### Thermodynamics

1. Equilibrium constant for the following reaction at 298 K
2 NH3 (g) + CO2 (g)    NH2CONH2 (aq.) + H2O (l) is
(∆rG° = −13.6 KJ mol−1)
1) 2.38             2) 2.4 × 102          3) 13.6        4) 4.76
Ans: (2)

2. The reaction is spontaneous at all the temperatures is
1) ∆H = −ve , ∆S = −ve, ∆G = −ve

2) ∆H = +ve, ∆S = +ve, ∆G = −ve
3) ∆H = −ve, ∆S = +ve, ∆G = −ve

4) ∆H = −ve, ∆S = −ve, ∆G = +ve
Ans: (3)
Explanation: If ∆H = −ve, ∆S = +ve, ∆G = −ve, the reaction is spontaneous at all the temperatures.

3. The correct mathematical relation in terms of ∆G, ∆H and ∆S is

Ans: (4)

4. For an insulated system ∆U = 0. So ∆S will be
1) −ve         2) +ve     3) 0         4) we can’t say
Ans: (2)
Explanation: ∆H = ∆U + ∆ nRT, ∆G = ∆H − T ∆S
∵  ∆U = 0 ∆S will be +ve.

1) The entropy of a pure and perfectly crystalline substance is zero at −273°C.
2) The energy is neither created nor destroyed.
3) Entropy is neither created nor destroyed.
4) Heat can’t flow from colder body to hotter body by its own.
Ans: (1)
Explanation: It is formula for 3rd law of thermodynamics.

6. Incorrect statement among the following is
1) Internal energy is state function.
2) Work is not a state function.
3) Temperature is a path function as well as extensive property.
4) Volume is state function as well as extensive property.
Ans: (3)
Explanation: Temperature is a state function and intensive property.

7. A swimmer coming out from a river is covered with film of water weighing 36 g. The amount of heat given by the sun to evoparate this water at 298 K is

1) 37.56 KJ mol−1          2) 34.46 KJ mol−1           3) 34.46 J mol−1           4) 37.56 J mol−1
Ans: (2)

8. The correct order of standard molar enthalpies of C (graphite), H2 gas, Cl2 gas, Br2 liquid is
1) Br2 (l) > Cl2 (g) > H2 (g) > C (graphite)
2) C (graphite) > H2 (g) > Cl2 (g) > Br2 (l)
3) C (graphite) > Br2 (l) > Cl2 (g) > H2 (g)
4) C (graphite) = H2 (g) = Cl2 (g) = Br2 (l)
Ans: (4)
Explanation: C (graphite), Br2, Cl2 and H2 are in their standard states. So thier standard molar enthalpies are also zero.

9. Entropy decreases in the case of
1) H2 (g) 2 H (g)
2) 2 NaHCO3 (s) Na2CO3 (s) + CO2 (g) + H2O (g)
3) A liquid crystallises to a solid
4) Temperature of a crystalline solid is raised from 0 K to 107 K
Ans: (3)
Explanation: After crystallisation, the molecules attain an ordered state and entropy decreases.

10. The bond enthalpy (∆H) of C − Cl in CCl4 is

1) 1304 KJ mol−1       2) 326 KJ mol−1      3) 1302 KJ mol−1          4) 258 KJ mol−1
Ans: (2)

11. The change in internal energy if heat is absorbed and work is done on the system is
1) ∆E = Q + W            2) ∆E = −W − Q           3) ∆E = W −Q               4) ∆E = Q − W
Ans: (1)
Explanation: Q = +ve, W = +ve.

12. 50% of N2O4 dissociates at 333 K. ∆rG° for this change will be (in KJ mol−1)
1) 763.8                       2) −76.38                        3) −7638                         4) −763.8
Ans: (4)

13. The entropy change is positive among the following reactions is
1)   2) Na+ (g) + Cl (g)  NaCl (s)
3) NaCl (l NaCl (s)                                      4) H2O (l H2O (g)
Ans: (4)
Explanation: Disorderness of gaseous H2O molecules i.e. entropy is more than liquid H2O molecules.

