JEE Chemistry 2023 Question with Solution

Given: Heat required to melt one mole of sodium chloride is $\Delta H = 30.4\,\text{kJ mol}^{-1}$ and entropy change is $\Delta S = 28.4\,\text{J K}^{-1}\text{mol}^{-1}$.

Find: The melting point $T$ of sodium chloride.

At the melting point, the process is at equilibrium, so

$$\Delta G = 0$$

Using

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

we get

$$\Delta H = T\Delta S$$

Convert enthalpy into consistent units:

$$\Delta H = 30.4\,\text{kJ mol}^{-1} = 30400\,\text{J mol}^{-1}$$

Now calculate the melting temperature:

$$T = \frac{\Delta H}{\Delta S} = \frac{30400}{28.4} = 1070.42\,\text{K}$$

Nearest integer:

$$T \approx 1070\,\text{K}$$

Therefore, the melting point of sodium chloride is $1070\,\text{K}$.

Given: A phase transition at the melting point with known $\Delta H$ and $\Delta S$.

Find: The melting temperature.

At any phase transition temperature, use the direct equilibrium relation

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

because

$$\Delta G = 0$$

Substitute

$$T = \frac{30400}{28.4} \approx 1070.42\,\text{K}$$

Hence, the correct numerical answer is 1070.