Question

Compounds ‘A’ and ‘B’ react according to the following chemical equation.

$A_{(g)} + 2B_{(g)} \rightarrow 2C_{(g)}$

Concentration of either ‘A’ or ‘B’ were changed keeping the concentration of one of the reactants constant and rate were measured as a function of initial concentration following results were obtained. Choose the correct option for the rate equation for this reaction.

Experiment	Initial concentration Of [A]/molL−1	Initial Concentration of [B]/molL−1	Initial rate of formation) [C] molL−1s−1
1.	0.30	0.30	0.10
2.	0.30	0.60	0.40
3.	0.60	0.30	0.20

• A. Rate = $k\lbrack A\rbrack^{2}\lbrack B\rbrack$
• B. Rate = $k\lbrack A\rbrack\lbrack B\rbrack^{2}$
• C. Rate = $k\lbrack A\rbrack\lbrack B\rbrack$
• D. $Rate = k\lbrack A\rbrack^{2}\lbrack B\rbrack^{0}$

Answer 1

Answer: (B)

Rate = $k\lbrack A\rbrack\lbrack B\rbrack^{2}$

Answer 2

Let order with respect to A and B are x and y

respectively.

$\therefore$ $Rate = k(A)^{x}(B)^{y}$

$0.1 = k(0.3)^{x}(0.3)^{y}$ ……. (i)

$0.4 = k(0.3)^{x}(0.6)^{y}$ …….. (ii)

$0.2 = k(0.6)^{x}(0.3)^{y}$ …….. (iii)

Dividing (ii) by (i)

$\frac{0.4}{0.1} = \frac{(0.6)^{y}}{(0.3)^{y}}$

$\therefore y = 2$

Dividing (iii) by (i)

$\frac{0.2}{0.1} = \frac{(0.6)^{x}}{(0.3)^{x}}$

$\therefore x = 1$

Rate law will be:

$Rate = k\lbrack A\rbrack^{1}\lbrack B\rbrack^{2}$