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Q.1. For the reaction
A(l) → 2B(g)
ΔU = 2.1 kcal, ΔS = 20 cal K^{−1} at 300 K.
Hence ΔG in kcal is ______. (2020)
Ans. (2.70)
For the reaction
A(l) → 2B(g)
Δn_{g} = 2 − 0 = 2
From the relation
ΔH = ΔU + Δn_{g}RT ...(1)
Substituting the values in Eq. (1), we get
ΔH = 2.1 × 10^{3} + 2 × 2 × 300 = 3300 cal
We know
ΔG = ΔH − T ΔS ...(2)
Substituting the values in Eq. (2), we get
ΔG = 3300  300 × 20
= 2700 cal =  2.7 kcal
Q.2. The standard heat of formation (Δ_{f}H^{0}_{298}) of ethane (in kJ mol^{–1}), if the heat of combustion of ethane, hydrogen and graphite are −1560, −393.5 and −286 kJ mol^{–1}, respectively is ______. (2020)
Ans. (192.5)
We know,
Standard heat of formation of ethane is Δ_{f}H
Given:
...(1)
...(2)
...(3)
On applying algebraic operations on equation, that is, 2 × Eq. (3) + 3 × Eq. (2) – Eq. (1), We get,
Q.3. The magnitude of work done by a gas that undergoes a reversible expansion along the path ABC shown in the figure is ________. (2020)
Ans. (4)
The magnitude of word done undergo a reversible expansion along the path ABC is the are of the trapezium ABC.
Thus, the word done is
Q.4. At constant volume, 4 mol of an ideal gas when heated from 300 K to 500 K changes its internal energy by 5000 J. The molar heat capacity at constant volume is _____. (2020)
Ans. (6.25)
We know
ΔU = nC_{V,m}ΔT ...(1)
Substituting the given values in Eq. (1), we get
Q.5. If enthalpy of atomization for Br_{2}(l) is x kJ mol^{–1} and bond enthalpy for Br_{2} is y kJ mol^{–1}, the relation between them (2020)
(1) is x = y
(2) Does not exist
(3) is x > y
(4) is x < y
Ans. (3)
Given, enthalpy of atomization of B_{2}(l) is x kJ mol^{–1} and bond enthalpy of B_{2}(l) is y kJ mol^{–1}
From the above two reactions
x = Δ_{vap} H + y
⇒ x > y.
Q.6. The true statement amongst the following is (2020)
(1) Both ΔS and S are functions of temperature.
(2) Both S and ΔS are not functions of temperature.
(3) S is not a function of temperature but ΔS is a function of temperature.
(4) S is a function of temperature but ΔS is not a function of temperature.
Ans. (1)
Both entropy (S) and change in entropy (ΔS) is the function of temperature.
Q.7. Consider the reversible isothermal expansion of an ideal gas in a closed system at two different temperatures T_{1} and T_{2} (T_{1} < T_{2}). The correct graphical depiction of the dependence of work done (w) on the final volume (V) is: (2019)
(1)
(2)
(3)
(4) 
Ans. (2)
⇒
w = nRT (InV_{2}  InV_{1})
w = nRT In V_{2}  nRT V_{1}
y = mx + c
So, slope of curve 2 is more than curve 1 and intercept of curve 2 is more negative than curve 1.
Q.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^{1} kg^{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) (2019)
(1) 7.90 kJ kg^{1 }K^{1}
(2) 2.64 kJ kg^{1} K^{1}
(3) 8.49 kJ kg^{1} K^{1}
(4) 9.26 kJ kg^{1} K^{1}
Ans. (4)
As we know,
Also,
Also,
Now,
∴ Total entropy change
ΔS = 9.26 kJ kg^{1} K^{1}
Q.9. A process has ΔH = 200 J mol^{1} and ΔS = 40 JK^{1} mol^{1}. Out of the values given below, choose the minimum temperature above which the process will be spontaneous: (2019)
(1) 20 K
(2) 12 K
(3) 5 K
(4) 4 K
Ans. (3)
ΔH = 200 J mol^{1}
ΔS = 40 JK^{1} mol^{1}
For spontaneous reaction,
ΔG < 0
ΔH  TΔS < 0; ΔH < TΔS
5 < T
So, minimum temperature is 5 K.
