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:

• A. 1, 0
• B. 0, 1
• C. 1, 2
• D. 2, 1

Answer 1

The Correct Option is C

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.