242 kJ mol–1 For an isolated system, ∆U = 0, what will be ∆S ? For the reaction at 298 K, 2A + B → C ∆H = 400 kJ mol–1 and ∆S = 0.2 kJ K–1 mol–1 At what temperature will the reaction become spontaneous considering ∆H and ∆S to be constant over the temperature range. For the reaction, 2 Cl(g) → Cl2(g), what are the signs of ∆H and ∆S ? For the reaction 2 A(g) + B(g) → 2D(g) ∆U 0 = –10.5 kJ and ∆S0 = –44.1 JK–1. Calculate ∆G0 for the reaction, and predict whether the reaction may occur spontaneously. 2020-21

THERMODYNAMICS 191 6.20 The equilibrium constant for a reaction is 10. What will be the value of ∆G0 ? 6.21 R = 8.314 JK–1 mol–1, T = 300 K. 6.22 Comment on the thermodynamic stability of NO(g), given 1 N2(g) + 1 O2(g) → NO(g) ; ∆rH0 = 90 kJ mol–1 2 2 NO(g) + 1 O2(g) → NO2(g) : ∆rH0= –74 kJ mol–1 2 Calculate the entropy change in surroundings when 1.00 mol of H2O(l) is formed under standard conditions. ∆f H0 = –286 kJ mol–1. 2020-21

192 CHEMISTRY EQUILIBRIUM UNIT 7 After studying this unit you will be Chemical equilibria are important in numerous biological and environmental processes. For example, equilibria able to involving O2 molecules and the protein hemoglobin play a crucial role in the transport and delivery of O2 from our • identify dynamic nature of lungs to our muscles. Similar equilibria involving CO molecules and hemoglobin account for the toxicity of CO. equilibrium involved in physical When a liquid evaporates in a closed container, and chemical processes; molecules with relatively higher kinetic energy escape the • state the law of equilibrium; liquid surface into the vapour phase and number of liquid molecules from the vapour phase strike the liquid surface • explain characteristics of and are retained in the liquid phase. It gives rise to a constant vapour pressure because of an equilibrium in which the equilibria involved in physical number of molecules leaving the liquid equals the number returning to liquid from the vapour. We say that the system and chemical processes; for has reached equilibrium state at this stage. However, this • write expressions is not static equilibrium and there is a lot of activity at the boundary between the liquid and the vapour. Thus, at equilibrium constants; equilibrium, the rate of evaporation is equal to the rate of • establish a relationship between condensation. It may be represented by Kp and Kc; H2O (l) H2O (vap) • explain various factors that The double half arrows indicate that the processes in both the directions are going on simultaneously. The mixture affect the equilibrium state of a of reactants and products in the equilibrium state is called an equilibrium mixture. reaction; • classify substances as acids or Equilibrium can be established for both physical processes and chemical reactions. The reaction may be fast bases according to Arrhenius, or slow depending on the experimental conditions and the nature of the reactants. When the reactants in a closed vessel Bronsted-Lowry and Lewis at a particular temperature react to give products, the concentrations of the reactants keep on decreasing, while concepts; those of products keep on increasing for some time after • classify acids and bases as weak which there is no change in the concentrations of either of the reactants or products. This stage of the system is the or strong in terms of their dynamic equilibrium and the rates of the forward and ionization constants; • explain the dependence of degree of ionization on concentration of the electrolyte and that of the common ion; • describe pH scale for representing hydrogen ion concentration; • explain ionisation of water and its duel role as acid and base; • describe ionic product (Kw ) and pKw for water; • appreciate use of buffer solutions; • calculate solubility product constant. 2020-21

