THERMODYNAMICS. Extensive properties Intensive properties

Transcription

1 Thermodynamics The branch of chemistry deals with the energy change associated with chemical reactions is called chemical thermodynamics. System and surrounding A system may be defined as the specified portion of matter or universe, which is selected for a particular study and is covered by a bounding surface. The rest of the universe which might be in a position to exchange energy and matter with the system is said to be its surroundings. System and the surroundings together constitute the universe. The wall that separates the system from the surroundings is called boundary. It may be real or imaginary. For example one mole of an ideal gas taken in a closed vessel is a system. The rest of the universe is its surroundings. 1. Open system: a system which can exchange energy as well as matter with the surroundings is called an open system. E.g. An open vessel containing a liquid at a given temperature and pressure. 2. Closed system: a system which can exchange energy but not matter with the surroundings is called a closed system. E.g. A closed vessel containing a definite quantity of material like a gas or a liquid. 3. Isolated system: a system which can exchange neither energy nor matter with the surroundings is called an isolated system. E.g. a perfect closed thermo flask containing a liquid. The state of a thermodynamic system is described by its measurable or macroscopic properties like pressure, temperature etc. we can describe the state of a gas by quoting its pressure [P], volume [ V], temperature [T], composition etc. variables like P, V, T are called state variables or state functions, because their values depend only on the state of the system and not on how it is reached. In order to completely define the state of o system it is not necessary to define all the properties of the system. Thermodynamic process: A thermodynamic system can be changed from one state to another by a process. They are: 1. Reversible process: it is a process which takes place infinitesimally slowly such that the driving force is only infinitesimally greater than the opposing force, its direction can be reversed and the system is always at equilibrium. 2. Irreversible process: it is a process which doesn't take place infinitesimally slowly. Such process can't be reversed by changing the variables. 3. Isothermal process: It is a process during which the temperature of the system remains constant, that is, T = Adiabatic process: It is a process during which the system does not exchange heat with the surroundin gs, that is, q = Isobaric process: It is a process during which the pressure of the system remains constant, that is, P = Isochoric process: It is a process during which the volume of the system remains constant, that is, V = Cyclic process: It is the process during which the system after undergoing a series of changes comes back to the initial state, that is, the series of changes take place in a cyclic manner. Extensive & Intensive Properties: Properties which depends on quantity of matter called extensive properties. eg. Mass, volume, heat capacity, enthalpy, entropy etc. The properties which do not depends on matter present and depends upon nature of substance called Intensive properties. eg. T, P, density, refractive index, viscosity, boiling point, ph, mole fraction etc. Extensive properties Intensive properties mass, volume, amount, heat capacity, internal energy, enthalpy, entropy, free energy pressure, temperature, density, surface tension, viscosity, refractive index, molarity, molar volume, specific heat capacity, molar heat capacity, molar energy State function A function that depends only on the state of the system and not on the process by which this state is achieved is known as a state function. Eg:- internal energy (U), enthalpy (H), entropy (S) and free energy (G and A) change etc. An Ideal Learner s School Of Chemistry 1

2 Path Function It is a thermodynamics property that depends upon how the process is carried out, that is, it depends upon the path taken by the system during the thermodynamic process. e.g. work, heat, etc. Internal Energy (intrinsic energy) (U): Every system is associated with a definite amount of energy under a given set of conditions called internal energy. It is the sum of many types of energies like vibrational energy, translational energy. etc. It is an extensive property and state function. Its absolute value cannot be determined but experimentally change in internal energy (ΔU) can be determined by ΔU = U 2 U 1 The SI unit of U is Joule. In a cyclic process, there is no net change in internal energy. Internal energy may change, when 1. Heat absorbed by the system or liberated from the system. 