Which of the following rate law corresponds to the data shown for the reaction?Expt No. [A]0moleL-1 [B]0moleL-1 Initital rate 1. 0.012 0.035 0.10 2. 0.024 0.070 0.80 3. 0.024 0.035 0.10 4. 0.012 0.070 0.80

Expt. 4 Expt. 1=[0.012]x[0.070]γ[0.012]x[0.035]γ=0.800.102y =8y = 3 Expt. 3 Expt. 1=[0.024]x[0.035]3[0.012]7[0.035]3=0.100.102x =1⇒x=0 Rate =k[A]0[B]3

Which of the following rate law corresponds to the data shown for the reaction?Expt No. [A]0moleL-1 [B]0moleL-1 Initital rate 1. 0.012 0.035 0.10 2. 0.024 0.070 0.80 3. 0.024 0.035 0.10 4. 0.012 0.070 0.80

Expt. 4 Expt. 1=[0.012]x[0.070]γ[0.012]x[0.035]γ=0.800.102y =8y = 3 Expt. 3 Expt. 1=[0.024]x[0.035]3[0.012]7[0.035]3=0.100.102x =1⇒x=0 Rate =k[A]0[B]3

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Which of the following rate law corresponds to the data shown for the reaction?
Expt No.
[A]₀moleL^-1
[B]₀moleL^-1
Initital rate

1.
0.012
0.035
0.10

2.
0.024
0.070
0.80

3.
0.024
0.035
0.10

4.
0.012
0.070
0.80

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Rate law = K[A]³

Rate law = K[B]³

Rate law = K[A][B]³

Rate law = K[A]³[B]

answer is B.

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Detailed Solution

$\begin{array}{l} \frac{Expt. 4}{Expt. 1} = \frac{[0 .012]^{x} [0 .070]^{γ}}{[0 .012]^{x} [0 .035]^{γ}} = \frac{0 .80}{0 .10} \\ 2^{y = 8} \\ y = 3 \\ \frac{Expt. 3}{Expt. 1} = \frac{[0 .024]^{x} [0 .035]^{3}}{[0 .012]^{7} [0 .035]^{3}} = \frac{0 .10}{0 .10} \\ 2^{x} = 1 \Rightarrow x = 0 \\ Rate = k [A]^{0} [B]^{3} \end{array}$