EQUILIBRIUM 193 reverse reactions become equal. It is due to and the atmospheric pressure are in this dynamic equilibrium stage that there is equilibrium state and the system shows no change in the concentrations of various interesting characteristic features. We observe species in the reaction mixture. Based on the that the mass of ice and water do not change extent to which the reactions proceed to reach with time and the temperature remains the state of chemical equilibrium, these may constant. However, the equilibrium is not be classified in three groups. static. The intense activity can be noticed at the boundary between ice and water. (i) The reactions that proceed nearly to Molecules from the liquid water collide against completion and only negligible ice and adhere to it and some molecules of ice concentrations of the reactants are left. In escape into liquid phase. There is no change some cases, it may not be even possible to of mass of ice and water, as the rates of transfer detect these experimentally. of molecules from ice into water and of reverse transfer from water into ice are equal at (ii) The reactions in which only small amounts atmospheric pressure and 273 K. of products are formed and most of the reactants remain unchanged at It is obvious that ice and water are in equilibrium stage. equilibrium only at particular temperature and pressure. For any pure substance at (iii) The reactions in which the concentrations atmospheric pressure, the temperature at of the reactants and products are which the solid and liquid phases are at comparable, when the system is in equilibrium is called the normal melting point equilibrium. or normal freezing point of the substance. The system here is in dynamic equilibrium and The extent of a reaction in equilibrium we can infer the following: varies with the experimental conditions such as concentrations of reactants, temperature, (i) Both the opposing processes occur etc. Optimisation of the operational conditions simultaneously. is very important in industry and laboratory so that equilibrium is favorable in the (ii) Both the processes occur at the same rate direction of the desired product. Some so that the amount of ice and water important aspects of equilibrium involving remains constant. physical and chemical processes are dealt in this unit along with the equilibrium involving 7.1.2 Liquid-Vapour Equilibrium ions in aqueous solutions which is called as ionic equilibrium. This equilibrium can be better understood if we consider the example of a transparent box 7.1 EQUILIBRIUM IN PHYSICAL carrying a U-tube with mercury (manometer). PROCESSES Drying agent like anhydrous calcium chloride (or phosphorus penta-oxide) is placed for a The characteristics of system at equilibrium few hours in the box. After removing the are better understood if we examine some drying agent by tilting the box on one side, a physical processes. The most familiar watch glass (or petri dish) containing water is examples are phase transformation quickly placed inside the box. It will be processes, e.g., observed that the mercury level in the right limb of the manometer slowly increases and solid liquid finally attains a constant value, that is, the liquid gas pressure inside the box increases and reaches a constant value. Also the volume of water in solid gas the watch glass decreases (Fig. 7.1). Initially there was no water vapour (or very less) inside 7.1.1 Solid-Liquid Equilibrium the box. As water evaporated the pressure in the box increased due to addition of water Ice and water kept in a perfectly insulated thermos flask (no exchange of heat between its contents and the surroundings) at 273K 2020-21

194 CHEMISTRY Fig.7.1 Measuring equilibrium vapour pressure of water at a constant temperature molecules into the gaseous phase inside the dispersed into large volume of the room. As a box. The rate of evaporation is constant. consequence the rate of condensation from However, the rate of increase in pressure vapour to liquid state is much less than the decreases with time due to condensation of rate of evaporation. These are open systems vapour into water. Finally it leads to an and it is not possible to reach equilibrium in equilibrium condition when there is no net an open system. evaporation. This implies that the number of water molecules from the gaseous state into Water and water vapour are in equilibrium the liquid state also increases till the position at atmospheric pressure (1.013 bar) equilibrium is attained i.e., and at 100°C in a closed vessel. The boiling point of water is 100°C at 1.013 bar pressure. rate of evaporation= rate of condensation For any pure liquid at one atmospheric pressure (1.013 bar), the temperature at H2O(l) H2O (vap) which the liquid and vapours are at equilibrium is called normal boiling point of At equilibrium the pressure exerted by the the liquid. Boiling point of the liquid depends on the atmospheric pressure. It depends on water molecules at a given temperature the altitude of the place; at high altitude the boiling point decreases. remains constant and is called the equilibrium 7.1.3 Solid – Vapour Equilibrium vapour pressure of water (or just vapour Let us now consider the systems where solids pressure of water); vapour pressure of water sublime to vapour phase. If we place solid iodine in a closed vessel, after sometime the vessel gets increases with temperature. If the above filled up with violet vapour and the intensity of colour increases with time. After certain time the experiment is repeated with methyl alcohol, intensity of colour becomes constant and at this stage equilibrium is attained. Hence solid iodine acetone and ether, it is observed that different sublimes to give iodine vapour and the iodine vapour condenses to give solid iodine. The liquids have different equilibrium vapour equilibrium can be represented as, pressures at the same temperature, and the I2(solid) I2 (vapour) Other examples showing this kind of liquid which has a higher vapour pressure is equilibrium are, more volatile and has a lower boiling point. Camphor (solid) Camphor (vapour) If we expose three watch glasses NH4Cl (solid) NH4Cl (vapour) containing separately 1mL each of acetone, ethyl alcohol, and water to atmosphere and repeat the experiment with different volumes of the liquids in a warmer room, it is observed that in all such cases the liquid eventually disappears and the time taken for complete evaporation depends on (i) the nature of the liquid, (ii) the amount of the liquid and (iii) the temperature. When the watch glass is open to the atmosphere, the rate of evaporation remains constant but the molecules are 2020-21