2. Work done by the system or work done on the system. 3. Matter enters or leaves the system. First Law of Thermodynamics: It is law of conservation energy. Energy can neither be created nor destroyed; it may be converted from one from into another. i.e. total energy of universe is a constant. We know that internal energy of a system can be changed, 1. Heat absorbed by the system or liberated from the system. 2. Work done by the system or work done on the system. Consider a system in state I where its internal energy is E 1. Let it absorb Q quantity of heat from the surroundings and W be the work done on the system, so that it changes to a state II where its internal energy is E 2 Then the change in internal energy E 2 -E 1 = E = Q+W i.e ΔE = Q + W This is the mathematical expression for first law of thermodynamics. According to IUPAC conventions heat absorbed by a system Q is +ve and heat given out by a system Q is ve. Also work done by a system W is ve and work done on a system W is +ve. Note : for isothermal process ΔT =0 thus ΔE =0. Hence Q= -W i.e heat given to a system is used in work done by the system. For adiabatic process Q = 0 thus ΔE = W or ΔE = -W i.e. work is done by the system on the cost of its internal energy. For isochoric process ΔV = 0 thus Q = ΔE. i.e. heat given to a system under constant volume is used up in i ncreasing internal energy. Work of expansion: Consider a uniform cylinder having area of cross section a cm 2, containing a gas closed with a weightless frictionless piston. P be the pressure acting on the piston. Let the volume of the gas be V1. Then the force acting on the piston = P.a. Let the gas expand to a volume V2 by pushing the piston up through a distance h by absorbing heat from the surroundings. Work done by the gas = -P.a.h But a.h = ΔV, the increase in volume. Therefore work of expansion = -PΔV or P(V2-V1) Thus the work of expansion is the product of external pressure and change in volume. For expansion it is ve and for compression it is +ve by convention. An Ideal Learner s School Of Chemistry 2

3 Work done in the reversible isothermal expansion of an ideal gas: Let us consider one mole of an ideal gas enclosed in a cylinder fitted with a weightless and frictionless piston. The cylinde r is not insulated. If the external pressure (P) is lowered by infinitesimal amount dp, i.e. it falls from P to P-dp, the gas will expand by an infinitesimal volume dv. i.e. the volume changes from V to V+dv. If the external pressure is lowered again by the same infinitesimal amount, dp, the gas will undergo the second infinitesimal expansion dv before the pressure again equals the new external pressure. The piston again comes to rest. This process is continued such that the external pressure is lowered by successive small amount and as a result, the gas undergoes a series o f small successive increments of volume dv at a time. Since during expansion, pressure decreases and volume increases, these two parameters are assigned opposite signe. Thus the work done by the gas in the infinitesimal expansion is given as dw = -(P-dp)dv = -Pdv + dpdv But the product dpdv is negligible as both dp and dv are infinitesimal. Therefore dw = -Pdv (1) The total work done (W) by the gas in the expansion from initial volume V 1 to final volume V 2 will be the sum of the Pdv terms. Thus (1) on integration, W = (2) For one mole of ideal gas PV = RT and hence P = RT/V Thus W = - i.e. W = -RT ln or W = RT log (3) Again for an ideal gas P 1 V 1 = P 2 V 2 Therefore V 2 /V 1 = P 1 /P 2 Hence W = RT log For n moles, W = nRT log Maximum work: We know that the magnitude of work done by a system on expansion depends on the magnitude of the opposing [external] pressure. The closer is the opposing pressure to the pressure of the gas, the greater is the work of expansion. Thus maximum work is obtained when the two opposing pressure differ only by an infinitesimal amount from each other. This is the condition required for a thermodynamically reversible process. Thus the condition for maximum work coincides with that for thermodynamic reversibility. Note : no work is done during free expansion. For isothermal process dt = 0 and de =0 If a process carried out at constant volume i.e. dv = 0, then de = q-pdv = q v For isothermal expansion of an ideal gas into vaccum, W=0, P ex =0, q=0 hence de=0 For isothermal irreversible change, q=-w=p ex dv, here de=0 For isothermal reversible change, Q=-W=2.303nRT log For adiabatic change q=0, de=w Enthalpy (heat content), H: Enthalpy is the sum of internal energy and product of pressure and volume. It is a state function and extensive property. Mathematically, H = E + PV When the system undergoes a change, its enthalpy changes, i.e. ΔH=ΔE+PΔV+VΔP An Ideal Learner s School Of Chemistry 3

