The values of pressure equilibrium constant recorded at different temperatures for the following equilibrium reaction have been given below: A ( g ) rightleftharpoons B ( g )+ C ( g ) begin array c c 1/T\,( K -1 ) & log 10 K p \\ 0.05 & 3.5 \\ 0.06 & 2.5 \\ 0.07 & 1.5 end array The magnitude of frac Δ H ° R calculated from the above data is \ \ \ \ .(Nearest integer).

Chemistry · Thermodynamics & Thermochemistry

JEE Main 2026 — 5 April, Morning Shift — Question 67

The values of pressure equilibrium constant recorded at different temperatures for the following equilibrium reaction have been given below: $\mathrm{A}(\mathrm{g}) \rightleftharpoons \mathrm{B}(\mathrm{g})+\mathrm{C}(\mathrm{g})$

\begin{array}{c c} \frac{1}{T}\,(\mathrm{K}^{-1}) & \log_{10} K_p \\ 0.05 & 3.5 \\ 0.06 & 2.5 \\ 0.07 & 1.5 \end{array}

The magnitude of $\frac{\Delta \mathrm{H}^{\circ}}{\mathrm{R}}$ calculated from the above data is $\_\_\_\_$ .(Nearest integer).

Answer: 230

Numerical answer — enter this value.

Step-by-step solution

Using formula

\begin{aligned} \log K_{P_2}-\log K_{P_1} &=\frac{\Delta H}{2.303R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)\\ 3.5-2.5 &=\frac{\Delta H}{2.303R}(0.06-0.05)\\ \frac{\Delta H}{R} &=\frac{2.303}{0.01}\\ \frac{\Delta H}{R} &=230.3\approx230 \end{aligned}

Answer key and solution verified before publishing.

Practise Thermodynamics & Thermochemistry

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2026
Paper: 5 April, Morning Shift
Subject: Chemistry
Chapter: Thermodynamics & Thermochemistry
Topic: Second law of Thermodynamics