Thermodynamics & Thermochemistry Exercises

Thermodynamics & Thermochemistry THERMODYNAMICS IInd & IIIrd LAW PART - I : SUBJECTIVE QUESTIONS Section (A) : Introduction about entropy A-1. The entropy of a gas increases on its expansion. Why ? A-2. Entropy of the solutions is higher than that of pure liquid. Why ? A-3. What are the signs of S for the system and for the surrounding in each of the following processes ? (a) Water boils in a teakettle on a hot stove. (b) Ice in an ice cube tray, left an a table melts. (c) A cup of coffee is reheated in a microwave oven. A-4. State the second law of thermodynamics. A-5. Write statement of IIIrd law of thermodynamics ? A-6. State the thermodynamic condition of spontaneous occurance of a process ? A-7. If H for a reaction has a positive value, how would you know the sign requirement of S for it so that the reaction is spontaneous ? Section (B) : Entropy Calculation B-1. The entropy of vaporization of benzene is 85 JK –1 mol–1. When 117 g benzene vaporizes at it’s normal boiling point, calculate the entropy change of surrounding. B-2. Calculate standard entropy change in the reaction Fe2O3 (s) + 3H2 (g) ⎯→ 2Fe (s) + 3H2O ( ) Given : Sºm(Fe2O3, S) = 87.4 , Sºm(Fe, S) = 27.3, Sºm(H2, g) = 130.7, Sºm(H2O, ) = 69.9 JK–1 mol–1. B-3. One mole of liquid iron at its boiling point was vaporized in an oven at 3500 K. If iron boils at 3133 K and enthalpy of vaporization is 349 KJ mol –1, determine Ssystem, Ssurrounding and Suniverse.(Oven is considered as surroundings). B-4. Calculate the entropy change in surroundings when 1.00 mol of H2O( ) is formed under standard conditions at 298 K. Given rHº = – 286 kJ mol–1. B-5. Order of increasing of entropy among given condition of substance is : (I) 1 mole of H2O(l) at 298 K and 0.101 M Pa (II) 1 mole of ice at 273 K and 0.101 M Pa (III) 1 mole of H2(g) at 298 K and 1 atm (IV) 1 mole of C2H6(g) at 298 K and 1 atm B-6. Oxygen & ozone are gases at standard temperature. Their molar entropies are in the sequence O 2 < O3. Using molecular properties, explain why ozone is more disordered than oxygen. Section (C) : Free energy C-1. What will be the states of a chemical reaction occuring at constant pressure and temperature conditions when : (i) G = 0 (ii) G > 0 (iii) G < 0 C-2. A gaseous reactant A forms two different product, in parallel reaction, B and C as follows : A ⎯→ B; Hº = –3kJ, Sº = 20JK–1 ; A ⎯→ C; Hº = –3.6 kJ, Sº = 10 JK–1 Discuss the relative stability of B and C on the basis of Gibb’s free energy change at 27ºC. Thermodynamics & Thermochemistry C-3. With the following informations, determine standard Gibb’s free energy of formation of N 2O4(g). 1 1 N2(g) + O2(g) ⎯→ NO(g) Gº = 86.6 kJ .....(i) 2 2 1 NO(g) + O2(g) ⎯→ NO2(g) Gº = –34.82 kJ .....(ii) 2 2NO2(g) ⎯→ N2O4(g) Gº = –5.77 kJ .....(iii) C-4. For the equilibrium : PCl5(g) is G0for the reaction ? PCl3(g) + Cl2(g) at 298 K, equilibrium constant, K = 1.8 × 10–7. What PART - II : ONLY ONE OPTION CORRECT TYPE Section (A) : Introduction about entropy A-1. In which state, matter have highest entropy (A) Solid (B) Liquid (C) Gas (D) Equal in all A-2. Predict which of the following reaction (s) has a positive entropy change ? I. Ag+(aq) + Cl–(aq) ⎯→ AgCl(s) II. NH4Cl(s) ⎯→ NH3(g) + HCl(g) III. 2NH3(g) ⎯→ N2(g) + 3H2(g) (A) I and II (B) III (C) II and III (D) II A-3. Mixing of non-reacting ideal gases is generally accompanied by (A) Decrease in entropy (B) Increase in entropy (C) Change in enthalpy (D) Increase in free energy A-4. Which of the following reactions is associated with the most negative change in entropy ? (A) 2SO2(g) + O2(g) ⎯→ 2SO3(g) (B) C2H4(g) + H2(g) ⎯→C2H6(g) (C) C(s, graphite) + O2 ⎯→ CO2(g) (D) 3C2H2(g) ⎯→ C6H6( ) A-5. For the gas - phase decomposition, PCl5(g) (A) H < 0, S < 0 (B) H > 0, S > 0 A-6. PCl3(g) + Cl2(g) : (C) H > 0, S < 0 (D) H < 0, S > 0 Which one of the following has Sº greater