JEE Chemistry 2025 Question with Solution

Given: Decomposition of $N_2O_4(g) \rightleftharpoons 2NO_2(g)$ with $\Delta H^\circ = 55.0 \, \text{kJ mol}^{-1} = 55000 \, \text{J mol}^{-1}$, $\Delta S^\circ = 175.0 \, \text{J mol}^{-1} \text{K}^{-1}$, and $T = 25^\circ \text{C} = 298 \, \text{K}$.

Find: The standard free energy change $\Delta G^\circ$ in $\text{J mol}^{-1}$.

Use the Gibbs free energy relation:

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

Substitute the given values:

$$\Delta G^\circ = 55000 - (298)(175)$$

Now calculate:

$$\Delta G^\circ = 55000 - 52150 = 2850 \, \text{J mol}^{-1}$$

Therefore, the standard free energy change is $2850 \, \text{J mol}^{-1}$.

Given: $\Delta H^\circ = 55.0 \, \text{kJ mol}^{-1}$, $\Delta S^\circ = 175.0 \, \text{J mol}^{-1} \text{K}^{-1}$, and $T = 298 \, \text{K}$.

Find: $\Delta G^\circ$.

First convert enthalpy into joules so that all units are consistent:

$$55.0 \, \text{kJ mol}^{-1} = 55.0 \times 10^3 \, \text{J mol}^{-1} = 55000 \, \text{J mol}^{-1}$$

Now apply:

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

So,

$$\Delta G^\circ = (55.0 \times 10^3) - (298)(175)$$

Next,

$$(298)(175) = 52150$$

Hence,

$$\Delta G^\circ = 55000 - 52150 = 2850 \, \text{J mol}^{-1}$$

Therefore, the required numerical value is 2850.