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

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(D) The balanced chemical equation is: $N_{2(g)} + 3 H_{2(g)} \longrightarrow 2 NH_{3(g)}$ The rate of reaction is given by: $	ext{Rate} = -\frac{d[N

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$0.044 \ mol \ dm^{-3} \ s^{-1}$

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(D) The balanced chemical equation is: $N_{2(g)} + 3 H_{2(g)} \longrightarrow 2 NH_{3(g)}$ The rate of reaction is given by: $	ext{Rate} = -\frac{d[N_{2}]}{dt} = -\frac{1}{3} \frac{d[H_{2}]}{dt} = \frac{1}{2} \frac{d[NH_{3}]}{dt}$ Given that the rate of formation of $NH_{3}$ is $\frac{d[NH_{3}]}{dt} = 0.088 \ mol \ dm^{-3} \ s^{-1}$. Substituting this into the rate expression: $-\frac{d[N_{2}]}{dt} = \frac{1}{2} 	imes \frac{d[NH_{3}]}{dt}$ $-\frac{d[N_{2}]}{dt} = \frac{1}{2} 	imes 0.088 \ mol \ dm^{-3} \ s^{-1} = 0.044 \ mol \ dm^{-3} \ s^{-1}$ Therefore,the rate of consumption of $N_{2}$ is $0.044 \ mol \ dm^{-3} \ s^{-1}$.

For the reaction,$N_{2(g)} + 3 H_{2(g)} \longrightarrow 2 NH_{3(g)}$,$NH_{3}$ is formed at a rate of $0.088 \ mol \ dm^{-3} \ s^{-1}$. Calculate the consumption rate of $N_{2(g)}$.

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