JEE MAIN - Chemistry (2025 - 23rd January Morning Shift - No. 22)
The standard enthalpy and standard entropy of decomposition of $\mathrm{N}_2 \mathrm{O}_4$ to $\mathrm{NO}_2$ are $55.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $175.0 \mathrm{~J} / \mathrm{K} / \mathrm{mol}$ respectively. The standard free energy change for this reaction at $25^{\circ} \mathrm{C}$ in J $\mathrm{mol}^{-1}$ is ________ (Nearest integer)
Answer
2850
Explanation
$\Delta \mathrm{H}_{\mathrm{rxn}}^{\circ}=55 \mathrm{~kJ} / \mathrm{mol}, \quad \mathrm{T}=298 \mathrm{~K}$
$$\begin{aligned} & \Delta \mathrm{S}_{\mathrm{rxn}}^{\mathrm{o}}=175 \mathrm{~J} / \mathrm{mol} \\ & \Delta \mathrm{G}_{\mathrm{rxn}}^{\mathrm{o}}=\Delta \mathrm{H}_{\mathrm{rxn}}^{\mathrm{o}}-\mathrm{T} \Delta \mathrm{~S}_{\mathrm{rxn}}^{\mathrm{o}} \\ & \Rightarrow \Delta \mathrm{G}_{\mathrm{rxn}}^{\mathrm{o}}=55000 \mathrm{~J} / \mathrm{mol}-298 \times 175 \mathrm{~J} / \mathrm{mol} \\ & \Rightarrow \Delta \mathrm{G}_{\mathrm{rxn}}^{\mathrm{o}}=55000-52150 \\ & \Rightarrow \Delta \mathrm{G}_{\mathrm{rxn}}^{\mathrm{o}}=2850 \mathrm{~J} / \mathrm{mol} \end{aligned}$$
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