Question

In a catalytic experiment involving the Haber process, $N_2+3H_2\to 2N{H}_3$, the rate of reaction was measured as $Rate=\frac{\Delta \left[N{H}_3\right]}{\Delta t}=2\times {10}^{-4}mol{L}^{-1}{s}^{-1}$.
If there were no sides reactions, what was the rate of reaction expressed in terms of (a) $N_2$, (b) $H_2$?

• A. (a) $-1\times {10}^{-4}mol{L}^{-1}{s}^{-1}$, (b) $3\times {10}^{-4}mol{L}^{-1}{s}^{-1}$
• B. (a) $-3\times {10}^{-4}mol{L}^{-1}{s}^{-1}$, (b) $1\times {10}^{-4}mol{L}^{-1}{s}^{-1}$
• C. (a) $-1\times {10}^{-3}mol{L}^{-1}{s}^{-1}$, (b) $-3\times {10}^{-3}mol{L}^{-1}{s}^{-1}$
• D. None of these

Answer 1

The correct option is A (a) $-1\times {10}^{-4}mol{L}^{-1}{s}^{-1}$, (b) $3\times {10}^{-4}mol{L}^{-1}{s}^{-1}$

(a) $N_2+3H_2\to 2N{H}_3$
$Rate=\frac{\Delta \left[N{H}_3\right]}{\Delta t}=2\times {10}^{-4}mol{L}^{-1}{s}^{-1}$
$-\frac{d{N}_2}{dt}=\frac{-1}{3}\frac{d{H}_2}{dt}=\frac{1}{2}\frac{dN{H}_3}{dt}$
$\frac{d{N}_2}{dt}=-1\times {10}^{-4}$
(b) $-\frac{d{H}_2}{dt}=-\frac{3}{2}\times 2\times {10}^{-1}$
$=3\times {10}^{-4}$