Consider the reaction: $N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)}$. The equality relationship between $\frac{d[NH_3]}{dt}$ and $\frac{d[H_2]}{dt}$ is
- A$\frac{d[NH_3]}{dt} = - \frac{d[H_2]}{dt}$
- B$\frac{d[NH_3]}{dt} = - \frac{1}{3} \frac{d[H_2]}{dt}$
- C$+ \frac{d[NH_3]}{dt} = - \frac{2}{3} \frac{d[H_2]}{dt}$
- D$+ \frac{d[NH_3]}{dt} = - \frac{3}{2} \frac{d[H_2]}{dt}$
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View SolutionFor the reaction $N_{2(g)} + 3H_{2(g)} \to 2NH_{3(g)}$ under certain conditions of temperature and partial pressure of the reactants,the rate of formation of $NH_3$ is $0.001 \ kg \ L^{-1} \ h^{-1}$. The rate of consumption of $H_2$ under the same condition is $..... \ kg \ L^{-1} \ h^{-1}$.
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