Question

For the reaction $N_{2}+3H_{2} \rightarrow 2NH_{3}$
if $\frac{\Delta\left[NH_{3}\right]}{\Delta t}=2\times10^{-4}\,mol\,l^{-1}s^{-1},$ then value of $\frac{-\Delta\left[H_{2}\right]}{\Delta t}$ would be

• A. $1 × 10^{-4}\, mol\, L^{-1}s^{-1}$
• B. $3 × 10^{-4}\, mol\, L^{-1}s^{-1}$
• C. $4 × 10^{-4}\, mol\, L^{-1}s^{-1}$
• D. $6 × 10^{-4}\, mol\, L^{-1}s^{-1}$

Answer 1

Answer: (B)

$N_{2}+3H_{2} \rightarrow 2NH_{3}$
$\frac{-\left[N_{2}\right]}{\Delta t}=\frac{1}{3} \frac{\Delta\left[H_{2}\right]}{\Delta t}=\frac{1}{2} \frac{\Delta\left[NH_{3}\right]}{\Delta t}$
$\therefore \frac{-\Delta\left[H_{2}\right]}{\Delta t}=\frac{3}{2}\times\frac{\Delta\left[NH_{3}\right]}{\Delta t}=\frac{3}{2}\times2\times10^{-4}$
$=3\times10^{-4}\,mol\,L^{-1}S^{-1}$