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

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(A) The rate of reaction is given by the expression: $-\frac{d[N_2]}{dt} = -\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt}$.Given that

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$4.44 	imes 10^{-3} \ mol \ dm^{-3} \ s^{-1}$

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(A) The rate of reaction is given by the expression: $-\frac{d[N_2]}{dt} = -\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt}$.Given that the rate of disappearance of $N_2$ is $-\frac{d[N_2]}{dt} = 2.22 	imes 10^{-3} \ mol \ dm^{-3} \ s^{-1}$.From the stoichiometric relationship,we have: $\frac{d[NH_3]}{dt} = 2 	imes (-\frac{d[N_2]}{dt})$.Substituting the given value: $\frac{d[NH_3]}{dt} = 2 	imes (2.22 	imes 10^{-3} \ mol \ dm^{-3} \ s^{-1}) = 4.44 	imes 10^{-3} \ mol \ dm^{-3} \ s^{-1}$.

For the reaction $N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)}$,the rate of disappearance of $N_{2(g)}$ is $2.22 \times 10^{-3} \ mol \ dm^{-3} \ s^{-1}$. What is the rate of appearance of $NH_{3(g)}$?

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