Q.10. The process with negative entropy change is: (2019)
(1) Dissociation of CaSO_{4}(s) to CaO(s) and SO_{3}(g)
(2) Sublimation of dry ice
(3) Dissolution of iodine in water
(4) Synthesis of ammonia from N_{2} and H_{2}
Ans. (4)
In the process of synthesis of ammonia from N_{2} and H_{2}, number of moles decreases which implies that the change in entropy will be negative.
N_{2}(g) + 3H_{2}(g) → 2NH_{3}(g)
Q.11. An ideal gas undergoes isothermal compression from 5 m^{3} to 1 m^{3} against a constant external pressure of 4 Nm^{2}. Heat released in this process is used to increase the temperature of 1 mole of Al. If molar heat capacity of Al is 24 J mol^{1} K^{1}, the temperature of Al increases by: (2019)
(1) 3/2 K
(2) 2 K
(3) 2/3 K
(4) 1 K
Ans. (3)
We know that,
w = P_{ext}(V_{f}  V_{i})
w = 4 Nm^{2}(1  5)m^{3}
w = 16 Nm ⇒ 16 J
For isothermal compression,
ΔU = q + w
⇒ q = w = 16 J (∵ ΔU = 0 for isothermal process)
From calorimetry,
Heat given = n C ΔT
∴ Change in temperature, ΔT = 2/3 K
Q.12. Two blocks of the same metal having same mass and at temperature T_{1}, and T_{2}, respectively, are brought in contact with each other and allowed to attain thermal equilibrium at constant pressure. The change in entropy, ΔS, for this process is: (2019)
(1)
(2)
(3)
(4)
Ans. (1)
Final temperature
Q.13. For the chemical reaction X ⇌ Y, the standard reaction Gibbs energy depends on temperature T (in K) as
The major component of the reaction mixture at T is: (2019)
(1) Y if T= 300 K
(2) Y if T= 280 K
(3) X if T = 350 K
(4) X if T = 315 K
Ans. (4)
At 315 K; ΔGº
ΔG° = 120  118.125 = positive
Since ΔG° is positive then K_{e}_{q} < 1.
Q.14. The standard reaction Gibbs energy for a chemical reaction at an absolute temperature T is given by ΔG° = A  BT where A and B are nonzero constants. Which of the following is true about this reaction? (2019)
(1) Exothermic if B < 0
(2) Endothermic if A > 0
(3) Endothermic if A < 0 and B > 0
(4) Exothermic if A > 0 and B < 0
Ans. (3)
MgO (s) + C(s) → Mg (s) + CO (g)
For a reaction to be spontaneous
ΔG < 0
ΔH°  TΔS° < 0
T > 2480.3 K
Q.15. For a diatomic ideal gas in a closed system, which of the following plots does not correctly describe the relation between various thermodynamic quantities? (2019)
(1)
(2)
(3)
(4)
Ans. (2)
ΔG° = A  BT
A and B are nonzero constants
∴ ΔG° = ΔH°  TΔS° = A  BT
∴ Reaction will be endothermic if ΔG° > 0.
Hence, A > O and B < O.