than zero. CaCO3 (s) (A) CaO (s) + CO2(g) (B) NaCl (aq) NaCl (s) (C) NaNO3(s) Na+ (aq) + NO3– (aq) (D) N2 (g) + 3H2 (g) 2NH3 (g) A-7. For which reaction from the following, will be maximum entropy change : 1 (A) Ca(s) + O2 (g) → CaO(s) (B) CaCO3 (s) → CaO(s) + CO2 (g) 2 (C) C(s) + O2 (g) → CO2 (g) (D) N2 (g) + O2 (g) → 2NO(g) A-8. Which of the following statement is true. The entropy of the universe (A) Increases and tends towards maximum value (B) Decreases and tends to be zero (C) Remains constant (D) Decreases and increases with a periodic rate Section (B) : Entropy Calculation B-1. An isolated system comprises the liquid in equilibrium with vapours. At this stage the molar entropy of the vapour is : (A) Less than that of liquid (B) more than that of liquid (C) Equal to zero (D) Equal to that of liquid Thermodynamics & Thermochemistry B-2. B-3. 5   When two mole of an ideal gas  Cp,m = R  heated from 300 K to 600 K at constant pressure. The 2   change in entropy of gas (S) is : 3 3 5 (A) R ln 2 (B) – R ln 2 (C) 5R ln 2 (D) R ln 2 2 2 2 The entropy change when two moles of ideal monoatomic gas is heated from 200 to 300ºC reversibly and isochorically ? 3 5 3  300   573   573   573  (A) R ln  (B) R ln  (C) 3R ln  (D) R ln      2 2 2 200 273 473        473  5   B-4. If one mole of an ideal gas  Cp,m = R  is expanded isothermally at 300 K until it’s volume is tripled, 2   then change in entropy of gas is : 5 (A) zero (B) infinity (C) R ln 3 (D) R ln 3 2 B-5. Calculate the total entropy change for the transition at 368 K of 1 mol of sulphur from the monoclinic to the rhombic solid state and H = – 401.7 J mol–1 for the transition. Assume the surroundings to be an ice-water bath at 0ºC : (A) – 1.09 JK–1 (B) 1.47 JK–1 (C) 0.38 JK–1 (D) None of these Section (C) : Free energy C-1. Which of the follwing is incorrect regarding Gibbs free energy (A) it is a state function (B) it is extensive property (C) it is macroscopic property (D) it is intensive property C-2. Which of the following conditions regarding a chemical process ensures its spontaneity at all temperature ? (A) H > 0, S < 0 (B) H < 0, S > 0 (C) H < 0, S < 0 (D) H > 0, S < 0 C-3. When reaction in standard state is at equilibrium, then : (A) Hº = 0 (B) Sº = 0 (C) equilibrium constant K = 0 (D) equilibrium constant K = 1 C-4. What is the free energy change (G) when 1.0 mole of water at 100ºC and 1 atm pressure is converted into steam at 100ºC and 1 atm pressure ? (A) 80 cal (B) 540 cal (C) 620 cal (D) Zero C-5. A reaction has H = – 33 kJ and S = – 58 J/K. This reaction would be : (A) spontaneous at all temperatures (B) non-spontaneous at all temperatures (C) spontaneous above a certain temperature (D) spontaneous below a certain temperature C-6. For a reaction A(g) B(g) at equilibrium. The partial pressure of B is found to be one fourth of the partial pressure of A. The value of G° of the reaction A → B is (A) RT n 4 (B) – RT n 4 (C) RT log 4 (D) – RT log 4 3 O2 (g) ⎯→ Fe2O3 (s) 2 and Gº = – 19 K cal for (2) 4 Fe2O3 (s) + Fe(s) ⎯→3 Fe3O4 (s) What is the Gibbs free energy of formation of Fe3O4(s) ? (A) + 229.6 kcal/mol (B) – 242.3 kcal/mol (C) – 727 kcal/mol (D) – 229.6 kcal/mol C-7. If Gº = – 177 K cal for C-8. (1) 2 Fe(s) + For a paticular reaction H° = – 76.6 KJ and S° = 226 JK–1. This reaction is : (A) Spontaneous at all temperatures (B) Non spontaneous at all temperatures (C) Spontaneous at temperature below 66°C (D) Spontaneous at temperature above 66°C Thermodynamics & Thermochemistry PART - III : MATCH THE COLUMN 1. 