For the reaction $2 A + B \to C$ , the values of initial rate at different reactant concentrations are given in the table below: The rate law for the reaction is:

$[A] (m o l L^{- 1})$ $[B] (m o l L^{- 1})$ $Initial Rate (m o l L^{- 1} s^{- 1})$

0.05 0.05 0.045

0.10 0.05 0.090

0.20 0.10 0.72

Rate =

k [A]^{2} [B]^{2}

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Rate =

k [A] [B]^{2}

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Rate =

k [A] [B]

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Rate =

k [A]^{2} [B]

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Solution

The correct option is B Rate = $k [A] [B]^{2}$
Let, $r = k [A]^{x} [B]^{y}$
$0.045 = k (0.05)^{x} (0.05)^{y}$ .... (1)
$0.090 = k (0.10)^{x} (0.05)^{y}$ ..... (2)
$0.72 = k (0.20)^{x} (0.10)^{y}$ ..... (3)
Dividing (2) by (1),
$2 = 2^{x} \Rightarrow x = 1$
Dividing (3) by (2),
$8 = (2^{1}) (2)^{y} \Rightarrow y = 2$
So, rate= $k [A] [B]^{2}$

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For the reaction 2A+B→C, the values of initial rate at different reactant concentrations are given in the table below: The rate law for the reaction is: [A](molL−1) [B](molL−1) Initial Rate(molL−1s−1) 0.05 0.05 0.045 0.10 0.05 0.090 0.20 0.10 0.72

The correct option is B Rate = k[A][B]2Let, r=k[A]x[B]y 0.045=k(0.05)x(0.05)y.... (1) 0.090=k(0.10)x(0.05)y..... (2) 0.72=k(0.20)x(0.10)y..... (3) Dividing (2) by (1), 2=2x⇒x=1 Dividing (3) by (2), 8=(21)(2)y⇒y=2 So, rate= k[A][B]2

For the reaction $2 A + B \to C$ , the values of initial rate at different reactant concentrations are given in the table below: The rate law for the reaction is:

$[A] (m o l L^{- 1})$ $[B] (m o l L^{- 1})$ $Initial Rate (m o l L^{- 1} s^{- 1})$

0.05 0.05 0.045

0.10 0.05 0.090

0.20 0.10 0.72