Easy INTEGER +4 / -1 PYQ · JEE Mains 2025

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 (integer) 2850

Solution

<p>$\Delta \mathrm{H}_{\mathrm{rxn}}^{\circ}=55 \mathrm{~kJ} / \mathrm{mol}, \quad \mathrm{T}=298 \mathrm{~K}$</p> <p>$$\begin{aligned} & \Delta \mathrm{S}_{\mathrm{rxn}}^{\mathrm{o}}=175 \mathrm{~J} / \mathrm{mol} \\ & \Delta \mathrm{G}_{\mathrm{rxn}}^{\mathrm{o}}=\Delta \mathrm{H}_{\mathrm{rxn}}^{\mathrm{o}}-\mathrm{T} \Delta \mathrm{~S}_{\mathrm{rxn}}^{\mathrm{o}} \\ & \Rightarrow \Delta \mathrm{G}_{\mathrm{rxn}}^{\mathrm{o}}=55000 \mathrm{~J} / \mathrm{mol}-298 \times 175 \mathrm{~J} / \mathrm{mol} \\ & \Rightarrow \Delta \mathrm{G}_{\mathrm{rxn}}^{\mathrm{o}}=55000-52150 \\ & \Rightarrow \Delta \mathrm{G}_{\mathrm{rxn}}^{\mathrm{o}}=2850 \mathrm{~J} / \mathrm{mol} \end{aligned}$$</p>

About this question

Subject: Chemistry · Chapter: Thermodynamics · Topic: Zeroth and First Law

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