Chemistry · Thermodynamics & Thermochemistry
JEE Main 2026 — 4 April, Evening Shift — Question 63
For the following reaction at and 2 atm pressure, is dissociated The magnitude of standard free energy change at this temperature is x . [Nearest integer] Given : ,
Answer: 2474
Numerical answer — enter this value.
Step-by-step solution
t=0 & 1 & - \\ t_{\text{eqm.}} & 1-2x & 2x & x \\ & 1-0.5 & 0.5 & 0.25 \\ & 0.5 & 0.5 & 0.25 \\ & \dfrac{0.5}{1.25}\times 2 & \dfrac{0.5}{1.25}\times 2 & \dfrac{0.25}{1.25}\times 2 \end{array}$$$ $$ \begin{aligned} \mathrm{K}_{\mathrm{P}} & =\frac{\mathrm{P}_{\mathrm{N}_{2} \mathrm{O}_{4}}^{2} \cdot \mathrm{P}_{\mathrm{O}_{2}}}{\mathrm{P}_{\mathrm{N}_{2} \mathrm{O}_{5}}^{2}}=0.4 \Delta \mathrm{G}^{\circ} & =-2.303 \mathrm{RT} \log \mathrm{~K}_{\mathrm{p}} & =-2.303 \times 8.314 \times 323 \log (4 / 10) & =2473.81 \mathrm{~J} / \mathrm{mole} \end{aligned} $$
Answer key and solution verified before publishing.
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- Exam
- JEE Main 2026
- Subject
- Chemistry
- Chapter
- Thermodynamics & Thermochemistry
- Topic
- Gibbs Free Energy - Relation with Equilibrium, Metallurgy and Electrochemistry