Question

For a first order reversible reaction $A\underset{K_b}{\stackrel{K_f}{\leftarrow }}B$, the initial concentration of $A$ and $B$ are $[A{]}_0$ and zero respectively. If concentrations at equilibrium are $[A{]}_{eq.}$ and $[B{]}_{eq.}$, derive an expression for the time taken by $B$ to attain concentration equal to $[B{]}_{eq/2}$.

Answer 1

The given reaction is

$K_f$

$A\rightleftharpoons B$

$K_b$

Initial concentration of $A=[A{]}_0$ & $B=0$

Concentration at equilibrium are $[A{]}_{eq}$ & $[B{]}_{eq}$ respectively.

Now,

Rate=$\frac{d\left[A\right]}{dt}=K_f\left[A\right]-K_b\left[B\right]$ $-\left(i\right)$

At, $t=0,\left[B\right]=0$

$\Rightarrow \left[B\right]=[A{]}_0-[A]$ $-\left(ii\right)$

Using $\left(ii\right)$ in $\left(i\right)$:-

$\therefore$ $\frac{d\left[A\right]}{dt}=K_f\left(\right[A\left]\right)-K_b\left(\right[A{]}_0-\left[A\right])$

$\frac{d\left[A\right]}{dt}=(K_f+K_b)\left[A\right]-K_b[A{]}_0$ $-\left(iii\right)$

At equilibrium, $\frac{d\left[A\right]}{dt}=0$ & $\left[A\right]=[A{]}_{eq}$

$\Rightarrow (K_f+K_b)[A{]}_{eq}-K_b[A{]}_0=0$

$\Rightarrow [A{]}_{eq}=\frac{K_b}{K_f+K_b}[A{]}_0$ $-\left(iv\right)$

Now, from $e{q}^n\left(i\right)$, at equilibrium, we have:-

$K_f[A{]}_{eq}-K_b[B{]}_{eq}=0$

$\Rightarrow [B{]}_{eq}=\frac{K_f}{K_b}[A{]}_{eq}$

$\Rightarrow [B{]}_{eq}=K_{eq}[A{]}_{eq}$ $-\left(v\right)$ $(\because K_{eq}=\frac{K_f}{K_b})$

Now, Integrating $e{q}^n\left(iii\right)$ after substituting $[A{]}_{eq}$ from $\left(iv\right)$:-

$\Rightarrow \left[A\right]=[A{]}_{eq}+C_1{e}^{-(K_f+K_b)t}$ $-\left(vi\right)$

;$C$ is integration constant

At, $t=0,\left[A\right]=[A{]}_0$

$\Rightarrow \left[A\right]=[A{]}_{eq}+([A{]}_0-[A{]}_{eq})e^{-(K_f+K_b)t}$ $-\left(vii\right)$

Now, when $\left[B\right]=[B{]}_{eq}/2$ (From $\left(ii\right)$) - $\left(viii\right)$

Putting $\left(viii\right)$ in $\left(vii\right)$:-

$\therefore$ $[A{]}_0-\frac{[B{]}_{eq}}{2}=[A{]}_{eq}+\left(\right[A{]}_0-\left[A{]}_{eq}\right)e^{-(K_f+K_b)t}$

$\Rightarrow e^{-}(K_b+K_f)t=\frac{2[A{]}_0-[B{]}_{eq}-2[A{]}_{eq}}{2\left(\right[A{]}_0-\left[A{]}_{eq}\right)}$

Taking natural log on both sides:-

$\Rightarrow -(K_f+K_b)t=ln\left(\frac{2[A{]}_0-[B{]}_{eq}-2[A{]}_{eq}}{2\left(\right[A{]}_0-\left[A{]}_{eq}\right)}\right)$

$\Rightarrow t=\frac{-1}{(K_f+K_b)}ln\left(\frac{2[A{]}_0-[B{]}_{eq}-2[A{]}_{eq}}{2\left(\right[A{]}_0-\left[A{]}_{eq}\right)}\right)$