Question:
Following data is for a reaction between reactants A and B:
\[ \begin{array}{ccc} 2 \times 10^{-3} & 0.1 \, \text{M} & 0.1 \, \text{M} \\ 4 \times 10^{-3} & 0.2 \, \text{M} & 0.1 \, \text{M} \\ 1.6 \times 10^{-2} & 0.2 \, \text{M} & 0.2 \, \text{M} \\ \end{array} \]
The order of the reaction with respect to A and B, respectively, are:
Following data is for a reaction between reactants A and B:
\[ \begin{array}{ccc} 2 \times 10^{-3} & 0.1 \, \text{M} & 0.1 \, \text{M} \\ 4 \times 10^{-3} & 0.2 \, \text{M} & 0.1 \, \text{M} \\ 1.6 \times 10^{-2} & 0.2 \, \text{M} & 0.2 \, \text{M} \\ \end{array} \]
The order of the reaction with respect to A and B, respectively, are:
\[ \begin{array}{ccc} 2 \times 10^{-3} & 0.1 \, \text{M} & 0.1 \, \text{M} \\ 4 \times 10^{-3} & 0.2 \, \text{M} & 0.1 \, \text{M} \\ 1.6 \times 10^{-2} & 0.2 \, \text{M} & 0.2 \, \text{M} \\ \end{array} \]
The order of the reaction with respect to A and B, respectively, are:
Updated On: Jan 13, 2026
- 1, 0
- 0, 1
- 1, 2
- 2, 1
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The Correct Option is C
Solution and Explanation
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):
\[\frac{4\times10^{-3}}{2\times10^{-3}}=\frac{k[0.2]^{x}[0.1]^{y}}{k[0.1]^{x}[0.1]^{y}}\]
\[2=2^{x}\]
\[x=1\]
Comparing the second and third rows (keeping [A] constant):
\[\frac{1.6\times10^{-2}}{4\times10^{-3}}=\frac{k[0.2]^{x}[0.2]^{y}}{k[0.2]^{x}[0.1]^{y}}\]
\[4=2^{y}\]
\[y=2\]
Therefore, the order of the reaction with respect to A is 1, and with respect to B is 2.
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