The reaction given below 2 NH_{3(g)} xrightar | vedclass

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(D) For the reaction,$2 NH_{3(g)} \xrightarrow{Pt} N_{2(g)} + 3 H_{2(g)}$ The rate of reaction is given by: $	ext{Rate} = -\frac{1}{2} \frac{d[NH_3]}

Vedclass · Accepted Answer

$7.5 	imes 10^{-6} \ mol \ dm^{-3} \ sec^{-1}$

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(D) For the reaction,$2 NH_{3(g)} \xrightarrow{Pt} N_{2(g)} + 3 H_{2(g)}$ The rate of reaction is given by: $	ext{Rate} = -\frac{1}{2} \frac{d[NH_3]}{dt} = \frac{d[N_2]}{dt} = \frac{1}{3} \frac{d[H_2]}{dt}$ Given that the rate of reaction is $2.5 	imes 10^{-6} \ mol \ dm^{-3} \ sec^{-1}$,we have: $	ext{Rate} = \frac{1}{3} \frac{d[H_2]}{dt} = 2.5 	imes 10^{-6} \ mol \ dm^{-3} \ sec^{-1}$ Therefore,the rate of formation of $H_{2(g)}$ is: $\frac{d[H_2]}{dt} = 3 	imes 2.5 	imes 10^{-6} \ mol \ dm^{-3} \ sec^{-1} = 7.5 	imes 10^{-6} \ mol \ dm^{-3} \ sec^{-1}$

The reaction given below $2 NH_{3(g)} \xrightarrow{Pt} N_{2(g)} + 3 H_{2(g)}$ has a rate of reaction of $2.5 \times 10^{-6} \ mol \ dm^{-3} \ sec^{-1}$. What is the rate of formation of $H_{2(g)}$?

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