4 When the process is carried out at constant pressure ΔP=0 and therefore VΔP=0 Then, ΔH=ΔE+PΔV But from first law of thermodynamics, ΔE=Q-PΔV or ΔE+PΔV=Q Thus ΔH=Q p i.e. heat absorbed by the system at constant pressure. For exothermic reaction (the reaction in which heat is evolved), ΔH = -ve whereas for endothermic reaction (the reaction in which heat is absorbed), ΔH = +ve. Relationship between ΔH and ΔE We have ΔH=ΔE+PΔV Therefore ΔH-ΔE=PΔV, the work of expansion. For solids and liquids the volume change is negligible, therefore, enthalpy and internal energy changes are practically same. So ΔH=ΔE, but for gaseous reactions these may be different. Let n A be the total number of moles of reactants in a reaction occupying a volume V A at constant pressure P and temperature T. if the gases are assumed to behave ideally, PV A =n A RT (1) Let n B be the total number of moles of product in a reaction occupying a volume V B. Then PV B =n B RT (2) Then, (2)-(1) ---> PV B -PV A = n B RT-n A RT P(V B -V A )=RT(n B -n A ) PΔV=ΔnRT Therefore ΔH=ΔE+ΔnRT Where Δn is the change in number of moles during the reaction. When Δn=0, ΔH=ΔE Heat capacity: Heat capacity of a system, between any two temperatures, is defined as the amount of heat required to raise the temperature of the system from the lower to higher temperature divided by the temperature difference. When the mass of the system is one gram, the heat capacity is called specific heat capacity of the system. When the mass of the system is one mole, the heat capacity is called molar heat capacity c. Thus the molar heat capacity of a system between temperatures T 1 and T 2 will be expressed as C= or C= Since heat capacity varies with temperature, hence the true molar heat capacity is defined by the differential equation C= Molar heat capacity at constant volume, Cv: From the first law of thermodynamics, ΔE=Q+W (1) At constant volume W=0, hence ΔE=Q (2) Therefore Cv =[ v = [ v i.e. C v = [ v (3) Thus the molar heat capacity of a system of mass one mole at constant volume is the increase of internal energy of the system per degree rise of temperature. Molar heat capacity at constant pressure, Cp: From the first law of thermodynamics, ΔE=Q+W hence Q= ΔE-W Or Q= ΔE-[-P ΔV]= ΔE+P ΔV (1) Therefore Cp =[ p = [ p (2) But ΔE+P ΔV= ΔH Therefore Cp= [ i.e. C p = [ p (3) p Thus the molar heat capacity of a system of mass one mole at constant pressure is the increase of enthalpy of the system per degree rise of temperature. An Ideal Learner s School Of Chemistry 4

5 Relationship Between Cp And Cv We have C v = [ v and C p = [ p Also H=E+PV Therefore dh=de+d[pv] Dividing by dt = + Or Cp = Cv+ For one mole of an ideal gas PV=RT Thus,Cp=Cv+ i.e. Cp=Cv+R or Cp-Cv=R Thus the difference between the molar heat capacity of a gas at constant pressure Cp and at constant volume Cv is equal to the gas constant R i.e.8.314j Bomb calorimeter: For chemical reactions, heat absorbed at constant volume, is measured in a bomb calorimeter. Here, a steel vessel (the bomb) is immersed in a water bath. The whole device is called calorimeter. The steel vessel is immersed in water bath to ensure tha t no heat is lost to the surroundings. A combustible substance is burnt in pure dioxy gen supplied in the steel bomb. Heat evolved during the reaction is transferred to the water around the bomb and its temperature is monitored. Since the bomb calorimeter is sealed, its volume does not change i.e., the energy changes associated with reactio ns are measured at constant volume. Under these conditions, no work is done as the reaction is carried out at constant volume in the bomb calorimeter. Even for reactions involving gases, there is no work done as ΔV = 0. Temperature change of the calorimete r produced by the completed reaction is then converted to q v, by using the known heat capacity of the calorimeter with the help of equation; q (water + calorimeter) = (m s + C) ΔT = q (surroundings) where, m = mass of water, s = specific heat capacity of water and C = heat capacity of the calorimeter. Enthalpy of Reaction Heat of a reaction is defined as the heat exchanged between the system and the surroundings during the complete transformation of reactants to products at a constant temperature and pressure for the given stoichiometrically balanced equation. During a chemical process some bonds break and others are formed so there is a net change in the energy of the system. This results in the heat exchange during a process. Since enthalpy is a state function, the change in enthalpy is equal to the difference in the enthalpies of the final state an d the initial state. i.e. ΔrH = ΣH (p) ΣH (R) Enthalpy of reaction expressed at the standard state conditions ( 298 K, 1 bar) is called standard enthalpy of reaction (Δ r H 0 ). Various Forms of Enthalpy of Reaction 1. Enthalpy of Formation (ΔH f ) It is heat change when one mole of compound is obtained from its constituent elements. Its unit is kj mol -1. Enthalpy of formation of an element in its standard state is taken as zero at all temperatures. 