Question

Compounds ‘A’ and ‘B’ react according to the following chemical equation.
\[\ce{A(g) + 2B(g) -> 2C(g)}\]
Concentration of either ‘A’ or ‘B’ were changed keeping the concentrations of one of the reactants constant and rates were measured as a function of initial concentration. Following results were obtained. Choose the correct option for the rate equations for this reaction.

Experiment	Initial concentration of [A]/mol L–¹	Initial concentration of [B]/mol L–¹	Initial rate of formation of [C]/mol L–¹ s–¹
1.	0.30	0.30	0.10
2.	0.30	0.60	0.40
3.	0.60	0.30	0.20

विकल्प

• Rate = k[A]²[B]

• Rate = k[A][B]²

• Rate = k[A][B]

• Rate = k[A]²[B]⁰ 

Answer 1

Rate = k[A][B]² 

Explanation:

Rate = k[A]^x[B]^y

When concentration of B is doubled keeping the concentration of A constant, the rate of formation of C increases by a factor of four. This indicates that the rate of reactions depends upon the square of concentration of B. When concentration of A is doubled, the rate of formation of C also doubles from the initial value. This shows that the rate depends on first power of concentration of A. Hence Rate= k[A]^x[B]^y.

Answer 2

Let order of A and B be x and y.

r = k [A]^x [B]^y

0.1 = k (0.3)^x (0.3)^y  ...(1)

0.4 = 0.1 = k (0.3)^x (0.6)^y   ...(2)

0.2 = 0.1 = k (0.6)^x (0.3)^y   ...(3)

Divide 2 by 1

`0.4/0.1 = (0.6)^"y"/(0.3)^"y"`

Divide 3 by 1

`0.2/0.1 = (0.6)^"x"/(0.3)^"x"`

Hence Rate law is

r = k [A]¹ [B]²