Considering the reaction aG + bH rightarrow t | vedclass

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(D) The rate law for the reaction is given by: $	ext{Rate} = k[G]^x[H]^y$. From the problem,when $[G]$ is doubled and $[H]$ is constant,the rate doub

Vedclass · Accepted Answer

$3$

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(D) The rate law for the reaction is given by: $	ext{Rate} = k[G]^x[H]^y$. From the problem,when $[G]$ is doubled and $[H]$ is constant,the rate doubles: $2 	imes 	ext{Rate} = k[2G]^x[H]^y = 2^x 	imes 	ext{Rate}$. Thus,$2^x = 2$,which implies $x = 1$. When both $[G]$ and $[H]$ are doubled,the rate increases by $8$ times: $8 	imes 	ext{Rate} = k[2G]^x[2H]^y = 2^x 	imes 2^y 	imes 	ext{Rate}$. Substituting $x = 1$: $8 = 2^1 	imes 2^y$,so $8 = 2 	imes 2^y$,which means $2^y = 4$. Therefore,$y = 2$. The overall order of the reaction is $x + y = 1 + 2 = 3$.

Initial $[A_2]$	Initial $[B_2]$	$R.O.R. (r) \ M s^{-1}$
$0.2$	$0.2$	$0.04$
$0.1$	$0.4$	$0.04$
$0.2$	$0.4$	$0.08$

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