2. Match the column: Columm-I (A) (Gsystem)T,P = 0 (B) Ssystem + Ssurrounding > 0 (C) Ssystem + S surrounding < 0 (D) (Gsystem)T,P > 0 (p) (q) (r) (s) Match the column : Columm-I (A) Reversible adiabatic compression (B) Reversible vaporisation of liquid (C) 2N(g) ⎯→ N2(g)  (D) → MgO(s) + CO2(g) MgCO3(s) ⎯⎯ Columm-II Process is in equilibrium Process is nonspontaneous Process is spontaneous System is unable to do useful work (p) (q) (r) (s) Columm-II Ssystem > 0 Ssystem < 0 Ssurrounding < 0 Ssurrounding = 0 PART - I : ONLY ONE OPTION CORRECT TYPE 1. Which of the following option the first compound has less entropy than second : (A) (i) aqueous solution of 1 M of MgCl2 (ii) aqueous solution of 1 M of NaCl (B) (i) Br2 liquid at 25ºC (ii) Br2 liquid at 20ºC (C) (i) HgO solid (ii) HgS solid (D) (i) Br2 liquid (ii) I2 solid 2. Isentropic process is (A) adiabatic and irreversible process (C) adiabatic and reversible process (B) isothermal and reversible process (D) isothermal and reversible for which Q = 0 3. Third law of thermodynamics states that : (A) the entropy of a perfectly crystalline pure substance at zero K is zero. (B) absolute entropy of hydrogen ion is zero at zero K. (C) net change in entropy in coverssion H2(g) (130 K) → H2(g) (200K) is zero. (D) entropy generally decrease in combustion reactions. 4. Select correct statements : S1 : For every chemical reaction at equilibrium, standard gibbs energy of reaction is zero S2 : At constant temperature and pressure, chemical reactions are spontaneous in the direction of decreasing gibbs energy. S3 : Spontaneity is related to change in entropy of universe. (A) S1 S2 S3 (B) only S1 (C) S2, S3 (D) S1, S3 PART - II : SINGLE AND DOUBLE VALUE INTEGER TYPE 1. The equilibrium constant for a reaction is 10. What will be the magnitude value of Gº ? R = 8.31 JK–1 mol–1, T = 314 K. (Approximate integer value in KJ mol–1) 2. For the reaction at 298 K A (g) + B (g) C (g) + D (g) If Hº = – 29.8 Kcal and Sº = – 0.1 Kcal K–1 then calculate equilibrium constant (K) : 3. Calculate the magnitude of free energy in KJ mol–1 when 1 mole of an ionic salt MX (s) is dissolved in water at 27°C. Given Lattice energy of MX = 780 kJ mol–1 Hydration energy of MX = – 775.0 kJ mol–1 Entropy change of dissolution at 27°C = 40 J mol–1 K–1 Thermodynamics & Thermochemistry 4. For the formation of C (g) at 300 K. A (g) + 3 B (g) ⎯→ 2C (g) Calculate the magnitude of Gº (Kcal) if given data : A B C 0 0 –10 Hfº (Kcal mol-1) –1 -1 40 30 45 Sfº (Cal K mol ) 5. The entropies of H2(g) and H(g) are 60 and 50 J mole–1 K–1 respectively at 300 K. Using the data given below calculate the bond enthalpy of H2 (g) in Kcal mole–1. H2 (g) ⎯→ 2H (g) ; Gº = 21.6 KJ mole–1 6. The free energy change for a reaction is –213.3 kJ mol–1 at 25ºC. If the enthalpy change of the reaction is –217.77 kJ mole–1. Calculate the magnitude of entropy change for the reaction in Joule mole –1. 7. Calculate the magnitude of standard entropy change for reaction X 10–7 at 300 K. 8. Calculate the magnitude of standard free energy of formation of ammonium chloride at 25ºC (approximate integer in Kcal mol–1), the equation showing the formation of NH 4Cl from its elements is ½N2(g) + 2H2(g) + ½Cl2(g) ⎯→ NH4Cl(s) For NH4Cl, Hºf is – 313 kJ mol–1, Also given that SH0 2 = 130.6 JK −1 mol−1 SN0 2 = 191.5 JK −1 mol−1 S0Cl2 = 223.0J K −1 mol−1 9. Y if Hº = 25 KJ and Keq is 0 SNH = 94.6 JK −1 mol−1 4 Cl 1 3 N2(g) + H2(g) ⎯→ NH3(g) ; H = – 30 kJ to be at equilibrium at 477ºC. If standard 2 2 entropy of N2 (g) and NH3 (g) are 60 and 50 J mole–1 K–1 respectively then calculate the standard entropy of H2(g) in Jmole–1K–1. For the reaction PART - III : ONE OR MORE THAN ONE OPTIONS CORRECT TYPE 1. 2. In which of the following entropy increases : (A) Fe(s) + O2(g) ⎯→ Fe2O3(s). (C) Crystallisation of sugar from solution (B) Melting of ice (D) Vaporisation of camphor When a liquid solidifies, generally, there is : (A) Decrease in enthalpy (C) Increase in enthalpy (B) Decrease in entropy (D) Increase in entropy 3. Which of the following is false about molar entropy ? (A) It is same for all type of gases (B) For the gas of comparable mass. It decreases with the increase in atomicity. (C) Under identical condition, it is greater for heavier gas. (D) For ideal gas of comparable molar mass it decreases with the increase in thermo molecular attractions. 