Q 1 :

Which of the following is not correct?            [2024]

  • G is zero for a reversible reaction.

     

  • G is negative for a spontaneous reaction.

     

  • G is positive for a spontaneous reaction.

     

  • G is positive for a non-spontaneous reaction.

     

(3)           

                  G=0reaction at equilibrium

                  G>0non spontaneous reaction

                  G<0spontaneous reaction

 



Q 2 :

Consider the following reaction at 298 K. 32O2(g)O3(g)Kp=2.47×10-29,ΔrGΘ for the reaction is __________ kJ.

(Given R=8.314 JK-1 mol-1)                   [2024]



(163)

log(K)=log10(2.47×10-29)log10(25×10-30)

           =log10(52×10-30)=2log105-30 log 10

                                                =2×0.7-30×1=-28.6

ΔG°=-2.303 RT log10K

ΔG°=-2.303×8.314JK mol×298K×log10(2.47×10-29)

         =-2.303×8.314×298×(-28.6)J mol-1=163.2kJmol-1

 



Q 3 :

For the reaction at 298 K, 2A + BC. H = 400 kJ mol-1 and S = 0.2 kJ mol-1 K-1. The reaction will become spontaneous above ________ K.         [2024]



(2000)

For reaction to be spontaneous,

ΔG<0

ΔH-TΔS<0

400kJ mol-1-T×0.2kJ mol-1K-1<0

400kJ mol-1<T×0.2kJ mol-1K-1

T>4000.2K

T>2000K

 



Q 4 :

When ΔHvap=30kJ/mol and Svap=75J mol-1 K-1, then the temperature of vapour, at one atmosphere is ___________ K.      [2024]



(400) 

When external pressure is 1 atm ,then vapours are formed at normal boiling point of the liquid. So in the question, normal boiling point of the liquid is asked.

ΔSvap=ΔHvapTbpt

75J mol-1K-1=30×1000Jmol-1Tbpt

Tbpt=400K



Q 5 :

Let us consider an endothermic reaction which is non-spontaneous at the freezing point of water. However, the reaction is spontaneous at boiling point of water. Choose the correct option.               [2025]

  • Both ΔH and ΔS are (–ve)

     

  • Both ΔH and ΔS are (+ve)

     

  • ΔH is (–ve) but ΔS is (+ve)

     

  • ΔH is (+ve) but ΔS is (–ve)

     

(2)

For reaction to be spontaneous ΔG=ΔHTΔS<0. As the reaction is endothermic, ΔH>0. If ΔS>0, then at low temperature ΔG becomes positive i.e. reaction becomes non-spontaneous, and at high temperature ΔG becomes negative i.e. the reaction becomes spontaneous.



Q 6 :

Match List-I with List-II.                                          [2025]

 

List-I (Partial Derivatives)

 

List-II (Thermodynamic Quantity)

A. (GT)P I. Cp
B. (HT)P II. − S
C. (GP)T III. Cv
D. (UT)V IV. V

 

Choose the correct answer from the options given below:

  • A-II, B-I, C-IV, D-III

     

  • A-II, B-III, C-I, D-IV

     

  • A-II, B-I, C-III, D-IV

     

  • A-I, B-II, C-IV, D-III

     

(1)

G=H-TS

dG=dH-TdS-SdT

dG=d(U+PV)-TdS-SdT

dG=dU+PdV+VdP-TdS-SdT

dG=qrev+Wrev+PdV+VdP-TdS-SdT

dG=TdS-PdV+PdV+VdP-TdS-SdT 

dG=VdP-SdT

At constant pressure, dP=0, so [GT]P=-S

At constant temperature, dT=0, so [GP]T=V

(B) H=U+PV

dH=dU+PdV+VdP

dH=qrev+Wrev+PdV+VdP

dH=qrev-PdV+PdV+VdP

dH=qrev+VdP

dH=nCdT+VdP

At constant pressure C=Cp and dP=0

dH=nCpdT

Taking 1 mole of substance, (n=1)

