Question

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

if $-\frac{d\left[N{H}_3\right]}{dt}=k_1\left[N{H}_3\right],\frac{d\left[N_2\right]}{dt}=k_2\left[N{H}_3\right],\frac{d\left[H_2\right]}{dt}=k_3\left[N{H}_3\right]$

then the relation between $k_1,k_2$ and $k_3$ is?

• A. $k_1=k_2=k_3$
• B. $k_1=3k_2=2k_3$
• C. $1.5k_1=3k_2=k_3$
• D. $2k_1=k_2=3k_3$

Answer 1

Answer: (C)

$1.5k_1=3k_2=k_3$

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

$Rate=-\frac{1}{2}\frac{d\left[N{H}_3\right]}{dt}=\frac{d\left[N_2\right]}{dt}=\frac{1}{3}\frac{d\left[H_2\right]}{dt}$

$\frac{1}{2}{k}_1\left[N{H}_3\right]=k_2\left[N{H}_3\right]=\frac{1}{3}{k}_3\left[N{H}_3\right]$

$1.5k_1=3k_2=k_3$.