Q.16. Given:
Based on the above thermochemical equations, find out which one of the following algebraic relationships is correct? (2019)
(1) x = y + z
(2) z = x + y
(3) y = 2z  x
(4) x = y  z
Ans. (1)
(ΔT_{f})x = (ΔT_{f})_{y}
K_{f}.m_{x} = K_{f}.m_{y}
m_{y} = 3.27 A; y ≈ 3A
Q.17. For silver, C_{p}(JK^{1} mol^{1}) = 23 + 0.01T. If the temperature (T) of 3 moles of silver is raised from 300 K to 1000 K at 1 atm pressure, the value of ΔH will be close to: (2019)
(1) 62 kJ
(2) 16 kJ
(3) 21 kJ
(4) 13 kJ
Ans. (1)
Given: n = 3
T_{1} = 300; T_{2} = 1000
C_{p} =23 + o.01T
The relation between ΔH and C is
After putting all variable values in eq. (i)
= 3[16100 + 4550] = 3 × 20650 = 61950 J
= 61.95 kJ
= 62 kJ
Q.18. Which one of the following equations does not correctly represent the first law of thermodynamics for the given processes involving an ideal gas ? (Assume nonexpansion work is zero) (2019)
(1) Cyclic process : q = w
(2) Adiabatic process: ΔU= w
(3) Isochoric process: ΔU = q
(4) Isothermal process: q = w
Ans. (2)
From first law of thermodynamics, ΔU = q + w
For adiabatic process, q = 0
∴ ΔU = w
For isothermal process, ΔU = 0
For cyclic process, ΔU = 0
For isochoric process, w = 0
Q.19. 5 moles of an ideal gas at 100 K are allowed to undergo reversible compression till its temperature becomes 200 K. If C_{v} = 28 J K^{1} mol^{1}, calculate ΔU and ΔPV for this process. (R = 8.0 J K^{1} mol^{1}) (2019)
(1) ΔU = 14 kJ; Δ(PV) = 18 kJ
(2) ΔU = 14 kJ; Δ(PV) = 0.8 kJ
(3) ΔU = 14 kJ; Δ(PV) = 4 kJ
(4) ΔU = 14 kJ; Δ(PV) = 8.0 kJ
Ans. (3)
ΔU = nCv ΔT = 5 × 28 × 100 = 14 KJ
Δ(PV) = nR (T_{2}  T_{1}) = 5 × 8 × 100 = 4 KJ
Q.20. Among the following, the set of parameters that represents path functions, is: (2019)
(a) q + w
(b) q
(c) w
(d) HTS
(1) (B) and (C)
(2) (B), (C) and (D)
(3)(A) and (D)
(4) (A), (B) and (C)
Ans. (1)
We know that heat and work are not state functions but q + w = ΔU is a state function. H TS (i.e. G) is also a state function.
Q.21. Consider the given plot of enthalpy of the following reaction between A and B, A + B → C + D.
Identify the incorrect statement. (2019)
(1) Activation enthalpy to form C is 5kJ mol^{1} less than that to form D.
(2) C is the thermodynamically stable product.
(3) D is kinetically stable product.
(4) Formation of A and B from C has highest enthalpy of activation.
Ans. (1)
As we can see from the graph that activation enthalpy to form D from A+B is 15  5 = 10 kJ mol^{1}, whereas, to form C from A + B is 20  5 = 15 kJ mol^{1}. Therefore, activation enthalpy to form C is 5 kJ more than that to form D.
Q.22. During compression of a spring the work done is 10 kJ and 2 kJ escaped to the surroundings as heat. The change in internal energy, ΔU (in kJ) is: (2019)
(1) 12
(2) 8
(3) 8
(4) 12
Ans. (3)
w= 10 kJ
q = 2 kJ
ΔU = q + w = 2 + 10 = 8 kJ
Q.23. A process will be spontaneous at all temperatures if: (2019)
(1) ΔH < 0 and ΔS < 0
(2) ΔH > 0 and ΔS < 0
(3) ΔH < 0 and ΔS > 0
(4) ΔH > 0 and ΔS > 0
Ans. (3)
A reaction is spontaneous if ΔG_{sys} is negative.
ΔG_{sys} = ΔH_{sys} TΔS_{sys}
A reaction will be spontaneous at all temperatures if ΔH_{sys} is negative and ΔS_{sys} is positive.
Q.24. The difference between ΔH and ΔU (ΔHΔU), when the combustion of one mole of heptane (I) is carried out at a temperature T, is equal to: (2019)
(1) 4 RT
(2) 3RT
(3) 4RT
(4) 3 RT
Ans. (1)
ΔH  ΔU = Δn_{g}RT
Δn_{g} = no. of moles of product in gaseous state  no. of moles of reactant in gaseous state.
