Question

Decomposition of $N{H}_3$ on gold surface follows zero order kinetics. If rate constant $K$ is $5\times {10}^{-4}M$ $s^{-1}$, rate of formation of $N_2$ will be:

• A. ${10}^{-3}M{s}^{-1}$
• B. $2.5\times {10}^{-4}M{s}^{-1}$
• C. $5\times {10}^{-4}M{s}^{-1}$
• D. zero

Answer 1

The correct option is C $5\times {10}^{-4}M{s}^{-1}$

$N_2+3H_2\to 2N{H}_3$

Rate $=\frac{-1}{2}\frac{d\left[N{H}_3\right]}{dT}=\frac{d\left[N_2\right]}{dT}=\frac{1}{3}\frac{d\left[N{H}_3\right]}{dT}=k=5\times {10}^{-4}M{s}^{-1}$

Rate of formation of $N_2$ $=\frac{d\left[N_2\right]}{dT}=k=5\times {10}^{-4}M{s}^{-1}$