Question

Formation of ammonia is given by the elementary reaction $N_{2}+3H_{2} \rightarrow 2NH_{3}$ . If the rate of formation of ammonia is $\frac{\Delta \left[N H_{3}\right]}{\Delta t}=4\times \text{1}\text{0}^{- 4}moll^{- 1}s^{- 1},$ 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)

$N_{2}+3H_{2}\rightleftharpoons2NH_{3}$

$\frac{- \Delta \left[N_{2}\right]}{\Delta t}=-\frac{1}{3}\frac{\Delta \left[H_{2}\right]}{\Delta t}=\frac{1}{2}\frac{\Delta \left[N H_{3}\right]}{\Delta t}$

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

$=6\times \text{1}\text{0}^{- 4}mol\,litre^{- 1}sec^{- 1}$