∵ Δn_{g} = 4
∴ ΔH  ΔU = 4RT
Q.25. An ideal gas is allowed to expand from 1 L to 10 L against a constant external pressure of 1 bar. The work done in kJ is: (2019)
(1) 9.0
(2) +10.0
(3) 0.9
(4) 2.0
Ans. (3)
w =  PΔV
= (1 bar) x (9 L)
= (10^{5} Pa) x (9 x 10^{3}) m^{3}
=  9 x 10^{2} N.m
=  900 J
=  0.9 kJ
Q.26. Enthalpy of sublimation of iodine is 24 cal g^{1} at 200°C. If specific heat of I_{2}(s) and I_{2}(vap) are 0.055 and 0.031 cal g^{1}K^{1} respectively, then enthalpy of sublimation of iodine at 250°C in cal g^{1} is: (2019)
(1) 2.85
(2) 5.7
(3) 22.8
(4) 11.4
Ans. (3)
I_{2}(s) → I_{2}(g).
Heat of reaction depend upon temperature i.e., it varies with temperature, as given by Kirchoff's equation,
where ΔC_{p} = C_{p} of product  C_{p} of reactant
∴ ΔC_{p} = 0.031  0.055 =  0.024 cal/g
ΔH_{(250)}  ΔH_{(200)} = 0.024(523  473)
ΔH_{(250)} = 24  50 x 0.024 = 22.8 cal/g
Q.27. The INCORRECT match in the following is: (2019)
(1) ΔG° < 0, K > 1
(2) ΔG° = 0, K = 1
(3) ΔG° > 0, K < 1
(4) ΔG° < 0, K < 1
Ans. (4)
ΔGº = RT Ink
∴ If K > 1 then ΔGº < 0
If K < 1 then ΔGº > 0
If K = 1 then ΔGº = 0
Q.28. Which of the following lines correctly show the temperature dependence of equilibrium constant K, for an exothermic reaction? (2018)
(1) A and B
(2) B and C
(3) C and D
(4) A and D
Ans. (2)
ΔGº = RT In K
ΔHº  TΔSº = RT In K
Therefore In K vs 1/T graph will be a straight line with slope equal to Since reaction is exothermic, therefore ΔH^{0} itself will be negative resulting in positive slope.
Q.29. The combustion of benzene (1) gives CO_{2}(g) and H_{2}O(I). Given that heat of combustion of benzene at constant volume is 3263.9 kJ mol^{1} at 25^{0} C; heat of combust ion (in kJ mol^{1}) of benzene at constant pressure will be: (2018)
(1) 4152.6
(2) 452.46
(3) 3260
(4) 3267.6
Ans. (4)
Δng = 67.5 = 1.5
ΔH = ΔU + ΔngRT
= 3263.9  1.5 x 8.134 x 10^{3} x 298
= 3267.6
Q.30. An ideal gas undergoes a cyclic process as shown in Figure.
ΔU_{BC} = −5 KJ mol^{−1}, q_{AB} = 2KJ mol^{−1}
W_{AB }= −5KJ mol^{−1}, W_{CA} = 3KJ mol^{−1}
Heat absorbed by the system during process CA is: (2018)
(1) 18 KJ mol^{1}
(2) 5 kJ mol^{1}
(3) +5 KJ mol^{1}
(4) 18 KJ mol^{1}
Ans. (3)
AB → isobaric
BC → Isochoric
CA → not defined
ΔU_{AB }= q + W
= 2 − 5 = −3
ΔU_{ABC}=ΔU_{AB}+ΔU_{BC}
=− 3 − 5= − 8 kJ
ΔU_{CBA }= +8
= Q + W
8 = Q + 3
Q = +5 kJ
Q.31. For which of the following processes, ΔS is negative? (2018)
(1) C(diamond) → C(graphite)
(2) N_{2}(g, 1 atm) → N_{2}(g, 5 atm)
(3) N_{2}(g, 273 K) → N_{2}(g, 300 K)
(4) H_{2}(g) → 2H(g)
Ans. (2)
N_{2} (g, 1 atm) → N_{2} (g, 5 atm)
for isothermal process
Therefore, ΔS < 0
Q.32. ΔU is equal to (2017)
(1) Isochoric work
(2) Isobaric work
(3) Adiabatic work
(4) Isothermal work
Ans. (3)
For adiabatic process, q = 0
∴ As per 1^{st} law of thermodynamics,
ΔU = W
Q.33. For a reaction, A(g) → A(l); ΔH = –3RT.
