The plot of log k f versus 1 / T for a reversible reaction A ( g ) rightleftharpoons P ( g ) is shown Pre-exponential factors for the forward and backward reactions are 10 15 s -1 and 10 11 s -1 , respectively. If the value of log K for the reaction at 500 K is 6 , the value of log k b at 250 K is \ \ \ \ . [ K= equilibrium constant of the reaction k f = rate constant of forward reaction k b = rate constant of backward reaction]

Chemistry · Chemical Equilibrium

JEE Advanced 2023 — Paper 1 — Question 40

The plot of $\log k_{f}$ versus $1 / \mathrm{T}$ for a reversible reaction $\mathrm{A}(\mathrm{g}) \rightleftharpoons \mathrm{P}(\mathrm{g})$ is shown Pre-exponential factors for the forward and backward reactions are $10^{15} \mathrm{~s}^{-1}$ and $10^{11} \mathrm{~s}^{-1}$ , respectively. If the value of $\log \mathrm{K}$ for the reaction at 500 K is 6 , the value of $\left|\log k_{b}\right|$ at 250 K is $\_\_\_\_$ . [ $K=$ equilibrium constant of the reaction $k_{f}=$ rate constant of forward reaction $k_{b}=$ rate constant of backward reaction]

Answer: 5

Numerical answer — enter this value.

Step-by-step solution

$\log \mathrm{k}_{\mathrm{f}}=9$ at $500 \mathrm{~K}, \therefore \mathrm{k}_{\mathrm{f}}=10^{9}$ $\log k_{b}=\frac{-\left(E_{a}\right)_{b}}{2.303 R} \frac{1}{T}+\log A_{b}$ $\mathrm{K}_{\text {eq }}=10^{6}=\frac{10^{9}}{\mathrm{k}_{\mathrm{b}}}$ $\mathrm{k}_{\mathrm{b}}=10^{3}$ $3=\frac{-\left(E_{a}\right)_{b}}{2.303 R} \times 0.02+11$ $\frac{\left(E_{a}\right)_{b}}{2.303 R}=\frac{8}{0.002}=4000$ $\log \mathrm{k}_{\mathrm{b}}=-4000 \times \frac{1}{250}+11$ $=-16+11$ $=-5$ $\left|\log \mathrm{k}_{\mathrm{b}}\right|=5$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2023
Paper: Paper 1
Subject: Chemistry
Chapter: Chemical Equilibrium
Topic: Equilibria Involving Physical Processes