The reaction cis-X $\underset{k_b}{\overset{k_f}{\rightleftharpoons }}$ trans-X is first order in both directions. At 25°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?

cis-X  $\underset{k_b}{\overset{k_f}{\rightleftharpoons }}$
Initial      $a$ $0;$            $k_{\left(aq\right)}=\frac{k_f}{k_b};$$k_b=\frac{3\times {10}^{-4}}{0.1}=3\times {10}^{-3}$                           
At time t      $a-x$              $x$ 
At eqm       $a-x_e$                $x_e$
As we know   $(k_f+k_b)$  =   $\frac{1}{t}in\left(\frac{x_e}{x_e-x}\right)$
Given    $x=\frac{x_e}{2}$
 $\therefore (k_f+k_b)=\frac{1}{2}in2$
Or      $\begin{array}{l}(3\times {10}^{-3}+3\times {10}^{-4})=\frac{0.693}{t}\\ t=210\sec \end{array}$