For the reaction 3 , A_{(g)} xrightarrow{K} B | vedclass

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Question

(D) The rate law for the reaction $3 \, A_{(g)} \rightarrow B_{(g)} + C_{(g)}$ is given by $-\frac{1}{3} \frac{d[A]}{dt} = K[A]^2$. Given $K = 10^{-14

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None of these

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(D) The rate law for the reaction $3 \, A_{(g)} ightarrow B_{(g)} + C_{(g)}$ is given by $-\frac{1}{3} \frac{d[A]}{dt} = K[A]^2$. Given $K = 10^{-14} \, L/mol \cdot min$ and $[A] = 0.5 \, M$. Substituting the values: $-\frac{d[A]}{dt} = 3 	imes K 	imes [A]^2$. $-\frac{d[A]}{dt} = 3 	imes 10^{-14} 	imes (0.5)^2 = 3 	imes 10^{-14} 	imes 0.25 = 7.5 	imes 10^{-15} \, M/min$. To convert the rate from $M/min$ to $M/sec$,divide by $60$: $-\frac{d[A]}{dt} = \frac{7.5 	imes 10^{-15}}{60} = 1.25 	imes 10^{-16} \, M/sec$. Since this value is not among the options,the correct choice is $D$.

Experiment	$[Cl_2] \ (M)$	$[NO] \ (M)$	Rate $(mol \ L^{-1} \sec^{-1})$
$I$	$0.05$	$0.05$	$1 \times 10^{-3}$
$II$	$0.15$	$0.05$	$3 \times 10^{-3}$
$III$	$0.05$	$0.15$	$9 \times 10^{-3}$

For the reaction $3 \, A_{(g)} \xrightarrow{K} B_{(g)} + C_{(g)}$,$K$ is $10^{-14} \, L/mol \cdot min$. If $[A] = 0.5 \, M$,then the value of $-\frac{d[A]}{dt}$ (in $M/sec$) is:

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