The correct statement for the reaction is: (2017)
(1) ΔH = ΔU = 0
(2) ΔH < ΔU
(3) ΔH > ΔU
(4) ΔH = ΔU ≠ 0
Ans. (3)
For the reaction A(g) → A(1)
The change in the number of moles of gaseous species Δn = 0  1 = 1
The enthalpy change ΔH = 3RT
ΔH = ΔU + ΔnRT
3RT = ΔU + (1)RT
The change in internal energy ΔU = 2RT
Hence, ΔH > ΔU
Q.34. The enthalpy change on freezing of 1 mol of water at 5°C to ice at –5°C is: (2017)
(Given Δ_{f}_{us} H = 6 kJ mol^{1} at 0ºC,
C_{p}(H_{2}O, ℓ) = 75.3 J mol^{1} K^{1},
Cp (H_{2}O, S) = 36.8 J mol^{1} K^{1})
(1) 6.00 kJ mol^{1}
(2) 5.81 kJ mol^{}^{1}
(3) 5.44 kJ mol^{1}
(4) 6.56 kJ mol^{1}
Ans. (4)
In order to calculate the enthalpy change for H_{2}O at 5ºC, we need to calculate the enthalpy change of all the transformation involved in the process.
(a) Energy change of 1 mol, H_{2}O(I), at 5ºC → 1 mol, H_{2}O(I), 0ºC
(b) Energy change of 1 mol, H_{2}O(I), at 0ºC → 1 mol, H_{2}O(s)(ice), 0ºC
(c) Energy change of 1 mol, Ice(s), at 0ºC → 1 mol, Ice(s), 5^{0}C
Total ΔH = C_{p}(H_{2}O(l)]ΔT +ΔH freezing + C_{p} [H_{2}O (s)] ΔT
= (75.3 J mol^{1} K^{1})(5)K + (6 x 10^{3} J mol^{1}) + (36.8 J mol^{1} K^{1}) (5) K
ΔH = 6.56 kJ mol^{1} (exothermic process)
So, ΔH = 6.56 kJ
Q.35. A gas undergoes change from state A to state B. In this process, the heat absorbed and work done by the gas is 5 J and 8 J, respectively. Now gas is brought back to A by another process during which 3 J of heat is evolved. In this reverse process of B to A: (2017)
(1) 6J of the work will be done by the gas
(2) 6J of the work will be done by the surrounding on gas.
(3) 10 J of the work will be done by the surrounding on gas.
(4) 10 J of the work will be done by the gas
Ans. (2)
Q.36. The heats of combustion of carbon and carbon monoxide are  393.5 and 283.5 kJ mol^{1}, respectively.
The heat of formation (in kJ) of carbon monoxide per mole is: (2016)
(1) 110.5
(2) 676.5
(3) 676.5
(4) 110.5
Ans. (4)
Q.37. A reaction at 1 bar is nonspontaneous at low temperature but becomes spontaneous at high temperature. Identify the correct statement about the reaction among the following: (2016)
(1) Both ΔH and ΔS are negative.
(2) Both ΔH and ΔS are positive.
(3) ΔH is positive while ΔS is negative.
(4) ΔH is negative while ΔS is positive.
Ans. (2)
ΔG=ΔH−T. ΔS
If ΔH & ΔS are both positive, then ΔG may be negative at height temperature. Hence reaction becomes spontaneous at higher temperature.
Q.38. If 100 mole of H_{2}O_{2} decompose at 1 bar and 300 K, the work done (kJ) by one mole of O_{2}(g) as it expands against 1 bar pressure is:
2H_{2}O_{2}(l) ⇌ 2H_{2}O(l) + O_{2}(g) (R = 8.3 J K^{–1} mol^{–1}) (2016)
(1) 498.00
(2) 62.25
(3) 124.50
(4) 249.00
Ans. (3)
2H_{2}O_{2}(l) ⇌ 2H_{2}O(l) + O_{2}(g)
W = –P_{ext} (ΔV) = –(nO_{2}) RT
∵ 100 mol H_{2}O_{2} on decomposition will give 50 mol O_{2}
⇒ W = –(50)(8.3)(300)J
=  124500 J
W = – 124.5 kJ
⇒ Work done by O_{2}(g) = 124.5 kJ
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