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 ^oC 

3) 35.2 K 

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:

Δ_vapH = 21.16 kJ mol^-1 = 21.16 x 10³ J/mol

Δ_vapS = 89.0 J K^-1 mol^-1

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 ΔG^o value for above process? What is the extra information required to do so?