Question

For the reaction $2NH_{3} \to N_{2}+3H_{2}$,
if $-\frac{d\left[NH_{3}\right]}{dt}=k_{1}\left[NH_{3}\right], \frac{d \left[N_{2}\right]}{dt}=k_{2}\left[NH_{3}\right], \frac{d\left[H_{2}\right]}{dt}=K_{3}\left[NH_{3}\right]$
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. $1.5 k_{1} = 3k_{2} = k_{3}$
• D. $2k_{1} = k_{2} = 3k_{3}$

Answer 1

Answer: (C)

$2NH_{3} \rightarrow N_{2}+3H_{2}$
Rate $= -\frac{1}{2}\frac{d[NH_{3}]}{dt}=\frac{d[N_{2}]}{dt}=\frac{1}{3}\frac{d[H_{2}]}{dt}$
$\frac{1}{2}k_{1}[NH_{3}]=k_{2}[NH_{3}]=\frac{1}{3}k_{3}[NH_{3}]$
$1.5k_{1}=3k_{2}=k_{3}$