4. Which of the following statements is/are correct. (A) Reversible adiabatic process is iso entropic process (B) Ssystem for irreversible adiabatic compression is greater than zero (C) Ssystem for free expension in zero (D) Ssurrounding for irreversible isothermal compression is greater than zero 5. The normal boiling point of a liquid `X` is 400 K. Which of the following statement is true about the process X(l) ⎯→ X(g)? (A) at 400 K and 1 atm pressure G = 0 (B) at 400 K and 2 atm pressure G = + ve (C) at 400 K and 0.1 atm presure G = – ve (D) at 410 K and 1 atm pressure G = + ve 6. For isothermal expansion in case of an ideal gas : (A) H = 0 (B) E = 0 (C) G = –T.S (D) Tfinal = Tinitial Thermodynamics & Thermochemistry PART - IV : COMPREHENSION Read the following passage carefully and answer the questions. Comperhension # 1 Entropy is a state function and its value depends on two or three variables temperature (T), Pressure (P) and volume (V). Entropy change for an ideal gas having number of moles (n) can be determined by the following equation. T  V  S = 2.303 nCV log  2  + 2.303 nR log  2  T  V1   1 T  P  S = 2.303 nCp log  2  + 2.303 nR log  1   T1   P2  Since free energy change for a process or a chemical equation is a deciding factor of spontaneity, which can be obtained by using entropy change (S) according to the expression, G = H – TS at a temperature T. 1. What would be the entropy change involved in thermodynamic expansion of 2 moles of a gas from a volume of 5 L to a volume of 50 L at 25°C [Given R = 8.3 J/mole – K] (A) 38.23 J/K (B) 26.76 J/K (C) 20J/K (D) 28.23J/K 2. An isobaric process having one mole of ideal gas has entropy change 23.03 J/K for the temperature range 27°C to 327°C. What would be the molar specific heat capacity (CV) ? 10 10 (A) J/K mol (B) – 8.3 J/ K mol log2 log2 (C) 10 × log2 J/K mol (D) 10 log2 + 8.3 J/K mol 3. For a reaction M2O(s) ⎯→ 2M(s) + 1 O2(g); H = 30 kJ/mol and S = 0.07 kJ/K-mol at 1 atm. Calculate 2 upto which temperature the reaction would not be spontaneous. (A) T > 428.6 K (B) T > 300.8 K (C) T < 300.8 K (D) T < 428.6 K Comprehension # 2 Dependence of spontaneity on temperature : For a process to be spontaneous, at constant temperature and pressure, there must be decrease in free energy of the system in the direction of the process, i.e. GP, T < 0. GP, T = 0 implies the equilibrium condition and GP, T > 0 corresponds to non-spontaneity. Gibbs-Helmholtz equation relates the free energy change to the enthalpy and entropy changes of the process as : ...(1) G P,T = H – TS The magnitude of H does not change much with the change in temperature but the entropy factor TS changes appreciably. Thus, spontaneity of a process depends very much on temperature. For endothermic process, both H and S are positive. The energy factor, the first factor of equation, opposes the spontaneity whereas entropy factor favours it. At low temperature the favourable factor TS will be small and may be less than H, G will have positive value indicating the nonspontaneity of the process. On raising temperature, the factor TS increases appreciably and when it exceeds H, G would become negative and the process would be spontaneous. For an exothermic process, both H and S would be negative. In this case the first factor of eq. 1 favours the spontaneity whereas the second factor opposes it. At high temperature, when TS > H, G will have positive value, showing thereby the non-spontaneity of the process. However, on decreasing temperature, the factor TS decreases rapidly and when TS < H, G becomes negative and the process occurs spontaneously. Thus, an exothermic process may be spontaneous at low temperature and non-spontaneous at high temperature. 