The following graph shows the relationship be | vedclass

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(C) For a second-order reaction,the integrated rate equation is given by: $\frac{1}{(a-x)} = \frac{1}{a} + Kt$.Comparing this with the equation of a s

Vedclass · Accepted Answer

$0.125$

Vedclass · Answer

(C) For a second-order reaction,the integrated rate equation is given by: $\frac{1}{(a-x)} = \frac{1}{a} + Kt$.Comparing this with the equation of a straight line $y = mx + c$,the slope $m = K$ and the intercept $c = \frac{1}{a}$.Given $	heta = 	an^{-1}(1/2)$,the slope is $K = 	an 	heta = 1/2 = 0.5 \ L \ mol^{-1} \ min^{-1}$.The intercept $OA = \frac{1}{a} = 2 \ L \ mol^{-1}$,which gives the initial concentration $a = 1/2 = 0.5 \ mol \ L^{-1}$.The rate of reaction at the start $(t=0)$ is given by $r = K[A]_0^2$.Substituting the values: $r = 0.5 	imes (0.5)^2 = 0.5 	imes 0.25 = 0.125 \ mol \ L^{-1} \ min^{-1}$.

Experiment	$[S]$	$[Nu]$	Rate
$1$	$0.1$	$0.1$	$2.2 \times 10^{-3}$
$2$	$0.2$	$0.1$	$4.4 \times 10^{-3}$
$3$	$0.1$	$0.2$	$4.4 \times 10^{-3}$

The following graph shows the relationship between $(a-x)^{-1}$ and time $t$ for a second-order reaction. If $\theta = \tan^{-1}(1/2)$ and $OA = 2 \ L \ mol^{-1}$,then the rate at the start of the reaction will be (in $mol \ L^{-1} \ min^{-1}$):

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