14. The enthalpy of formation of NH3 is −46 KJ mol−1. The enthalpy change for the reaction
2 NH3 (g)  2 N2 (g) + 3 H2 (g) is
1) 46 KJ mol−1              2) 92 KJ mol−1             3) −92 KJ mol−1         4) −23 KJ mol−1
Ans: (2)

15. In the dissociation of CH4 (g)
1) 3 bonds have equal energy.
2) 2 bonds have equal energy.
3) bond energies are different for all the 4 bonds.
4) bond energies are euqal for all the 4 bonds.
Ans: (3)
Explanation: Bond energies for all the 4 bonds (C − Cl) are different.

16. The heat of combustion of solid benzoic acid at constant volume is −321.3 KJ mol−1 at 300 K. The heat of a combustion at constant pressure is
1) −321.3 − 150 R        2) −321.3 − 300 R         3) −321.3 + 300 R       4) −321.3 + 450 R
Ans: (1)

17. If H+ + OH-  H2O + 57.3 K.J., then the heat of neutralization of 1 gram mole of H2SO4 with NaOH will be
1) 28.7 K.J.                    2) 57.3 K.J.                  3) 114.6 K.J.                   4) 14.3 K.J.
Ans: (3)
Explanation: 1 mole of H2SO4 = 2 equivalents of H2SO4 = 2 moles of H2O = 2 × 57.3 = 114.6 KJ.

18. If one mole of NH3 and 1 mole of HCl are mixed in a closed container to form NH4Cl vapour, then
1) ∆H > ∆U                   2) ∆H < ∆U                  3) ∆H = ∆U                    4) ∆H = ∆U = 0
Ans: (2)

19. Bond dissociation enthalpy of H2, Cl2 and HCl are 434, 242 and 431 K.J. mol-1 respectively. Enthalpy of formation of HCl is
1) 93 KJ mol-1      2) −245 KJ mol-1      3) 245 KJ mol-1    4) -93 KJ mol-1
Ans: (4)

20. The decomposition of lime stone is non spontaneous at 298 K. The ∆H° and ∆S° values for the reaction are 176 KJ and 160 JK-1 mol-1 respectively. At what temperature, the decomposition becomes spontaneous?
1) above 827°C      2) below 500°C          3) at 500°C     4) at 1000° K
Ans: (1)

21. The enthalpy (H) of water, when it is super cooled to −4°C is
1) same as ice at 0°C          2) less than ice at −4°C
3) same as ice at −4°C       4) more than ice at −4°C
Ans: (2)
Explanation: When ice is cooled upto -4°C, heat is released and enthalpy is decreased.

22. The enthalpy and entropy change for the reaction Br2 (l) + Cl2 (g)  2 BrCl (g) are 30 KJ mol−1 & 105 JK−1 mol−1respectively. The temperature at which the reaction will be in equilibrium is
1) 450 K          2) 300 K        3) 285.7 K       4) 273 K
Ans: (3)

23. When Zn dust is added to aq. solution of CuSO4, 3.175 g of Cu metal and 20 J of heat is evolved. The ∆H for this reaction is
1) 100 J        2) 20 J       3) 200 J        4) 400 J
Ans: (4)

24. 1 mole of perfect gas expands isothermally to 10 times its original volume. The change in entropy is
1) 100 R       2) 2.303 R    3) 10 R           4) 4.606 R
Ans: (2)

25. If the bond enthalpies of H - H, Br - Br and H - Br are 433, 192 and 364 KJ mol−1 respectively, the ∆H° for the reaction
H2 (g) + Br2 (g)  2 HBr (g) is
1) +261 KJ      2) -261 KJ        3) +103 KJ     4) -103 KJ
Ans: (4)
Explanation:rH = Σ B.E. of reactants - Σ B.E. of products
= (433 + 192) - 2(364) = -103 KJ.

Posted Date : 15-10-2020