Question

Formation of ammonia is given by the elementary reaction $N_2+3H_2\to 2N{H}_3$ . If the rate of formation of ammonia is $\frac{\Delta \left[N{H}_3\right]}{\Delta t}=4\times \text{1}{\text{0}}^{-4}mol{l}^{-1}{s}^{-1},$ then calculate value of $\frac{-\Delta \left[H_2\right]}{\Delta t}$

• A. $3\times \text{1}{\text{0}}^{-4}mol{l}^{-1}{s}^{-1}$
• B. $4\times \text{1}{\text{0}}^{-4}mol{l}^{-1}{s}^{-1}$
• C. $6\times \text{1}{\text{0}}^{-4}mol{l}^{-1}{s}^{-1}$
• D. $\text{6.5}\times \text{1}{\text{0}}^{-4}mol{l}^{-1}{s}^{-1}$

Answer 1

Answer: (C)

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

$\frac{-\Delta \left[N_2\right]}{\Delta t}=-\frac{1}{3}\frac{\Delta \left[H_2\right]}{\Delta t}=\frac{1}{2}\frac{\Delta \left[N{H}_3\right]}{\Delta t}$

$\therefore \frac{-\Delta \left[H_2\right]}{\Delta t}=\frac{3}{2}\times \frac{\Delta \left[N{H}_3\right]}{\Delta t}=\frac{3}{2}\times 4\times \text{1}{\text{0}}^{-4}$

$=6\times \text{1}{\text{0}}^{-4}mollitr{e}^{-1}se{c}^{-1}$