For the reaction N_{2(g)} + 3 H_{2(g)} righta | vedclass

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(B) For the reaction $N_{2(g)} + 3 H_{2(g)} \rightarrow 2 NH_{3(g)}$,the rate of reaction is given by:Rate $= -\frac{d[N_2]}{dt} = -\frac{1}{3} \frac{

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$\frac{d[H_2]}{dt} = 3 \frac{d[N_2]}{dt}$

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(B) For the reaction $N_{2(g)} + 3 H_{2(g)} \rightarrow 2 NH_{3(g)}$,the rate of reaction is given by:Rate $= -\frac{d[N_2]}{dt} = -\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt}$Considering the terms for $N_2$ and $H_2$:$-\frac{d[N_2]}{dt} = -\frac{1}{3} \frac{d[H_2]}{dt}$Multiplying both sides by $-1$:$\frac{d[N_2]}{dt} = \frac{1}{3} \frac{d[H_2]}{dt}$Rearranging the equation to express $\frac{d[H_2]}{dt}$ in terms of $\frac{d[N_2]}{dt}$:$\frac{d[H_2]}{dt} = 3 \frac{d[N_2]}{dt}$

For the reaction $N_{2(g)} + 3 H_{2(g)} \rightarrow 2 NH_{3(g)}$,what is the relation between $\frac{d[N_2]}{dt}$ and $\frac{d[H_2]}{dt}$?

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