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Chemistry Question on Chemical Kinetics

Following data is for a reaction between reactants A and B:

2×10−30.1 M0.1 M 4×10−30.2 M0.1 M 1.6×10−20.2 M0.2 M \begin{array}{ccc} 2 \times 10^{-3} & 0.1 \, \text{M} & 0.1 \, \text{M} \\\ 4 \times 10^{-3} & 0.2 \, \text{M} & 0.1 \, \text{M} \\\ 1.6 \times 10^{-2} & 0.2 \, \text{M} & 0.2 \, \text{M} \\\ \end{array}
The order of the reaction with respect to A and B, respectively, are:

A

1, 0

B

0, 1

C

1, 2

D

2, 1

Answer

1, 2

Explanation

Solution

Let the rate law be given by: Rate = k[A]x[B]y, where x and y are the orders of the reaction with respect to A and B respectively.

From the given data:

Comparing the first and second rows (keeping [B] constant):

4×10−32×10−3=k[0.2]x[0.1]yk[0.1]x[0.1]y\frac{4\times10^{-3}}{2\times10^{-3}}=\frac{k[0.2]^{x}[0.1]^{y}}{k[0.1]^{x}[0.1]^{y}}

2=2x2=2^{x}

x=1x=1

Comparing the second and third rows (keeping [A] constant):

1.6×10−24×10−3=k[0.2]x[0.2]yk[0.2]x[0.1]y\frac{1.6\times10^{-2}}{4\times10^{-3}}=\frac{k[0.2]^{x}[0.2]^{y}}{k[0.2]^{x}[0.1]^{y}}

4=2y4=2^{y}

y=2y=2

Therefore, the order of the reaction with respect to A is 1, and with respect to B is 2.