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(C) For a reversible first-order reaction $A \underset{K_b}{\stackrel{K_f}{\rightleftharpoons}} B$,the rate law is given by: $-\frac{d[A]}{dt} = K_f[A

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$t = \frac{\ln 2}{(K_{f} + K_{b})}$

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(C) For a reversible first-order reaction $A \underset{K_b}{\stackrel{K_f}{ightleftharpoons}} B$,the rate law is given by: $-\frac{d[A]}{dt} = K_f[A] - K_b[B]$.At equilibrium,$-\frac{d[A]}{dt} = 0$,so $K_f[A_{	ext{eq}}] = K_b[B_{	ext{eq}}]$. Since $[B_{	ext{eq}}] = A_0 - A_{	ext{eq}}$,we have $K_b = \frac{K_f A_{	ext{eq}}}{A_0 - A_{	ext{eq}}}$.The integrated rate equation for this system is $(K_f + K_b)t = \ln \left( \frac{A_0 - A_{	ext{eq}}}{A_t - A_{	ext{eq}}} ight)$.Given $A_t = \frac{A_0 + A_{	ext{eq}}}{2}$,we substitute this into the expression:$A_t - A_{	ext{eq}} = \frac{A_0 + A_{	ext{eq}}}{2} - A_{	ext{eq}} = \frac{A_0 - A_{	ext{eq}}}{2}$.Substituting this into the integrated rate equation:$(K_f + K_b)t = \ln \left( \frac{A_0 - A_{	ext{eq}}}{(A_0 - A_{	ext{eq}})/2} ight) = \ln(2)$.Therefore,$t = \frac{\ln 2}{K_f + K_b}$.

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