[HT]P=Cp

(D) dU=qrev+Wrev

dU=qrev-PdV

dU=nCdT-PdV

At constant volume, C=Cv and dV=0

dU=nCvdT

[UT]V=Cv



Q 7 :

The effect of temperature on spontaneity of reactions are represented as:

  ΔH ΔS Temperature Spontaneity
(A) + any T Non spontaneous
(B) + + low T spontaneous
(C) low T Non spontaneous
(D) + any T spontaneous

 

The incorrect combinations are:                                          [2025]

  • (A) and (D) only

     

  • (B) and (D) only

     

  • (A) and (C) only

     

  • (B) and (C) only

     

(4)

A reaction is spontaneous if ΔG is negative.

S.no ΔH ΔS T ΔG=ΔHTΔS Spontaneity
(A) + any + Non-spontaneous
(B) + + low + Non-spontaneous
(C) low Spontaneous
(D) + any Spontaneous

 

 



Q 8 :

Which of the following graphs correctly represents the variation of thermodynamic properties of Haber’s process?            [2025]

  •  

  •  

  •  

  •  

(1)

(I)  N2(g)+3H2(g)2NH3(g),  ΔH°=-46.1kJ mol-1

As number of gaseous moles decrease, ΔS°<0.  

Both ΔH° and ΔS° are negative and almost remain constant with change of temperature.

(II)  ΔG°=-RTlnK

-ΔG°T=RlnK

For exothermic reaction, value of equilibrium constant decreases with increase in temperature, so -ΔG°T decreases with increase in temperature.

(III)  As ΔH° is negative -ΔH°T is positive and it decreases with increase in temperature.



Q 9 :

Let us consider a reversible reaction at temperature, T. In this reaction, both ΔH and ΔS were observed to have positive values. If the equilibrium temperature is Te, then the reaction becomes spontaneous at :                    [2025]
 

  • T=Te

     

  • Te>T

     

  • T>Te

     

  • Te=5T

     

(3)

When reaction is at equilibrium, then:

ΔG=0

ΔH-TeΔS=0

Te=ΔHΔS

For reaction to be spontaneous:

ΔG<0

ΔH-TΔS<0

TΔS>ΔH

T>ΔHΔS

T>Te



Q 10 :

Standard entropies of X2, Y2 and XY5 are 70, 50 and 110 JK-1mol-1 respectively. The temperature in Kelvin at which the reaction

12X2+52Y2XY5,  ΔH=-35 kJ mol-1

will be at equilibrium is _____ (Nearest integer).                                [2025]



(700)

12X2+52Y2XY5 

ΔS°=Sproducts-Sreactants=SXY5-[12SX2+52SY2]

=110JK-1mol-1-[12×70+52×50]JK-1mol-1

=-50JK-1mol-1

ΔG°=ΔH°-TΔS°=-35kJ mol-1-T×(-50JK-1mol-1)

         =-35000J mol-1+T×50JK-1mol-1

Taking ΔG°=0 for equilibrium, -35000J mol-1+T×50JK-1mol-1=0

T=3500050K=700K

(Note: At equilibrium ΔG=0, ΔG° is not necessarily zero at equilibrium, but as per data provided in the question, the only way to solve the question is by taking ΔG°=0).
 



Q 11 :

The standard enthalpy and standard entropy of decomposition of N2O4 to NO2 are 55.0 kJ mol-1 and 175.0 J/K/mol respectively. The standard free energy change for this reaction at 25°C in J mol-1 is _____ (Nearest integer).                     [2025]



(2850)

ΔG°=ΔH°-TΔS°

ΔG°=55kJ mol-1-298K×175JK-1mol-1

ΔG°=55000J mol-1-298K×175JK-1mol-1

ΔG°=55000J mol-1-52150J mol-1 

ΔG°=2850J mol-1

 



Q 12 :

Which of the following relations are correct?

