Question

For the reaction $N_2+3H_2\to 2N{H}_3$, the rate of disappearance of $H_2\text{is}0.01\text{mol}{\text{L}}^{-1}{\text{min}}^{-1}$. The rate of appearance of $N{H}_3$ would be:

• A. $0.01{\text{mol L}}^{-1}{\text{min}}^{-1}$
• B. $0.02{\text{mol L}}^{-1}{\text{min}}^{-1}$
• C. $0.007{\text{mol L}}^{-1}{\text{min}}^{-1}$
• D. $0.002{\text{mol L}}^{-1}{\text{min}}^{-1}$

Answer 1

The correct option is C $0.007{\text{mol L}}^{-1}{\text{min}}^{-1}$

For the reaction: $N_2+3H_2\to 2N{H}_3$
Rate of reaction $=\frac{-d\left[N_2\right]}{dt}=\frac{-1}{3}\frac{d\left[H_2\right]}{dt}=\frac{1}{2}\frac{d\left[N{H}_3\right]}{dt}$
$\therefore \frac{d\left[N{H}_3\right]}{dt}=\frac{2}{3}\times \frac{d\left[H_2\right]}{dt}=\frac{2}{3}\times 0.01=0.00666\approx 0.007\frac{\text{mol}}{\text{L min}}$