2. Enthalpy of Combustion (ΔHc.) It is the Enthalpy change taking place when one mole of a compound undergoes complete combustion in presence of oxygen. Enthalpy of combustion is always negative. 3. Enthalpy of Solution (symbol : Δ sol H 0 ): Enthalpy of solution of a substance is the enthalpy change when one mole of it dissolves in a specified amount of solvent. The enthalpy of solution at infinite dilution is the enthalpy change observed on dissolving the substance in an infinite amou nt of solvent when the interactions between the ions (or solute molecules) are negligible. Enthalpy of solution of ionic compound in water is determined by lattice enthalpy and enthalpy of hydration. If the lattice enthalpy is very high, the dissolution of the compound may not take place at all. Δ sol H 0 = Δ lattice H 0 + Δ hyd H 0 An Ideal Learner s School Of Chemistry 5

6 4. Lattice Enthalpy : The lattice enthalpy of an ionic compound is the enthalpy change which occurs when one mole of an ionic compound dissociates into its ions in gaseous state. 5. Enthalpy of Hydration It is the enthalpy change when one mole of anhydrous substances combines with requisite number of moles of water to form the hydrate. 6. Enthalpy of Atomisation: It is the enthalpy change occurring when one mole of the molecule breaks into its atoms. In case of diatomic molecules, like dihydrogen the enthalpy of atomization is also the bond dissociation enthalpy. In some cases, the enthalpy of atomization is same as the enthalpy of sublimation. 7. Enthalpy of Neutralisation It is the enthalpy change that takes place when 1 g-equivalent of an acid (or base) is neutralized by 1 g-equivalent of a base (or acid) in dilute solution. Enthalpy of neutralisation of strong acid and strong base is always constant, i.e., 57.1 kj. [Enthalpy of neutralisation of strong acid and weak base or weak acid and strong base is not constant and numerically less than 57.1 kj due to the fact that here the heat is used up in ionisation of weak acid or weak base. This is known as enthalpy of ionisation of weak acid / or base.] 8. Bond enthalpy: It is amount of energy released when gaseous atoms combines to form one mole of bonds between them or heat absorbed when one mole of bonds between them are broken to give free gaseous atoms. The energy required to break the particular bond in a gaseous molecule is called bond dissociation enthalpy. It is definite in quantity and expressed in kj mol -1. In diatomic molecule, bond dissociation enthalpy = Bond enthalpy In polyatomic molecule, bond dissociation enthalpy Bond Enthalpy. Here bond energy is taken as the average of the bond dissociation energies of the various similar bonds. 9. Enthalpy of Fusion: It is the change in enthalpy accompanied by the conversion of one mole of solid substance in standard state to its liquid state at its melting point. Enthalpy of fusion is positive 10. Enthalpy of vaporization: It is the change in enthalpy accompanied by the conversion of one mole of liquid substance in standard state to its gaseous state at its boiling point. 11. Enthalpy of sublimation: It is the change in enthalpy accompanied by the conversion of one mole of solid substance in standard state to its gaseous state at a temperature below its melting point. Hess s Law of Constant heat Summation This law states that enthalpy change in a process remains constant whether the reaction takes place in one step or in several steps. Suppose in a reaction, the reactant A changes into product B in one step and the enthalpy change during this process is H. now suppose the same process is carried out in two steps, the reactant A changes to C and the enthalpy change during this step is H 1. Then C changes to the product B and the enthalpy change during this step is H 2. Then by Hess s law, the enthalpy change for the reaction is the same whether the reaction follows path l or path ll i.e. ΔH = ΔH 1 + ΔH 2 An Ideal Learner s School Of Chemistry 6

