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}C

3) 35.2^{ }K

4) -35.2 ^{o}C

**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_{vap}H = 21.16 kJ mol^{-1} = 21.16 x 10^{3} J/mol

Δ_{vap}S =
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 ^{o}C.

### 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?