Question

Consider the chemical reaction,
$N_2\left(g\right)+3H_2\left(g\right)\to 2N{H}_3\left(g\right)$
The rate of this reaction can be expressed in terms of time derivative of concentration of $N_2\left(g\right),H_2\left(g\right)$ and $N{H}_3\left(g\right)$. Identify the correct relationship amongst the rate expressions

• A. $Rate=-d\left[N_2\right]/dt=-\frac{1}{3}d\left[H_2\right]/dt=\frac{1}{2}d\left[N{H}_3\right]/dt$
• B. $Rate=-d\left[N_2\right]/dt=-3d\left[H_2\right]/dt=2d\left[N{H}_3\right]/dt$
• C. $Rate=-d\left[N_2\right]/dt=\frac{1}{3}d\left[H_2\right]/dt=\frac{1}{2}d\left[N{H}_3\right]/dt$
• D. $Rate=-d\left[N_2\right]/dt=-d\left[H_2\right]/dt=d\left[N{H}_3\right]/dt$

Answer 1

The correct option is A $Rate=-d\left[N_2\right]/dt=-\frac{1}{3}d\left[H_2\right]/dt=\frac{1}{2}d\left[N{H}_3\right]/dt$

For, $N_2+3H_2\rightleftharpoons 2N{H}_3$
rate of reaction $=-\frac{d\left[N_2\right]}{dt}=-\frac{1}{3}\frac{d\left[H_2\right]}{dt}=\frac{1}{2}\frac{d\left[N{H}_3\right]}{dt}$
where, $-\frac{d\left[N_2\right]}{dt}$ is rate of consumption of $N_2$ (-ve sign)
$-\frac{d\left[H_2\right]}{dt}$ is rate of consumption of $H_2$ (-ve sign)
$+\frac{d\left[N{H}_3\right]}{dt}$ is rate of formation of $N{H}_3$ (+ve sign)
Individual rates become equal when each of these is divided by their respective stoichiometric coefficient.