Question

For the reaction A + B $\to$ Products, the following initial rates were obtained at various initial concentrations of reactants:

Sl. No.	[A] (mol L-1)	[B] (mol L-1)	Initial rate (mol L-1 s-1)
1	0.1	0.1	0.05
2	0.2	0.1	0.10
3	0.1	0.2	0.05

Hint:

To determine the order of a reaction, compare the changes in concentration and rate between different experiments, while keeping other concentrations constant.

Answer 1

Given Data

The data for the reaction is given below:

Sl. No.	[A] (mol L-1)	[B] (mol L-1)	Initial Rate (mol L-1 s-1)
1	0.1	0.1	0.05
2	0.2	0.1	0.10
3	0.1	0.2	0.05

Rate Law

The rate law for this reaction is assumed to be:

\[ \text{Rate} = k[A]^m[B]^n \]

where \(m\) is the order with respect to A and \(n\) is the order with respect to B.

(i) Order with respect to A

Compare experiments 1 and 2 (where [B] is constant):

\[ \frac{\text{Rate}_2}{\text{Rate}_1} = \frac{0.10}{0.05} = 2 \quad \text{and} \quad \frac{[A]_2^m}{[A]_1^m} = \frac{(0.2)^m}{(0.1)^m} = 2^m \]

So,

\[ 2^m = 2 \implies m = 1 \]

(ii) Order with respect to B

Compare experiments 1 and 3 (where [A] is constant):

\[ \frac{\text{Rate}_3}{\text{Rate}_1} = \frac{0.05}{0.05} = 1 \quad \text{and} \quad \frac{[B]_3^n}{[B]_1^n} = \frac{(0.2)^n}{(0.1)^n} = 2^n \]

So,

\[ 2^n = 1 \implies n = 0 \]

Overall Order of Reaction

\[ \text{Overall order} = m + n = 1 + 0 = 1 \]

✅ Therefore, the reaction is first-order with respect to A and zero-order with respect to B.