MOLAR ENTROPY -  ENTHALPY - GIBBS FREE ENERGY

6) For the vaporization process of HI, the molar entropy and molar enthalpy are respectively, 89.0 J K-1 mol-1 and 21.16 kJ mol-1. The temperature at which the value of ΔG becomes zero for this process is: 

1) 237800 K 

2) 273.8 o

3) 35.2

4) -35.2 oC  

Logic:

During vaporization, both liquid and vapor of HI are at equilibrium with each other and hence ΔG is zero.

The relation between ΔG, ΔH and ΔS is given by:

ΔG = ΔH - TΔS

At equilibrium:

0 = ΔH - T ΔS

or

TΔS = ΔH

or

T = ΔH/ΔS

From above equation it is possible to calculate the temperature at which ΔG becomes zero.

Solution:

ΔvapH = 21.16 kJ mol-1 = 21.16 x 103 J/mol

ΔvapS = 89.0 J K-1 mol-1

calculation of temperature of vaporization

Note: Entropy is usually expressed in J K-1 mol-1

Conclusion:

Take care of units. You may be tempted to tick for option 3. It is given in oC.

Homework:

1) What is special about this temperature? What is it known as?

2) Is it possible to calculate ΔGo value for above process? What is the extra information required to do so?

 

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Author: Aditya vardhan Vutturi
www.adichemistry.com