For the 2^{nd} order reaction R rightarrow P, | vedclass

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(B) For a $2^{nd}$ order reaction,the integrated rate law is $\frac{1}{[R]} = kt + \frac{1}{[R]_0}$.Comparing this with the equation of a straight lin

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Initial concentration of $R$ is $2 \ M$

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(B) For a $2^{nd}$ order reaction,the integrated rate law is $\frac{1}{[R]} = kt + \frac{1}{[R]_0}$.Comparing this with the equation of a straight line $y = mx + c$,we get the slope $k = 	an(	heta) = 	an(	an^{-1}(0.5)) = 0.5 \ L \ mol^{-1} \ min^{-1}$ and the intercept $\frac{1}{[R]_0} = 2 \ L \ mol^{-1}$.Thus,$[R]_0 = \frac{1}{2} = 0.5 \ M$. This makes option $B$ incorrect and option $C$ correct.The half-life is $t_{1/2} = \frac{1}{k[R]_0} = \frac{1}{0.5 	imes 0.5} = \frac{1}{0.25} = 4 \ min$. So,option $A$ is correct.For $75\%$ completion,$[R] = [R]_0 - 0.75[R]_0 = 0.25[R]_0 = 0.125 \ M$.Using the rate law: $\frac{1}{0.125} = 0.5 	imes t + 2 \implies 8 = 0.5t + 2 \implies 0.5t = 6 \implies t = 12 \ min$.Since $12 \ min = \frac{12}{60} \ hours = 0.2 \ hours$,option $D$ is correct.Therefore,the incorrect statement is $B$.

$Exp. \ No.$	$[A]$	$[B]$	$Rate$
$1.$	$1.0$	$0.15$	$4.2 \times 10^{-6}$
$2.$	$2.0$	$0.15$	$8.4 \times 10^{-6}$
$3.$	$1.0$	$0.20$	$5.6 \times 10^{-6}$

For the $2^{nd}$ order reaction $R \rightarrow P$,the following graph holds true. Identify the incorrect statement.

Similar Questions

$A$ study of chemical kinetics of the reaction $A + B \to$ Products,gave the following data at $25 \ ^oC$.

$Exp. \ No.$ $[A]$ $[B]$ $Rate$

$1.$ $1.0$ $0.15$ $4.2 \times 10^{-6}$

$2.$ $2.0$ $0.15$ $8.4 \times 10^{-6}$

$3.$ $1.0$ $0.20$ $5.6 \times 10^{-6}$

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