7 Applications of Hess s law: 1. Determination of enthalpy of formation of substances which cannot be measured experimentally. There is large number of compounds such as methane benzene etc which cannot be synthesized directly from their elements. The enthalpies of formation of such substances can be determined by an indirect method based on Hess s law. 2. Determination of enthalpies of transition of all allotropic modifications. 3. Predicting enthalpies of various reactions. Limitations of first law of thermodynamics: The first law of thermodynamics is concerned with the conversion of one form of energy into another form. It simply says that in any transformation the energy of universe remains constant. However, it fails to predict (1). Whether a process can occur or not. (2). The direction in which a process can occur. Need for the second law: The second law of thermodynamics helps us to determine the direction in which energy can be transformed. It also helps us to predict whether a given process can occur spontaneously. It also helps us to know the equilibrium conditions. The law states that it is impossible to convert heat into work without compensation. Spontaneous & Non Spontaneous Processes: A process which has a natural tendency to occur without proper initiation is called a spontaneous process or feasible or probable processes. Evaporation of water from an open vessel, burning of carbon in domestic oven etc is examples of spontaneous process. A process which can neither take place by itself or by initiation is called non Spontaneous. Various statements of the Second Law of Thermodynamics 1. Heat cannot pass spontaneously from a colder to a warmer body. (Clausius) 2. No change is possible in which the sole result is the absorption of heat from a reservoir and its complete conversion into work. (Kelvin) 3. The entropy of the universe is always Increasing in the course of every spontaneous or natural change. 4. In an adiabatic process the entropy either increases or remains unchanged. 5. It is impossible to convert heat into work without compensation. Entropy (S) Entropy is the measurement of randomness or disorder of the molecules of a system. It is a state function and extensive property. Units :JK -1 mol -1 The change in entropy during a process is mathematically given as Δ r S = Σ S (products) Σ S (reactants) = q rev / T = ΔH / T Where, q rev heat absorbed by the system in reversible manner T = temperature Δ S > 0, Increase in randomness, heat is absorbed Δ S < 0, Decrease in randomness, heat is evolved. Entropy of even elementary substances are not zero. Relationship between internal energy and entropy de = dq + dw...(first law of thermodynamics) de = dq PdV For reversible process, dq = dq rev = TdS de = TdS PdV This is also known as the combined form of first law and second law for reversible processes. Relationship between enthalpy and entropy H = E + PV Differentiating the equation we get dh = de + PdV + VdP...(1) From the combined form of the first law and the second law of Thermodynamics de = TdS PdV An Ideal Learner s School Of Chemistry 7

8 Substituting the above expression in Equation (1) we get dh = TdS PdV + PdV + VdP dh = TdS + VdP Entropy Change During Phase Transition The change of matter from one state to another state is called phase transition. The entropy changes at the time of phase transition: ΔS transition = 1. Entropy of fusion ΔS fusion : It is the change in entropy when one mole of a solid substance converted into its liquid state at its melting point (T f ) ΔS fusion = 2. Entropy of vaporization ΔS vap : It is the change in entropy when one mole of a liquid substance converted into its gaseous state at its boiling point (T b ) ΔS vap = 3. Entropy of sublimation ΔS sub : It is the change in entropy when one mole of a solid substance converted into its gaseous state at a temperature below its melting point. ΔS sub = Entropy Criterion of Spontaneous Process A process is a spontaneous if and only if the entropy of the universe increases. For a process to be Spontaneous (ΔS universe > 0 or ΔS syst + ΔS surr > 0) If ΔS (total) is +ve, the process is spontaneous. If ΔS (total) is ve, the process is non-spontaneous. At equilibrium state, ΔS = 0, Limitations of ΔS criterion and need for another term: We cannot find entropy change of surroundings during chemical changes. So we need another parameter for spontaneity viz Gibbs energy of system (G). Gibbs Free Energy (G) : It is defined as maximum amount of energy available to a system during the proces s that can be converted into useful work. It is a measure of capacity to do useful work. Mathematically, G = H TS Since H, T and S are state functions; therefore, G must also be a state function. Change in Gibbs energy during the process is given by Gibbs Helmholtz equation. ΔG = ΔH TΔS SΔT If the temperature is constant ΔT = 0, therefore, ΔG = ΔH TΔS Where, ΔG = Gibbs free energy change, ΔH enthalpy change, ΔS entropy change. The Gibbs energy criterion of spontaneity ΔG > 0, process is non-spontaneous ΔG < 0, 0, process is spontaneous ΔG = 0, process is in equilibrium state Now an exothermic reaction which is non-spontaneous at high temperature may become spontaneous at low temperature. Similarly, endothermic reactions which are non-spontaneous at low temperature may become spontaneous at high temperature. An Ideal Learner s School Of Chemistry 8

