Consider the kinetic data given in the follow | vedclass

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(A) The rate law is given by $r = K [A]^{n_1} [B]^{n_2} [C]^{n_3}$.Comparing experiments $1$ and $2$,$[A]$ and $[C]$ are constant,but $[B]$ changes an

Vedclass · Accepted Answer

$6.75$

Vedclass · Answer

(A) The rate law is given by $r = K [A]^{n_1} [B]^{n_2} [C]^{n_3}$.Comparing experiments $1$ and $2$,$[A]$ and $[C]$ are constant,but $[B]$ changes and the rate remains constant,so $n_2 = 0$.Comparing experiments $1$ and $3$,$[A]$ and $[B]$ are constant,$[C]$ doubles,and the rate doubles,so $n_3 = 1$.Comparing experiments $1$ and $4$,$[B]$ and $[C]$ are constant,$[A]$ increases by $1.5$ times,and the rate increases by $1.5$ times,so $n_1 = 1$.Thus,the rate law is $r = K [A] [C]$.Using experiment $1$: $6.0 	imes 10^{-5} = K 	imes 0.2 	imes 0.1$,which gives $K = 3.0 	imes 10^{-3} \ L \ mol^{-1} \ s^{-1}$.Now,for $[A] = 0.15 \ mol \ dm^{-3}, [B] = 0.25 \ mol \ dm^{-3}, [C] = 0.15 \ mol \ dm^{-3}$:$r = (3.0 	imes 10^{-3}) 	imes 0.15 	imes 0.15 = 6.75 	imes 10^{-5} \ mol \ dm^{-3} \ s^{-1}$.Comparing this with $Y 	imes 10^{-5}$,we get $Y = 6.75$.

Experiment No.	$[A] \ (mol \ dm^{-3})$	$[B] \ (mol \ dm^{-3})$	$[C] \ (mol \ dm^{-3})$	Rate of reaction $(mol \ dm^{-3} \ s^{-1})$
$1$	$0.2$	$0.1$	$0.1$	$6.0 \times 10^{-5}$
$2$	$0.2$	$0.2$	$0.1$	$6.0 \times 10^{-5}$
$3$	$0.2$	$0.1$	$0.2$	$1.2 \times 10^{-4}$
$4$	$0.3$	$0.1$	$0.1$	$9.0 \times 10^{-5}$

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