The plot of log k_f versus 1 / T for a revers | vedclass

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(B) For the reaction $A_{(g)} \rightleftharpoons P_{(g)}$,the Arrhenius equation for the forward reaction is $\log k_f = \frac{-E_f}{2.303 RT} + \log

Vedclass · Accepted Answer

$5$

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(B) For the reaction $A_{(g)} \rightleftharpoons P_{(g)}$,the Arrhenius equation for the forward reaction is $\log k_f = \frac{-E_f}{2.303 RT} + \log A_f$.From the given plot,at $\frac{1}{T} = 0.002 \ K^{-1}$,$\log k_f = 9$.Substituting the values: $9 = \frac{-E_f}{2.303 R}(0.002) + \log(10^{15}) \Rightarrow 9 = \frac{-E_f}{2.303 R}(0.002) + 15$.$\frac{E_f}{2.303 R} = \frac{6}{0.002} = 3000$.For the equilibrium constant $K = \frac{k_f}{k_b}$,we have $\log K = \log \left(\frac{A_f}{A_b}\right) - \frac{E_f - E_b}{2.303 RT}$.At $500 \ K$,$\log K = 6$ and $\frac{A_f}{A_b} = \frac{10^{15}}{10^{11}} = 10^4$,so $\log(10^4) = 4$.$6 = 4 - \frac{E_f - E_b}{2.303 R(500)}$ $\Rightarrow 2 = \frac{E_b - E_f}{2.303 R(500)}$ $\Rightarrow E_b - E_f = 1000(2.303 R)$.Since $\frac{E_f}{2.303 R} = 3000$,$E_f = 3000(2.303 R)$.Thus,$E_b = 1000(2.303 R) + 3000(2.303 R) = 4000(2.303 R)$.For the backward reaction at $250 \ K$,$\log k_b = \log A_b - \frac{E_b}{2.303 RT} = \log(10^{11}) - \frac{4000(2.303 R)}{2.303 R(250)} = 11 - 16 = -5$.Therefore,$|\log k_b| = |-5| = 5$.

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