Question

Let us consider the following mechanism:
$C{H}_{3}CN+H^{+}\rightleftharpoons C{H}_{3}CN{H}^{+}$ (fast)
$C{H}_{3}CN{H}^{+}+H_{2}O\to$ Product (slow)
What would be the rate law?

Answer 1

Given reaction are

${CH}_{3}CN+H^{+}\rightleftharpoons {CH}_3{CNH}^{+}$ $\left(fast\right)$ $\left(1\right)$

${CH}_{3}CN{H}^{+}{H}_{2}O\rightleftharpoons Product$ $\left(slow\right)$ $\left(2\right)$

Rate of complex reaction is detemined the slowest step in its mechanism. This slowest step is also known as rate setemining step(rds). The reason behind this is that the activation energy for the slowest step is highest among all steps.

From equation $\left(2\right)$ we get the rate as

$rate=K\left[{CH}_{3}CN{H}^{+}\right]$ $\left(3\right)$

From euation $\left(1\right)$ we get

$Keq=\frac{\left[{CH}_{3}CN{H}^{+}\right]}{\left[H^{+}\right]\left[{CH}_{3}CN\right]}$ $\left(4\right)$

From $\left(3\right)$ and $\left(4\right)$ we have

$rate=KKeq\left[H^{+}\right]\left[{CH}_{3}CN\right]$

Hence the rate law will be

$rate=K^{\prime }\left[H^{+}\right]\left[{CH}_{3}CN\right]$ $\{where{K}^{\prime }=KKeq\}$