EQUILIBRIUM 195 7.1.4 Equilibrium Involving Dissolution of pressure of the gas above the solvent. This Solid or Gases in Liquids amount decreases with increase of temperature. The soda water bottle is sealed Solids in liquids under pressure of gas when its solubility in water is high. As soon as the bottle is opened, We know from our experience that we can some of the dissolved carbon dioxide gas dissolve only a limited amount of salt or sugar escapes to reach a new equilibrium condition in a given amount of water at room required for the lower pressure, namely its temperature. If we make a thick sugar syrup partial pressure in the atmosphere. This is how solution by dissolving sugar at a higher the soda water in bottle when left open to the temperature, sugar crystals separate out if we air for some time, turns ‘flat’. It can be cool the syrup to the room temperature. We generalised that: call it a saturated solution when no more of solute can be dissolved in it at a given (i) For solid liquid equilibrium, there is temperature. The concentration of the solute only one temperature (melting point) at in a saturated solution depends upon the 1 atm (1.013 bar) at which the two phases temperature. In a saturated solution, a can coexist. If there is no exchange of heat dynamic equilibrium exits between the solute with the surroundings, the mass of the two molecules in the solid state and in the solution: phases remains constant. Sugar (solution) Sugar (solid), and (ii) For liquid vapour equilibrium, the vapour pressure is constant at a given the rate of dissolution of sugar = rate of temperature. crystallisation of sugar. (iii) For dissolution of solids in liquids, the Equality of the two rates and dynamic solubility is constant at a given nature of equilibrium has been confirmed with temperature. the help of radioactive sugar. If we drop some radioactive sugar into saturated solution of (iv) For dissolution of gases in liquids, the non-radioactive sugar, then after some time concentration of a gas in liquid is radioactivity is observed both in the solution proportional to the pressure and in the solid sugar. Initially there were no (concentration) of the gas over the liquid. radioactive sugar molecules in the solution These observations are summarised in but due to dynamic nature of equilibrium, Table 7.1 there is exchange between the radioactive and non-radioactive sugar molecules between the Table 7.1 Some Features of Physical two phases. The ratio of the radioactive to non- Equilibria radioactive molecules in the solution increases till it attains a constant value. Process Conclusion Gases in liquids Liquid Vapour tepmH2Opceornasttuarnet at given H2O (l) H2O (g) When a soda water bottle is opened, some of Melting point is fixed at the carbon dioxide gas dissolved in it fizzes Solid Liquid constant pressure out rapidly. The phenomenon arises due to H2O (s) H2O (l) difference in solubility of carbon dioxide at Concentration of solute different pressures. There is equilibrium Solute(s) Solute in solution is constant between the molecules in the gaseous state (solution) at a given temperature and the molecules dissolved in the liquid under pressure i.e., Sugar(s) Sugar (solution) CO2 (gas) CO2 (in solution) Gas(g) Gas (aq) [gas(aq)]/[gas(g)] is This equilibrium is governed by Henry’s CO2(g) CO2(aq) constant at a given law, which states that the mass of a gas temperature dissolved in a given mass of a solvent at [CO2(aq)]/[CO2(g)] is any temperature is proportional to the constant at a given temperature 2020-21

