The rate constant,k of a zero order reaction | vedclass

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(C) For a zero order reaction,the rate of reaction $r$ is equal to the rate constant $k$.Given $r = k = y 	imes 10^{-4} \ mol \ L^{-1} \ s^{-1}$.The

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$3 y 	imes 10^{-4}$

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(C) For a zero order reaction,the rate of reaction $r$ is equal to the rate constant $k$.Given $r = k = y 	imes 10^{-4} \ mol \ L^{-1} \ s^{-1}$.The stoichiometric equation is $2 NH_{3(g)} ightarrow N_{2(g)} + 3 H_{2(g)}$.The rate of reaction is expressed as $r = -\frac{1}{2} \frac{d[NH_3]}{dt} = \frac{d[N_2]}{dt} = \frac{1}{3} \frac{d[H_2]}{dt}$.Therefore,the rate of formation of hydrogen is $\frac{d[H_2]}{dt} = 3 	imes r$.Substituting the value of $r$,we get $\frac{d[H_2]}{dt} = 3 	imes (y 	imes 10^{-4}) = 3y 	imes 10^{-4} \ mol \ L^{-1} \ s^{-1}$.

The rate constant,$k$ of a zero order reaction $2 NH_{3(g)} \xrightarrow[1130 \ K]{Pt} N_{2(g)} + 3 H_{2(g)}$ is $y \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$. The rate of formation of hydrogen (in $mol \ L^{-1} \ s^{-1}$) is

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