9 Variation of G with Temperature and Pressure G = H TS Differential of this equation gives dg = dh TdS SdT...(1) According to combined form of first law and second law of thermodynamics, for reversible process dh = TdS + VdP Substituting the value in equation (1) we get dg = TdS +VdP TdS SdT or dg = SdT +VdP Standard Free Energy Change ( G 0 ) : It is defined as free energy change measured at 298 K and 1 atm Pressure. Standard Free energy of formation: (( G f 0 ) It is free energy change when 1 mole of compound is formed from its constituting elements in their standard state. G f 0 = ƩG f 0 (products) - ƩG f 0 (reactants) Gibbs Energy Change and Work In case of galvanic cell, Gibbs energy change r G,is related to the electrical work done by the cell. Thus r G = -nfe cell If reactants and products are in their standard states r G = -nfe 0 cell, Here E 0 cell is the standard cell potential. Gibbs Energy Change and Equilibrium Constant r G 0 = -RT lnk r G 0 = RT log K Third law of Thermodynamics This law was formulated by Nernst in According to this law, Every system is associated with a finite positive entropy and that entropy tends to zero at absolute zero (0K), and is applicable only in the case of perfectly crystalline substance. 1. Give enthalpy (H) of all elements in their standard state. In standard state enthalpies of all elements is zero. 2. From thermodynamic point to which system the animals and plants belong? Open system. 3. Predict the sign of S for the following reactions. CaCO 3 (s) CaO (s) + CO 2 (g) S is positive (entropy increases) 4. For the reaction 2Cl (g) Cl 2(g), What will be the sign of H and S? H: ( ve) because energy is released in bond formation and S: ( ve) because atoms combines to form molecules. 5. In a process, 701J heat is absorbed and 394J work is done by system. What is change in Internal energy for process? q = 701J, w = 394J, so E = q + w = = 307J. 6. Given: N 2(g) + 3H 2(g) 2NH 3(g), r H 0 = 92.4KJ.mol 1. What is the standard enthalpy of formation of NH 3(g). Enthalpy of formation of NH 3(g) = = KJ.mol -1 An Ideal Learner s School Of Chemistry 9

10 7. Calculate entropy change in surroundings when 1.0 mol of H 2 O(l) is formed under standard conditions? Given H 0 = 286KJmol 1. q (rev.) = H 0 = 286 KJmol -1 = Jmol -1 S = = = 959 J/K/mol 8. A real crystal has more entropy than an Ideal Crystal. Why? A real crystal has some disorder due to presence of defects in their structural arrangement, and Ideal crystal does not have any disorder. 9. Under what condition, the heat evolved/absorbed in a reaction is equal to its free energy change? In G = H T. S, when reaction is carried out at OK or S = 0, then G = H. 10. Predict the entropy change in- (i) A liquid crystallizes into solid (ii) Temperature of a crystallize solid raised from OK to 115K (i) Entropy decreases because molecules attain an ordered state. (ii) entropy increase because from OK to 115K particles begin to move. 11. What is bond energy? Why is it called enthalpy of atomization? It is the amount of energy required to dissociate one mole of bonds present between atoms in gas phase. As molecules dissociates into atoms in gas phase so bond energy of diatomic molecules is called enthalpy of atomization. 12. Calculate entropy change for the following process. H 2 O (s) H 2 O (l), is 6.0 KJ mol -1 at 0 0 C H 2 O (s) H 2 O (l) at 0 0 C H f = 6.0 KJ mol -1 = 6000J mol -1 T f = = 273K Thus S f = H f /T = 6000J mol -1 /273K = 21.98JK -1 mol For oxidation of iron, 4Fe (s) + 3O 2 (g) 2Fe 2 O 3 (s) S is 549.4J.K -1 mol -1, at 298K. Inspite of ve entropy change of this reaction, Why the reaction is spontaneous? ( r H 0 = 1648x10 3 J.mol -1 ) S surr = - r H 0 / T = - (-1648x10 3 J.mol -1 ) / 298K = 5530JK -1 mol -1. & S (system) = QJK -1 mol -1. r S (total) = = J.K -1 mol -1 Since rs (total) is +ve, so the reaction is spontaneous. 