Question

For the reaction 2A + B → C, the values of initial rate at different reactant concentration are given in the table below. The rate law for the reaction is:

[A] (mol L-1)	[B] (mol L-1)	Initial Rate (mol L-1 S-1)
0.05	0.05	0.045
0.10	0.05	0.090
0.20	0.10	0.720

Options

• Rate = K [A] [B]²

• Rate = K [A]² [B]²

• Rate = K [A] [B]

• Rate = K [A]² [B]

Answer 1

Rate = K [A] [B]² 

Explanation:

Let the rate equation for the given reaction is,

rate = K[A]^m [B]ⁿ

From 1^st value we get,

0.045 = K [0.05]^m [0.05]ⁿ   ...(i)

From 2^nd value we get,

0.090 = K [0.10]^m [0.05]ⁿ   ...(ii)

From 3^rd value we get,

0.720 = K [0.20]^m [0.10]ⁿ   ...(iii)

On dividing equation (ii) by (i) we get,

`0.090/0.045 = (0.10/0.05)^"m"`

⇒ 2 = 2^m

⇒  2^m = 2¹

∴ m = 1   ...(iv)

On dividing equation (iii) by (ii) we get

`0.720/0.090 = (2/1)^"m" xx (10/5)^"n"`

⇒ 8 = 2 × 2ⁿ

⇒ 2ⁿ = 4

⇒ 2ⁿ = 2²

∴ n = 2

Therefore the rate law for the reaction is Rate = K [A] [B]²