JEE Advanced 2023 Paper 1 · Q10 · Arrhenius Equation & Temperature

Question

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 ___.

SolveKar AI · Accepted Answer

At $500$ K with $1/T = 0.002$: $\log k_f = 9$ (from plot), so $k_f = 10^9$ and $K_{eq} = 10^6$, giving $k_b = k_f/K = 10^3$ at $500$ K. From Arrhenius: $\log k_b = \log A_b - E_{a,b}/(2.303 R T)$. At $500$ K: $3 = 11 - E_{a,b}/(2.303 R\cdot 500) \Rightarrow E_{a,b}/(2.303R) = 4000$. At $250$ K: $\log k_b = 11 - 4000/250 = 11 - 16 = -5$, so $|\log k_b| = 5$.

JEE Advanced 2023 Paper 1 · Q10 · Arrhenius Equation & Temperature

Solution