Medium
For the decomposition reaction $2NH_3(g) \rightarrow N_2(g) + 3H_2(g)$,if the rate constant $k = 2.6 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$,calculate the rate of formation of $N_2$ and $H_2$.
The rate of reaction is given by: $\text{Rate} = -\frac{1}{2} \frac{d[NH_3]}{dt} = \frac{d[N_2]}{dt} = \frac{1}{3} \frac{d[H_2]}{dt}$.
Since the reaction is zero order,$\text{Rate} = k = 2.6 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$.
For $N_2$: $\frac{d[N_2]}{dt} = \text{Rate} = 2.6 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$.
For $H_2$: $\frac{1}{3} \frac{d[H_2]}{dt} = \text{Rate}$ $\Rightarrow \frac{d[H_2]}{dt} = 3 \times \text{Rate} = 3 \times (2.6 \times 10^{-4}) = 7.8 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$.
Since the reaction is zero order,$\text{Rate} = k = 2.6 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$.
For $N_2$: $\frac{d[N_2]}{dt} = \text{Rate} = 2.6 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$.
For $H_2$: $\frac{1}{3} \frac{d[H_2]}{dt} = \text{Rate}$ $\Rightarrow \frac{d[H_2]}{dt} = 3 \times \text{Rate} = 3 \times (2.6 \times 10^{-4}) = 7.8 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$.
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