196 CHEMISTRY 7.1.5 General Characteristics of Equilibria Fig. 7.2 Attainment of chemical equilibrium. Involving Physical Processes Eventually, the two reactions occur at the For the physical processes discussed above, same rate and the system reaches a state of following characteristics are common to the equilibrium. system at equilibrium: Similarly, the reaction can reach the state of (i) Equilibrium is possible only in a closed equilibrium even if we start with only C and D; system at a given temperature. that is, no A and B being present initially, as the equilibrium can be reached from either direction. (ii) Both the opposing processes occur at the same rate and there is a dynamic but The dynamic nature of chemical stable condition. equilibrium can be demonstrated in the synthesis of ammonia by Haber’s process. In (iii) All measurable properties of the system a series of experiments, Haber started with remain constant. known amounts of dinitrogen and dihydrogen maintained at high temperature and pressure (iv) When equilibrium is attained for a physical and at regular intervals determined the process, it is characterised by constant amount of ammonia present. He was value of one of its parameters at a given successful in determining also the temperature. Table 7.1 lists such concentration of unreacted dihydrogen and quantities. dinitrogen. Fig. 7.4 (page 191) shows that after a certain time the composition of the mixture (v) The magnitude of such quantities at any remains the same even though some of the stage indicates the extent to which the reactants are still present. This constancy in physical process has proceeded before composition indicates that the reaction has reaching equilibrium. reached equilibrium. In order to understand the dynamic nature of the reaction, synthesis 7.2 EQUILIBRIUM IN CHEMICAL of ammonia is carried out with exactly the PROCESSES – DYNAMIC same starting conditions (of partial pressure EQUILIBRIUM and temperature) but using D2 (deuterium) in place of H2. The reaction mixtures starting Analogous to the physical systems chemical either with H2 or D2 reach equilibrium with reactions also attain a state of equilibrium. the same composition, except that D2 and ND3 These reactions can occur both in forward are present instead of H2 and NH3. After and backward directions. When the rates of equilibrium is attained, these two mixtures the forward and reverse reactions become equal, the concentrations of the reactants and the products remain constant. This is the stage of chemical equilibrium. This equilibrium is dynamic in nature as it consists of a forward reaction in which the reactants give product(s) and reverse reaction in which product(s) gives the original reactants. For a better comprehension, let us consider a general case of a reversible reaction, A+B C+D With passage of time, there is accumulation of the products C and D and depletion of the reactants A and B (Fig. 7.2). This leads to a decrease in the rate of forward reaction and an increase in he rate of the reverse reaction, 2020-21

EQUILIBRIUM 197 Dynamic Equilibrium – A Student’s Activity Equilibrium whether in a physical or in a chemical system, is always of dynamic nature. This can be demonstrated by the use of radioactive isotopes. This is not feasible in a school laboratory. However this concept can be easily comprehended by performing the following activity. The activity can be performed in a group of 5 or 6 students. Take two 100mL measuring cylinders (marked as 1 and 2) and two glass tubes each of 30 cm length. Diameter of the tubes may be same or different in the range of 3-5mm. Fill nearly half of the measuring cylinder-1 with coloured water (for this purpose add a crystal of potassium permanganate to water) and keep second cylinder (number 2) empty. Put one tube in cylinder 1 and second in cylinder 2. Immerse one tube in cylinder 1, close its upper tip with a finger and transfer the coloured water contained in its lower portion to cylinder 2. Using second tube, kept in 2nd cylinder, transfer the coloured water in a similar manner from cylinder 2 to cylinder 1. In this way keep on transferring coloured water using the two glass tubes from cylinder 1 to 2 and from 2 to 1 till you notice that the level of coloured water in both the cylinders becomes constant. If you continue intertransferring coloured solution between the cylinders, there will not be any further change in the levels of coloured water in two cylinders. If we take analogy of ‘level’ of coloured water with ‘concentration’ of reactants and products in the two cylinders, we can say the process of transfer, which continues even after the constancy of level, is indicative of dynamic nature of the process. If we repeat the experiment taking two tubes of different diameters we find that at equilibrium the level of coloured water in two cylinders is different. How far diameters are responsible for change in levels in two cylinders? Empty cylinder (2) is an indicator of no product in it at the beginning. Fig.7.3 Demonstrating dynamic nature of equilibrium. (a) initial stage (b) final stage after the equilibrium is attained. 2020-21