4. When CaCO3 is heated to a high temperature, it undergoes decomposition into CaO and CO 2 whereas it is quite stable at room temperature. The most likely explanation of it, is (A) The enthalpy of reaction (H) overweighs the term TS at high temperature. (B) The term TS overweighs the enthalpy of reaction at high temperature. (C) At high temperature, both enthalpy of reaction and entropy change become negative. (D) None of these. Thermodynamics & Thermochemistry 5. For the reaction at 25°C, X2O4( ) ⎯→ 2XO2(g) H = 2.1 Kcal and S = 20 cal K–1. The reaction would be (A) spontaneous (B) non-spontaneous (C) at equilibrium (D) unpredictable 6. For the reaction at 298 K, 2A + B ⎯→ C H = 100 kcal and S = 0.050 kcal K–1. If H and S are assumed to be constant over the temperature range, above what temperature will the reaction become spontaneous ? (A) 1000 K (B) 1500 K (C) 2000 K (D) 2500 K 7. A reaction has a value of H = – 40 kcal at 400K. Above 400 K, the reaction is spontaneous, below this temperature, it is not. The values of G and S at 400 K are respectively. (A) 0, – 0.1 cal K–1 (B) 0,100 cal K–1 –1 (C) – 10 kcal, – 100 cal K (D) 0, – 100 cal K–1 8. The enthalpy change for a certain reaction at 300 K is – 15.0 K cal mol–1. The entropy change under these conditions is – 7.2 cal K–1 mol–1. The free energy change for the reaction and its spontaneous/non-spontaneous character will be (A) – 12.84 kcal mol–1, spontaneous (B) 12.84 kcal mol–1, non-spontaneous –1 (C) – 17.16 kcal mol , spontaneous (D) None of these * Marked Questions may have more than one correct option. PART - I : (ADVANCED) / PROBLEMS (PREVIOUS YEARS) 1. The direct conversion of A to B is difficult, hence it is carried out by the following shown path: S(A ⎯⎯→ C) = 50 ; S(C ⎯→ D) = 30; S(B ⎯→ D) = + 20 The entropy change for the process A ⎯→ B is (A) 100 (B) – 60 (C) – 100 2. N2 + 3H2 2 NH3 (D) + 60 K = 4 × 106 at 298 K = 41 at 400 K Which statements is correct ? (A) If N2 is added at equlibrium condition, the equilibrium will shift to the forward direction because according to IInd law of thermodynamics the entropy must increases in the direction of spontaneous reaction. (B) The condition for equilibrium is 2GNH3 = 3GH2 + GN2 where G is Gibbs free energy per mole of the gaseous species measured at that partial pressure. (C) Addition of catalyst does not change Kp but changes H. (D) At 400 K addition of catalyst will increase forward reaction by 2 times while reverse reaction rate will be changed by 1.7 times. 3. 4. The value of log10K for a reaction A B is : –1 (Given : rHº298 K = –54.07 kJ mol , rSº298 K = 10 JK–1 mol–1 and R = 8.314 JK–1 mol–1 ; 2.303 × 8.314 × 298 = 5705) (A) 5 (B) 10 (C) 95 (D) 100 For the process H2O( ) (1 bar, 373 K) ⎯→ H2O(g) (1 bar, 373 K), the correct set of thermodynamic parameters is : (A) G = 0, S = +ve (B) G = 0, S = –ve (C) G = +ve, S = 0 (D) G = –ve, S = +ve Thermodynamics & Thermochemistry 5. Statement-1 : For every chemical reaction at equilibrium, standard Gibbs energy of reaction is zero. Statement-2 : At constant temperature and pressure, chemical reactions are spontaneous in the direction of decreasing Gibbs energy. (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (C) Statement-1 is True, Statement-2 is False (D) Statement-1 is False, Statement-2 is True 6. Statement-1 : There is a natural asymmetry between converting work to heat and converting heat to work. Statement-2 : No process is possible in which the sole result is the absorption of heat form a reservoir and its complete conversion into work. (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (C) Statement-1 is True, Statement-2 is False (D) Statement-1 is False, Statement-2 is True 7.