For the reaction $A + B \to C + D$ , doubling the concentration of both the reactants increases the reaction rate $8$ times an ddoubling the concentration of only $B$ simply dubles the reaction rate. Find the rate law for the above equation.

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Solution

$A + B \to C + D$
$R a t e_{1} = K [A]^{m} [B]^{n}$
When $[B]$ is doubled
$R a t e_{2} = k [A]^{m} [B]^{n}$
$2^{n} k R a t e_{1}$ $n = 1$
$R a t e_{2} = 2 R a t e_{1}$
$R a t e_{3} = k [2 A]^{m} [2 B]^{1}$
$R a t e_{3} = 2^{m} \times 2 k [A] [B]$
$R a t e_{3} = 8 R a t e_{1}$
So Rate law $= k [A]^{2} [B]$

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For the reaction A+B→C+D, doubling the concentration of both the reactants increases the reaction rate 8 times an ddoubling the concentration of only B simply dubles the reaction rate. Find the rate law for the above equation.

A+B→C+DRate1=K[A]m[B]nWhen [B] is doubledRate2=k[A]m[B]n2nkRate1 n=1Rate2=2Rate1Rate3=k[2A]m[2B]1Rate3=2m×2k[A][B]Rate3=8Rate1So Rate law =k[A]2[B]

For the reaction $A + B \to C + D$ , doubling the concentration of both the reactants increases the reaction rate $8$ times an ddoubling the concentration of only $B$ simply dubles the reaction rate. Find the rate law for the above equation.