14. For the reaction at 298K, 2A+B C, H = 400 KJ mol -1, S = 0.2 KJ mol -1 K -1. At what temperature will the reaction become spontaneous, considering H, S be constant at the temp. H = 400 KJ mol -1, S = 0.2 KJK -1 mol -1. G = H T. S O = x T ( G = 0 at equilibrium) T = 400/.2 =2000K so reaction will be spontaneous above 2000K. 15. The equilibrium constant for a reaction is 10. What will be the value of G 0? R= 8.314J.K -1 mol -1 T = 300K. G 0 = RT logk = x x 300 x log 10 = x 300 x 1 = J rg 0 = KJ.mol What do you understand by state function? Neither q nor w is a state function but q + w is a state function? Explain. The property whose value depends upon state of system and is independent of path. q + w = U, which is a state function as value of U does not depends upon path. 17. Calculate the temperature above which the reduction reaction becomes spontaneou s: PbO (s) + C (s) Pb (s) + CO (g), given [ H = KJ mol -1, S = 190J.K -1 mol -1 ]. G = H T. S, at equilibrium G = 0, H = T S T = H/ S = x 10 3 / 190 = K So the reaction will be spontaneous above K, as above this temperature G will be ve. An Ideal Learner s School Of Chemistry 10

11 18. Why is the entropy of a substance taken as zero at 0K? Calculate the r G for the reaction? N 2 (g) + 3H 2 (g) 2NH 3(g) at 298K The value of equilibrium constant (K) is 6.6x10 5, R = 8.314JK -1 mol -1. Because at O K there is complete order in the system. G 0 = RT logk = x x 298 x log6.6 x 10 5 = [log6.6 + log 10 5 ] = [ ] = J = J G 0 = KJ mol Why standard entropy of an elementary substance is not zero whereas standard enthalpy of formation is taken as zero? A substance has a perfectly ordered arrangement only at absolute zero. Hence, entropy is zero only at absolute zero. Enthalpy of formation is the heat change involved in the formation of one mole of the substance from its elements. An element formed from it means no heat change. 20. The equilibrium constant for a reaction is one or more if G for it is less than zero. Explain G = RT ln K, thus if G is less than zero. i.e., it is negative, then lnk will be positive and hence K will be greater than one. 21. Many thermodynamically feasible reactions do not occur under ordinary conditions. Why? Under ordinary conditions, the average energy of the reactants may be less than threshold energy. They require some activation energy to initiate the reaction. 1. The volume of gas is reduced to half from its original volume. The specific heat will be. (iii) ΔS (system) decreases but ΔS (surroundings) increases. (i) reduce to half (ii) be doubled (iv) ΔS (system) decreases and ΔS (surroundings) also (iii) remain constant (iv) increase four times decreases. 2. Δ f U 0 of formation of CH4 (g) at certain temperature is 393 kj mol 1. The value of Δ f H 0 is 5. The enthalpies of elements in their standard states are taken as zero. The enthalpy of formation of a compound (i) zero (ii) < Δ f U 0 (iii) > Δ f U 0 (iv) equal to Δ f U 0 (i) is always negative (ii) is always positive 3. In an adiabatic process, no transfer of heat takes place (iii) may be positive or negative (iv) is never negative between system and surroundings. Choose the correct option for free expansion of an ideal gas under adiabatic condition from the following. 6. Enthalpy of sublimation of a substance is equal to (i) enthalpy of fusion + enthalpy of vapourisation (i) q = 0, ΔT 0, w = 0 (ii) q 0, ΔT = 0, w = 0 (ii) enthalpy of fusion (iii) enthalpy of vapourisation (iii) q = 0, ΔT = 0, w = 0 (iv) q = 0, ΔT < 0, w 0 (iv) twice the enthalpy of vapourisation 4. The entropy change can be calculated by using the expression ΔS =q rev /T. When water freezes in a glass beaker, choose the correct statement amongst the following : (i) ΔS (system) decreases but ΔS (surroundings) remains the same. 7. Which of the following is not correct? (i) ΔG is zero for a reversible reaction (ii) ΔG is positive for a spontaneous reaction (iii) ΔG is negative for a spontaneous reaction (iv) ΔG is positive for a non-spontaneous reaction (ii) ΔS (sys) increases but ΔS (surr) decreases. 1. (iii) 2. (ii) 3. (iii) Justification : free expansion w = 0 adiabatic process q = 0 ΔU = q + w = 0, this means that internal energy remains constant. Therefore, ΔT = 0. In ideal gas there is no intermolecular attraction. Hence when such a gas expands under adiabatic conditions into a vaccum no heat is absorbed or evolved since no external work is done to separate the molecules 4. (iii) Justification : Freezing is exothermic process. The heat released increases the entropy of surrounding. 5. (iii) 6. (i) 7. (ii) 22. One mole of acetone requires less heat to vapourise than 1 mol of water. Which of the two liquids has higher enthalpy of An Ideal Learner s School Of Chemistry 11

