• # question_answer The reaction cis$-X\text{trans}-X$ is first order in both directions. At $\text{25}{}^\circ \text{C,}$ the equilibrium constant is 0.10 and the rate constant ${{k}_{f}}=3\times {{10}^{-4}}{{s}^{-1}}$. In an experiment starting with the pure cis-form, how long would it take for half of the equilibrium amount of the trans-isomer to be formed? A) 150 s                             B) 200 s C) 240 s                 D) 210 s

Solution :

$cis-Xtrans-X$
${{K}_{(aq)}}=\frac{{{k}_{f}}}{{{k}_{b}}};{{k}_{b}}=\frac{3\times {{10}^{-4}}}{0.1}=3\times {{10}^{-3}}$
 Initial a 0 At time t $a-x$ x At equilibrium $a-{{x}_{e}}$ ${{x}_{e}}$
As we know,
$\left( {{k}_{f}}+{{k}_{b}} \right)=\frac{1}{t}\ln \,\left( \frac{{{x}_{e}}}{{{x}_{e}}-x} \right)$
Given $x=\frac{{{x}_{e}}}{2}$
$\therefore$      $\,\left( {{k}_{f}}+{{k}_{b}} \right)=\frac{1}{2}\ln 2$
or         $\left( 3\times {{10}^{-3}}+3\times {{10}^{-4}} \right)=\frac{0.693}{t}$
$t=210s$

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