The rate constant for the reaction A longrigh | vedclass

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(C) The units of the rate constant $k = 2 	imes 10^{-4} \ L \ mol^{-1} \ min^{-1}$ indicate that the reaction is of second order.The rate law is $Rat

Vedclass · Accepted Answer

$0.5 \ M$

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(C) The units of the rate constant $k = 2 	imes 10^{-4} \ L \ mol^{-1} \ min^{-1}$ indicate that the reaction is of second order.The rate law is $Rate = k[A]^2$.First,convert the rate constant to $sec^{-1}$ units: $k = (2 	imes 10^{-4} \ L \ mol^{-1} \ min^{-1}) / 60 = (1 / 30) 	imes 10^{-4} \ L \ mol^{-1} \ sec^{-1}$.Given $Rate = (1 / 12) 	imes 10^{-5} \ M \ sec^{-1}$,we have $(1 / 12) 	imes 10^{-5} = (1 / 30) 	imes 10^{-4} 	imes [A]^2$.Simplifying: $[A]^2 = [(1 / 12) 	imes 10^{-5}] / [(1 / 30) 	imes 10^{-4}] = (30 / 12) 	imes 10^{-1} = 2.5 	imes 0.1 = 0.25$.Thus,$[A] = \sqrt{0.25} = 0.5 \ M$.

$P_{A_0} \text{ (atm)}$	$0.1$	$0.025$
$t_{1/2} \text{ (sec)}$	$100$	$50$

The rate constant for the reaction $A \longrightarrow B$ is $2 \times 10^{-4} \ L \ mol^{-1} \ min^{-1}$. The concentration of $A$ at which the rate of the reaction is $(1 / 12) \times 10^{-5} \ M \ sec^{-1}$ is :-

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