Question

For a reaction $2NH_3 \longrightarrow N_2 + 3H_2$, it is observed that $-\frac{d[NH_3]}{dt} = k_1[NH_3]$; $+\frac{d[N_2]}{dt} = k_2[NH_3]$ and $+\frac{d[H_2]}{dt} = k_3[NH_3]$

The correct relation between $k_1$, $k_2$ and $k_3$ is:

• A. k_1 = k_2 = k_3
• B. 2k_1 = 3k_2 = 6k_3
• C. 3k_1 = 6k_2 = 2k_3
• D. 6k_1 = 3k_2 = 2k_3

Answer 1

Answer: (C)

3k_1 = 6k_2 = 2k_3

Answer 2

The reaction is $2NH_3 \longrightarrow N_2 + 3H_2$. The rate of the reaction can be expressed in terms of the rate of change of concentration of each species, divided by its stoichiometric coefficient.

Rate $= -\frac{1}{2}\frac{d[NH_3]}{dt} = +\frac{d[N_2]}{dt} = +\frac{1}{3}\frac{d[H_2]}{dt}$.

We are given:

$-\frac{d[NH_3]}{dt} = k_1[NH_3]$

$+\frac{d[N_2]}{dt} = k_2[NH_3]$

$+\frac{d[H_2]}{dt} = k_3[NH_3]$

Substitute:

Rate $= \frac{1}{2}k_1[NH_3] = k_2[NH_3] = \frac{1}{3}k_3[NH_3]$.

Therefore, $\frac{1}{2}k_1 = k_2 = \frac{1}{3}k_3$.

Multiply by 6: $3k_1 = 6k_2 = 2k_3$.