JEE Chemistry 2024 Question with Solution

Given: $\Delta H_{\text{vap}} = 30 \, \text{kJ/mol}$ and $\Delta S_{\text{vap}} = 75 \, \text{J mol}^{-1}\text{K}^{-1}$.

Find: The temperature of vaporization at $1 \, \text{atmosphere}$.

At equilibrium for vaporization:

$$\Delta G = \Delta H - T\Delta S = 0$$

So,

$$T = \frac{\Delta H_{\text{vap}}}{\Delta S_{\text{vap}}}$$

Convert enthalpy into joules:

$$30 \, \text{kJ/mol} = 30000 \, \text{J/mol}$$

Substitute the values:

$$T = \frac{30000}{75} = 400 \, \text{K}$$

Therefore, the temperature of vaporization is $400 \, \text{K}$.

Given: $\Delta H_{\text{vap}} = 30 \, \text{kJ/mol}$ and $\Delta S_{\text{vap}} = 75 \, \text{J mol}^{-1}\text{K}^{-1}$.

Find: The vaporization temperature.

At the boiling point, liquid and vapour are in equilibrium, so $\Delta G = 0$. Hence the direct relation is:

$$T = \frac{\Delta H}{\Delta S}$$

Using $\Delta H = 30 \times 10^3 \, \text{J/mol}$,

$$T = \frac{30 \times 10^3}{75} = 400 \, \text{K}$$

Therefore, the correct numerical answer is 400.