12 vapourisation? Water 23. Standard molar enthalpy of formation, Δ f H 0 is just a special case of enthalpy of reaction, ΔrH 0. Is the ΔrH 0 for the following reaction same as Δ f H 0? Give reason for your answer. CaO(s) + CO 2 (g) CaCO 3 (s); Δ f H 0 = kj mol 1 No, since CaCO 3 has been formed from other compounds and not from its constituent elements. 24. The enthalpy of atomisation for the reaction CH 4 (g) C(g) + 4H (g) is 1665 kj mol 1. What is the bond energy of C H bond? 1665 / 4kJ mol 1 = kj mol Given that ΔH = 0 for mixing of two gases. Explain whether the diffusion of these gases into each other in a closed container is a spontaneous process or not? It is spontaneou s process. Althou gh enthalpy change is zero but randomness or disorder (i.e., ΔS ) increases. Therefore, in equation ΔG = ΔH TΔS, the term TΔS will be negative. Hence, ΔG will be negative. 26. Heat has randomising influence on a system and temperature is the measure of average chaotic motion of particles in the system. Write the mathematical relation which relates these three parameters. ΔS = q rev / T 27. Increase in enthalpy of the surroundings is equal to decrease in enthalpy of the system. Will the temperature of system and surroundings be the same when they are in thermal equilibrium? Yes 28. The standard molar entropy of H 2 O (l) is 70 J K 1 mol 1. Will the standard molar entropy of H 2 O(s) be more, or less than 70 J K 1 mol 1? Less, because ice is more ordered than H 2 O (l). 29. Identify the state functions and path functions out of the following : enthalpy, entropy, heat, temperature, work, free energy. State Functions : Enthalpy, Entropy, Temperature, Free energy Path Functions : Heat, Work 30. The molar enthalpy of vapourisation of acetone is less than that of water. Why? Because of strong hydrogen bonding in water, its enthalpy of vapourisation is more. 31. Which quantity out of Δ r G and Δ r G 0 will be zero at equilibrium? Δ r G will always be zero. Δ r G 0 is zero for K = 1 because ΔG 0 = RT lnk, ΔG 0 will be non-zero for other values of K. 32. Predict the change in internal energy for an isolated system at constant volume. For isolated system, there is no transfer of energy as heat or as work i.e., w=0 and q=0. According to the first law of thermodynamics. ΔU = q + w = = 0 ΔU = Although heat is a path function but heat absorbed by the system under certain specific conditions is independent of path. What are those conditions? Explain. At constant volume By first law of thermodynamics: q = ΔU + ( w) ( w) = pδv q = ΔU + pδv ΔV = 0, since volume is constant. q V = ΔU + 0 q V = ΔU = change in internal energy At constant pressure q p = ΔU + pδv But, ΔU + pδv = ΔH q p = ΔH = change in enthalpy. So, at a constant volume and at constant pressure heat change is a state function because it is equal to change in internal energy and change in enthalpy respectively which are state functions. An Ideal Learner s School Of Chemistry 12

13 34. Expansion of a gas in vacuum is called free expansion. Calculate the work done and the change in internal energy when 1 litre of ideal gas expands isothermally into vacuum until its total volume is 5 litre? ( w) = p ext (V 2 V 1 ) = 0 (5 1) = 0 For isothermal expansion q = 0 By first law of thermodynamics q = ΔU + ( w) 0 = ΔU + 0 so ΔU = Heat capacity (C p ) is an extensive property but specific heat (c) is an intensive property. What will be the relation between C p and c for 1 mol of water? For water, heat capacity = 18 specific heat or Cp = 18 c Specific heat = c = 4.18 J g 1 K 1 Heat capacity = C p = J K 1 = 75.3 J K The difference between C P and C V can be derived using the empirical relation H = U + PV. Calculate the difference between C P and C V for 10 moles of an ideal gas. C P C V = nr = J 37. If the combustion of 1g of graphite produces 20.7 kj of heat, what will be molar enthalpy change? Give the significance of sign also. Molar enthalpy change of graphite = enthalpy change for 1 g carbon molar mass of carbon = 20.7 kj g 1 12g mol 1 ΔH = kj mol 1 Negative value of ΔH exothermic reaction. 38. The enthalpy of vapourisation of CCl 4 is 30.5 kj mol 1. Calculate the heat required for the vapourisation of 284 g of CCl 4 at constant pressure. (Molar mass of CCl 4 = 154 g mol 1 ). q p = ΔH = 30.5 kj mol 1 Heat required for vapourisation of 284 g of CCl 4 = (284g / 154g mol -1 ) x 30.5 kj mol 1 = 56.2 kj 39. An ideal gas is allowed to expand against a constant pressure of 2 bar from 10 L to 50 L in one step. Calculate the amount of work done by the gas. If the same expansion were carried out reversibly, will the work done be higher or lower than the earlier case? (Given that 1 L bar = 100 J ) w = p ex (V f V i ) = 2 40 = 80 L bar = 8 kj The negative sign shows that work is done by the system on the surrounding. Work done will be more in the reversible expansion because internal pressure and exernal pressure are almost same at every step. An Ideal Learner s School Of Chemistry 13