* For an ideal gas, consider only P-V work in going from an initial state X to the final state Z. The final state Z can be reached by either of the two paths shown in the figure. Which of the following choice(s) is (are) correct? [take S as change in entropy and w as work done]. (A) Sx→z = Sx→y + Sy→z (B) wx→z = wx→y + wy→z (C) wx→y→z = wx→y (D) Sx→y→z = Sx→y 8. For the process, H2O( ) ⎯→ H2O(g) at T = 100C and 1 atmosphere pressure, the correct choice is : (A) Ssystem > 0 and Ssurroundings > 0 (C) Ssystem < 0 and Ssurroundings > 0 (B) Ssystem > 0 and Ssurroundings < 0 (D) Ssystem < 0 and Ssurroundings < 0 9. Match the thermodynamic processes given under Column-I with the expressions given under Column-II. Column-I Column-II (A) Freezing of water at 273 K and 1 atm (P) q=0 Expansion of 1 mol of an ideal gas into a vacuum under isolated (B) (Q) w=0 conditions Mixing of equal volumes of two ideal gases at constant (C) (R) Ssys < 0 temperature and pressure in an isolated container Reversible heating of H2(g) at 1 atm from 300 K to 600 K, (D) (S) U = 0 followed by reversible cooling to 300 K at 1 atm (T) G = 0 10. One mole of an ideal gas at 300 K in thermal contact with surroundings expands isothermally from 1.0 L to 2.0 L against a constant pressure of 3.0 atm. In this process, the change in entropy of surroundings (Ssurr) in JK–1 is : (1 L atm = 101.3 J) (A) 5.763 (B) 1.013 (C) –1.013 (D) –5.763 11.* For a reaction taking place in a container in equilibrium with its surroundings, the effect of temperature on its equilibrium constant K in terms of change in entropy is described by (A) With increase in temperature, the value of K for endothermic reaction increases because unfavourable change in entropy of the surroundings decreases (B) With increase in temperature, the value of K for exothermic reaction decreases because favourable change in entropy of the surroundings decreases Thermodynamics & Thermochemistry (C) With increase in temperature, the value of K for endothermic reaction increases because the entropy change of the system is negative (D) With increase in temperature, the value of K for exothermic reaction decreases because the entropy change of the system is positive For a reaction, A P, the plots of [A] and [P] with time at temperature T1 and T2 are given below. 10 T1 T2 -1 [A]/(mol L ) -1 10 [P]/(mol L ) 12.* 5 T2 T1 5 Time Time If T2 > T1, the correct statement(s) is (are) (Assume H and S are independent of temperature and ratio of lnK at T1 to lnK at T2 is greater than T2 . Here H, S, G and K are enthalpy, entropy, Gibbs energy and equilibrium constant, respectively.) T1 (A) H < 0, S < 0 (C) G < 0, S < 0 13. (B) G < 0, H > 0 (D) G < 0, S > 0 The surface of copper gets tarnished by the formation of copper oxide. N 2 gas was passed to prevent the oxide formation during heating of copper at 1250 K. However, the N 2 gas contains 1 mole % of water vapour as impurity. The water vapour oxidises copper as per the reaction given below : 2Cu(s) + H2O(g) ⎯→ Cu2O(s) + H2(g) pH2 is the minimum partial pressure of H2 (in bar) needed to prevent the oxidation at 1250 K. The value of ln pH2 is_______. ( ) (Given: total pressure = 1 bar, R (universal gas constant) = 8 J K–1 mol–1, ln(10) = 2.3, Cu(s) and Cu2O(s) are mutually immiscible. 1 At 1250 K : 2Cu(s) + O2(g) ⎯→ Cu2O(s) ; G° = –78,000 J mol–1 2 1 H2(g) + O2(g) ⎯→ H2O(g) ; G° = –1,78,000 J mol–1; G is the Gibbs energy) 2 PART - II : (MAIN) / PROBLEMS (PREVIOUS YEARS) 1. Identify the correct statement regarding a spontaneous process : (1) Exothermic processes are always spontaneous. (2) Lowering of energy in the reaction process is the only criterion for spontaneity. (3) For a spontaneous process in an isolated system, the change in entropy is positive. (4) Endothermic processes are never spotaneous. 2. In conversion of lime-stone to lime, CaCO3(s) ⎯→ CaO(s) + CO2(g) the values of Hº and Sº are + 179.1 kJ mol–1 and 160.2 J/K respectively at 298 K and 1 bar. Assuming that H° and S° do not change with temperature, temperature above which conversion of limestone to lime will be spontaneous is : (1) 845 K (2) 1118 K (3) 1008 (4) 1200 K 3. Standard entropy of X2, Y2 and XY3 are 60, 40 and 50 JK–1 mol–1, respectively. 