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Question

(A) For the reaction $A_2 + B_2 \rightarrow 2 AB$,the rate of reaction $R$ is defined as $R = k[A_2]^x [B_2]^y$.The rate of formation of $AB$ is given

Vedclass · Accepted Answer

$1.25 	imes 10^{-2}$

Vedclass · Answer

(A) For the reaction $A_2 + B_2 ightarrow 2 AB$,the rate of reaction $R$ is defined as $R = k[A_2]^x [B_2]^y$.The rate of formation of $AB$ is given as $\frac{d[AB]}{dt} = 2R$.From the table:$1.25 	imes 10^{-4} = k(0.1)^x(0.1)^y \dots (i)$$2.5 	imes 10^{-4} = k(0.2)^x(0.1)^y \dots (ii)$$5.0 	imes 10^{-4} = k(0.2)^x(0.2)^y \dots (iii)$Dividing $(ii)$ by $(i)$,we get $2^x = 2$,so $x = 1$.Dividing $(iii)$ by $(ii)$,we get $2^y = 2$,so $y = 1$.Thus,the rate law is $R = k[A_2][B_2]$.Using values from $(i)$: $1.25 	imes 10^{-4} = k(0.1)(0.1) = k(0.01)$.$k = \frac{1.25 	imes 10^{-4}}{0.01} = 1.25 	imes 10^{-2} \ L \ mol^{-1} s^{-1}$.

$[A_2]_0$	$[B_2]_0$	Initial rate of formation of $AB$ $(mol \ L^{-1} s^{-1})$
$0.1 \ M$	$0.1 \ M$	$2.5 \times 10^{-4}$
$0.2 \ M$	$0.1 \ M$	$5.0 \times 10^{-4}$
$0.2 \ M$	$0.2 \ M$	$1.0 \times 10^{-3}$

Experiment	$1$	$2$	$3$
$[HI] \ (mol \ L^{-1})$	$0.005$	$0.01$	$0.02$
Rate $(mol \ L^{-1} \ s^{-1})$	$7.5 \times 10^{-4}$	$3.0 \times 10^{-3}$	$1.2 \times 10^{-2}$

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