Question

In a catalytic experiment involving the Haber process, $N_2+3H_2\to 2N{H}_3$, the rate of formation of Ammonia 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, then the expression in terms of $N_2$ is:

• A. $1\times {10}^{-4}mol{L}^{-1}{s}^{-1}$
• B. $2\times {10}^{-4}mol{L}^{-1}{s}^{-1}$
• C. $3\times {10}^{-4}mol{L}^{-1}{s}^{-1}$
• D. $4\times {10}^{-4}mol{L}^{-1}{s}^{-1}$

Answer 1

Answer: (A)

$1\times {10}^{-4}mol{L}^{-1}{s}^{-1}$

The rate of consumption of nitrogen is one half the rate of formation of ammonia.

$\frac{\Delta \left[N_2\right]}{\Delta t}=\frac{1}{2}\times \frac{\Delta \left[N{H}_3\right]}{△t}=\frac{1}{2}\times 2\times {10}^{-4}mol{L}^{-1}{s}^{-1}=1\times {10}^{-4}mol{L}^{-1}{s}^{-1}$