1 3 For the reaction, X2 + Y2 ⎯→ XY3 H = – 30 kJ. To be at equilibrium the temperature will be : 2 2 (1) 500 K (2) 750 K (3) 1000 K (4) 1250 K Thermodynamics & Thermochemistry 4. For a particular reversible reaction at temperature T, H and S were found to be both +ve. If Te is the temperature at equilibrium, the reaction would be spontaneous when. (1) Te > T (2) T > Te (3) Te is 5 times T (4) T = Te 5. The entropy change involved in the isothermal reversible expansion of 2 moles of an ideal gas from a volume of 10 dm3 to a volume of 100 dm3 at 27°C is : (1) 38.3 J mol–1 K–1 (2) 35.8 J mol–1 K–1 (3) 32.3 J mol–1 K–1 (4) 42.3 J mol–1 K–1 6. In view of the signs of rGº for the following reactions : PbO2 + Pb ⎯→ 2PbO, rGº < 0 ; SnO2 + Sn ⎯→ 2SnO, rGº > 0, which oxidation states are more characteristics for lead and tin ? (1) For lead +2, for tin +2 (2) For lead +4, for tin +4 (3) For lead +2, for tin +4 (4) For lead +4, for tin +2 7. The incorrect expression among the following is : Gsystem V (1) =–T (2) In isothermal process, wreversible = –nRT ln f Stotal Vi (3) lnK = 8. Hº −TSº RT The following reaction is performed at 298 K. 2NO(g) + O2(g) 2NO2(g) The standard free energy of formation of NO(g) is 86.6 kJ/mol at 298 K. What is the standard free energy of formation of NO2(g) at 298 K ? (KP = 1.6 × 1012) (1) R(298) in (1.6 × 1012) – 86600 (2) 86600 + R(298) ln (1.6 x 1012) (3) 86600 – 9. (4) K = e–Gº/RT ln (1.6  1012 ) R (298) (4) 0.5 [2 × 86,600 – R (298) ln (1.6 × 1012)] The combustion of benzene (I) gives CO 2(g) and H2O(l). Given that heat of combustion of benzene at constant volume is –3263.9 kJ mol–1 at 25ºC; heat of combustion (in kJ mol–1) of benzene at constant pressure will be : (R = 8.314 JK–1 mol–1) (1) 3260 (2) –3267.6 (3) 4152.6 (4) –452.46 MAIN ONLINE PROBLEMS 1. The molar heat capacity (Cp) of CD2O is 10 cals at 1000 K. The change in entropy associated with cooling of 32 g of CD2O vapour from 1000 K to 100 K at constant pressure will be : (D = deuterium, at. mass = 2 u) (1) 23.03 cal deg–1 (2) – 23.03 cal deg–1 (3) 2.303 cal deg–1 (4) – 2.303 cal deg–1 2. Then entropy (Sº) of the following substances are : CH4(g) 186.2 JK–1 mol–1, O2(g) 205.0 JK–1 mol–1 –1 –1 CO2(g) 213.6 JK mol , H2O( ) 69.9 JK–1 mol–1 The entropy change (Sº) for the reaction CH4(g) + 2O2(g) → CO2(g) + 2H2O( ) is : (1) –312.5 JK–1 mol–1 3. (3) –108.1 JK–1 mol–1 (4) –37.6 JK–1 mol–1 For the reaction, A(g) + B(g) → C(g) + D(g), Hº and Sº are, respectively, –29.8 kJ mol–1 and –0.100 kJ K–1 mol–1 at 298 K. The equilibrium constant for the reaction at 298 K is: (1) 1 4. (2) –242.3 JK–1 mol–1 (2) 10 (3) 1.0 × 10–10 (4) 1.0 × 1010 A reaction at 1 bar is non-spontaneous at low temperature but becomes spontaneous at high temperature. Identify the correct statement about the reaction among the following: (1) Both H and S are positive. (3) H is positive while S is negative. (2) H is negative while S is positive. (4) Both H and S are negative. Thermodynamics & Thermochemistry 5. fG° at 500 K for substance 'S' in liquid state and gaseous state are +100.7 kcal mol –1 and +103 kcal mol–1, respectively. Vapour pressure of liquid 'S' at 500 K is approximately equal to : (R = 2 cal K–1 mol–1) (1) 0.1 atm (2) 1 atm (3) 10 (4) 100 atm 6. Given (i) 2Fe2O3(s) → 4Fe(s) + 3O2(g) ; r G° = + 1487.0 kJ mol–1 (ii) 2CO(g) + O2(g) → 2CO2(g) ; rG° = –514.4 kJ mol–1 Free energy change, r G° for the reaction 2Fe2O3(s) + 6CO(g) → 4Fe(s) + 6CO2(g) will be (1) –112.4 kJ mol–1 (2) –56.2 kJ mol–1 (3) –168.2 kJ mol–1 (4) –208.0 kJ mol–1 7. For which of the following processes, S is negative ? (1) C(diamond) → C(graphite) (2) N2(g, 1 atm) → N2(g, 5 atm) (3) N2(g, 273 K) → N2(g, 300 K) (4) H2(g) → 2H(g) 8. The entropy change associated with the conversion of 1 kg of ice at 273 K to water vapours at 383 K is: (Specific heat of water liquid and water vapour are 4.2 kJ K –1kg–1 and 2.0 kJ K–1 kg–1; heat of liquid fusion and vapourisation of water are 334 kJ kg –1 and 2491 kJ kg–1, respectively). (log 273 = 2.436, log 373 = 2.572, log 383 = 2.583) (1) 8.49 kJ kg–1 K–1 (2) 9.26 kJ kg–1 K–1 (3) 2.64 kJ kg–1 K–1 (4) 7.90 kJ kg–1 K–1 9. A process has H = 200 J mol–1 and S = 40 JK–1 mol–1. Out of the values given below, choose the minimum temperature above which the process will be spontaneous : (1) 5 K (2) 12 K (3) 4 K (4) 20 K 10. The process with negative entropy change is: (1) Sublimation of dry ice (2) Dissolution of iodine in water (3) Dissociation of CaSO4(s) to CaO(s) and SO3(g) (4) Synthesis of ammonia from N2 and H2 11. Two blocks of the same metal having same mass and at temperature T1 and T2, respectively, are brought in contact with each other and allowed to attain thermal equilibrium at constant pressure. The change in entropy, S, for this process is: 1    T1 + T2   (T1 + T2 ) 2   (1) 2CP In  (2) 2CP In    4T1T2   T1T2     (T1 + T2 )2   T + T2  (3) CP In  (4) 2CP In  1   4 T T 1 2  2T1T2    12. For the chemical reaction X K) as rG°(in kJ mol–1) = 120 – (1) Y is T = 280 K 13. Y, the standard reaction Gibbs energy depends on temperature T (in 3 T . The major component of the reaction mixture at T is : 8 (2) Y is T = 300 K (3) X if T = 315 K (4) X if T = 350 K The standard reaction Gibbs energuy for a chemical reaction at an absolute temperature T is given by G° = A – BT, where A and B are non-zero constants. Which of the following is TRUE about this reaction ? (1) Endothermic if A > 0 (2) Exothermic if B < 0 (3) Exothermic if A > 0 and B < 0 (4) Endothermic if A < 0 and B > 0 Thermodynamics & Thermochemistry 14. For the equilibrium 2H2O (1) 80 kJ mol–1 15. H3O+ + OH–, the value of G° at 298 K is approximately : (2) –100 kJ mol–1 (3) –80 kJ mol–1 (4) 100 kJ mol–1 The reaction, MgO(s) + C(s) → Mg(s) + CO(g), for which rH° = +491.1 kJ mol–1 and rS° = 198.0 JK–1 mol–1, is not feasible at 298 K. Temperature above which reaction will be feasible is: (1) 2380.5 K (2) 1890.0 K (3) 2040.5 K (4) 2480.3 K Thermodynamics & Thermochemistry EXERCISE - 1 PART – I A-1. Because larger space creates more disorder. A-2. Dissolution of solute makes the motion of solute particles free. Also the increased number of particles in mixed state increases disorderness. A-3. (a) Ssys is positive; Ssurr is negative, (c) Ssys is positive; Ssurr is negative. A-4. Entropy of the universe is constantly increasing. Suniverse = Ssystem + Ssurrounding > 0 for a spontaneous process. A-5. At 0 K, entropy of a perfect crystalline pure substance is taken to be zero. A-6. GT,P = (–)ve A-7. G = H – TS For spontaneous reaction G = (–) ve  S Should be positive B-1. – 1.5 × 85 J/K B-2. – 215.2 JK–1 mol–1 B-3. Ssystem = 111.4 JK–1, B-4. Entropy change in surroundings = 959.73 JK–1. B-5. II < I < III 0 Ssurr = –99.71 JK–1, Suniv = +11.69 JK–1 (ii) non-spontaneous (iii) Spontaneous. PART - II A-1. (C) A-2. (C) A-3. (B) A-4. (D) A-5. (B) A-6. (C) A-7. (B) A-8. (A) B-1. (B) B-2. (C) B-3. (C) B-4. (D) B-5. (C) C-1. (D) C-2. (B) C-3. (D) C-4. (D) C-5. (D) C-6. (A) C-7. (B) C-8. (A) PART - III 1. (A) → (p, s); (B) → (r) ; (C) → (q, s) ; (D) → (q, s). 2. (A) → (s); (B) → (p, r); (C) → (q); (D) → (p, r) Thermodynamics & Thermochemistry EXERCISE – 2 PART - I 1. (C) 2. (C) 3. (A) 4. (C) PART - II 1. 6 2. 1 3. 7 kJ mol–1. 4. 8 6. 15 7. 51 8. 48 9. 40 4. 5. 8 (ABD) 5. (ABC) 4. (B) 5. (A) 4. (A) 5. (D) 10. (C) 11. (AB) 5. (1) PART - III 1. (BD) 6. (ABCD) 2. (AB) 3. (ABD) PART - IV 1. (A) 2. (B) 3. (D) 6. (C) 7. (D) 8. (A) EXERCISE – 3 PART - I 1. (D) 2. (B) 3. (B) 6. (A) 7. (AC) 8. (B) 9. (A-R,T); (B-P,Q,S); (C-P,Q,S); (D-P,Q,S,T) 12. (AC) 13. –14.6 PART - II MAIN OFFLINE PROBLEMS 1. (3) 2. (2) 3. (2) 4. (2) 6. (3) 7. (3) 8. (4) 9. (2) MAIN ONLINE PROBLEMS 1. (2) 2. (2) 3. (1) 4. (1) 5. (1) 6. (2) 7. (2) 8. (2) 9. (1) 10. (4) 11. (3) 12. (3) 13. (1) 14. (1) 15. (4)