Q 1 :    

In which of the following processes entropy increases?

A.  A liquid evaporates to vapour.

B. Temperature of a crystalline solid lowered from 130 K to 0 K.

C. 2NaHCO3(s)Na2CO3(s)+CO2(g)+H2O(g)

D. Cl2(g)2Cl(g)

Choose the correct answer from the options given below:                 [2024]

  • A and C

     

  • A, B and D

     

  • A, C and D

     

  • C and D

     

(3)

A.  Molecules in gaseous phase have greater freedom of movement, i.e.,have greater randomness than those in liquid phase, hence entropy increases.

B.  On decreasing temperature of a solid, entropy decreases.

C.  As number of gaseous products increases, randomness increases.

D.  The products have larger number of gaseous particles, hence entropy increases.



Q 2 :    

For irreversible expansion of an ideal gas under isothermal condition, the correct option is           [2021]

  • ΔU0,ΔStotal=0

     

  • ΔU=0,ΔStotal=0

     

  • ΔU0,ΔStotal0

     

  • ΔU=0,ΔStotal0

     

(4)

For reversible and irreversible expansion for an ideal gas under isothermal condition, ΔU=0, but ΔStotal i.e., ΔSsys+ΔSsurr is not zero for irreversible process.
 

 



Q 3 :    

For the reaction, 2Cl(g)Cl2(g), the correct option is               [2020]

  • ΔrH>0 and ΔrS>0

     

  • ΔrH>0 and ΔrS<0

     

  • ΔrH<0 and ΔrS>0

     

  • ΔrH<0 and ΔrS<0

     

(4)

In the reaction, 2Cl(g)Cl2(g), the randomness decreases as 2 moles of Cl(g) are converted to 1 mole of Cl2(g); thus, ΔrS<0.
And this is an exothermic reaction, thus, ΔrH<0.
 

 



Q 4 :    

In which case change in entropy is negative?                 [2019]

  • 2H(g)H2(g)

     

  • Evaporation of water

     

  • Expansion of a gas at constant temperature

     

  • Sublimation of solid to gas

     

(1)

If Δng<0 then ΔS<0

 



Q 5 :    

For a given reaction, ΔH=35.5kJmol-1 and ΔS=83.6JK-1mol-1. The reaction is spontaneous at (Assume that ΔH and ΔS do not vary with temperature.)                  [2017]

  • T>425 K

     

  • all temperatures

     

  • T>298 K

     

  • T<425 K

     

(1)

For a spontaneous reaction, ΔG<0 i.e., ΔH-TΔS<0

T>ΔHΔST>35.5×100083.6=424.6425 K

  T>425 K
 

 



Q 6 :    

For a sample of perfect gas when its pressure is changed isothermally from pi to pf, the entropy change is given by         [2016]

  • ΔS=nRln(pfpi)

     

  • ΔS=nRln(pipf)

     

  • ΔS=nRTln(pfpi)

     

  • ΔS=RTln(pipf)

     

(2)

For an ideal gas undergoing reversible expansion, when temperature changes from Ti to Tf and pressure changes from pi to pf,

ΔS=nCplnTfTi+nRlnpipf

For an isothermal process, Ti=Tf so, ln 1=0

  ΔS=nRlnpipf



Q 7 :    

Consider the following liquid-vapour equilibrium. 

Liquid Vapour

Which of the following relations is correct?                     [2016]

  • dlnPdT2=-ΔHvT2

     

  • dlnPdT=ΔHvRT2

     

  • dlnGdT2=ΔHvRT2

     

  • dlnPdT=-ΔHvRT

     

(2)

This is Clausius--Clapeyron equation.
 

 



Q 8 :    

Which of the following statements is correct for the spontaneous adsorption of a gas?            [2014]

  • ΔS is negative and, therefore ΔH should be highly positive.

     

  • ΔS is negative and therefore, ΔH should be highly negative.

     

  • ΔS is positive and therefore, ΔH should be negative.

     

  • ΔS is positive and therefore, ΔH should also be highly positive.

     

(2)

Using Gibbs'-Helmholtz equation,

ΔG=ΔH-TΔS

During adsorption of a gas, entropy decreases i.e., ΔS<0

For spontaneous adsorption, ΔG should be negative, which is possible when ΔH is highly negative.



Q 9 :    

For the reaction, X2O4(l)2XO2(g)

ΔU=2.1 kcalΔS=20 cal K-1 at 300 K 

Hence, ΔG is                                                                           [2014]

  • 2.7 kcal

     

  • - 2.7 kcal

     

  • 9.3 kcal

     

  • - 9.3 kcal

     

(2)

ΔH=ΔU+ΔngRT

Given, ΔU=2.1 kcal, Δng=2,

      R=2×10-3 kcal, T = 300 K

 ΔH=2.1+2×2×10-3×300=3.3 kcal

Again, ΔG=ΔH-TΔS

Given, ΔS=20×10-3 kcal K-1

On putting the values of ΔH and ΔS in the equation, we get

ΔG=3.3-300×20×10-3=3.3-6×103×10-3=-2.7 kcal