For the reaction $2 NH_{3(g)} \rightarrow N_{2(g)} + 3 H_{2(g)}$,the rate of disappearance of $NH_3$ is $1.2 \times 10^{-3} \ mol \ L^{-1} \ s^{-1}$. What is the rate of formation of $N_2$ and $H_2$?
(N/A) The rate expression for the reaction is given by: $-\frac{1}{2} \frac{\Delta[NH_3]}{\Delta t} = \frac{\Delta[N_2]}{\Delta t} = \frac{1}{3} \frac{\Delta[H_2]}{\Delta t}$
Given that the rate of disappearance of $NH_3$ is $-\frac{\Delta[NH_3]}{\Delta t} = 1.2 \times 10^{-3} \ mol \ L^{-1} \ s^{-1}$.
For $N_2$: $\frac{\Delta[N_2]}{\Delta t} = \frac{1}{2} \times (1.2 \times 10^{-3}) = 0.6 \times 10^{-3} \ mol \ L^{-1} \ s^{-1}$.
For $H_2$: $\frac{1}{3} \frac{\Delta[H_2]}{\Delta t} = \frac{1}{2} \times (1.2 \times 10^{-3})$.
$\frac{\Delta[H_2]}{\Delta t} = \frac{3}{2} \times 1.2 \times 10^{-3} = 1.8 \times 10^{-3} \ mol \ L^{-1} \ s^{-1}$.
Given that the rate of disappearance of $NH_3$ is $-\frac{\Delta[NH_3]}{\Delta t} = 1.2 \times 10^{-3} \ mol \ L^{-1} \ s^{-1}$.
For $N_2$: $\frac{\Delta[N_2]}{\Delta t} = \frac{1}{2} \times (1.2 \times 10^{-3}) = 0.6 \times 10^{-3} \ mol \ L^{-1} \ s^{-1}$.
For $H_2$: $\frac{1}{3} \frac{\Delta[H_2]}{\Delta t} = \frac{1}{2} \times (1.2 \times 10^{-3})$.
$\frac{\Delta[H_2]}{\Delta t} = \frac{3}{2} \times 1.2 \times 10^{-3} = 1.8 \times 10^{-3} \ mol \ L^{-1} \ s^{-1}$.
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