A.   ΔU=q+PΔV

B.   ΔG=ΔH-TΔS

C.   ΔS=qrevT

D.   ΔH=ΔU-ΔnRT

Choose the most appropriate answer from the options given below:             [2023]

  • B and C only

     

  • B and D only

     

  • C and D only

     

  • A and B only

     

(1)

(A)  ΔU=q+w

         ΔU=q-PΔV

(B)  ΔG=ΔH-TΔS

(C)  ΔS=qrevT

(D)  ΔH=ΔV+ΔngRT

So correct option are B and C only.



Q 13 :

For independent processes at 300 K

Process ΔH/kJ mol-1 ΔS/J K-1
A –25 –80
B –22 40
C 25 –50
D 22 20

 

The number of non-spontaneous processes from the following is _________.          [2023]



(2)

Process Sign of ΔG Spontaneity of Process
A (−)ve Spontaneous
B (−)ve Spontaneous
C (+)ve Non spontaneous
D (+)ve Non spontaneous

 



Q 14 :

Consider the following reaction approaching equilibrium at 27°C and 1 atm pressure

A+Bkr=102kf=103C+D

The standard Gibbs energy change (ΔrGθ) at 27°C is (–) ____kJ mol-1 (Nearest integer).

(Given: R=8.3 J K-1mol-1 and ln10=2.3)                  [2023]



(6)

Keq=KfKb=103102=10

ΔG°=-RT ln Keq

         =-8.3×300 ln (10)

         =-8.3×300×2.3

          =-5.72×103J mol-1

          =-5.72kJ mol-1

           6 kJ mol-1



Q 15 :

Consider the graph of Gibbs free energy G vs extent of reaction. The number of statement/s from the following which are true with respect to points (A), (B) and (C) is ______.
  [2023]

A. Reaction is spontaneous at (A) and (B)
B. Reaction is at equilibrium at point (B) and non-spontaneous at point (C)
C. Reaction is spontaneous at (A) and non-spontaneous at (C)
D. Reaction is non-spontaneous at (A) and (B)



(2)

Reaction is spontaneous in a direction in which Gibbs free energy decreases. Hence, two statements i.e. B and C are correct.

 



Q 16 :

For complete combustion of ethene,

C2H4(g)+3O2(g)2CO2(g)+2H2O(l)

the amount of heat produced as measured in a bomb calorimeter is -1406 kJ mol-1 at 300 K.

The minimum value of TΔS needed to reach equilibrium is (–) _____ kJ. (Nearest integer)

Given: R=8.3 J K-1mol-1                    [2023]



(1411)

For equilibrium ΔG=0

ΔH-TΔS=0

TΔS=ΔH=ΔE+ΔngRT

ΔH=-1406+[-2]×8.3×10-3×300

      =-1410.98-1411 kJ/mol



Q 17 :

Use the following data : 

Substance

ΔfH (500 K)kJ mol⁻¹ 

S (500 K)J K⁻¹ mol⁻¹

AB(g)

32

222

A2(g)

6

146

B2(g)

x

280

One mole each of A2(g) and B2(g) are taken in a 1 L closed flask and allowed to establish the equilibrium at 500 K. 

A2(g)+B2(g)2AB(g)

The value of x (in kJ mol⁻¹) is __________. (Nearest integer)

(Given : log K = 2.2 R=8.3 J K⁻¹ mol⁻¹)    [2026]



(70)

A2+B2500K2AB    logK=2.2

ΔH°=(2×32)-(6+x)=(58-x) kJ

ΔS°=(2×222)-(146+280)=18 J

ΔG°=-RTlnK

ΔG°=-8.314×500×2.2×2.3031000

ΔG°=-21.06

ΔH°-TΔS°=-21.06

58-x-500(181000)=-21.06

x=70.06 kJ/mol