Question

Consider the reaction A+2B⇌CA+2B⇌C whose rate at 25 ∘C∘C was measured using three different sets of initial concentrations as listed in the following table: Trial [A][A] (MM) [B][B] (MM) Rate (M/sM/s) 1 0.50 0.050 1.5×10−2 2 0.50 0.100 3.0×10−2 3 1.00 0.050 6.0×10−2

Part A What is the rate law for this reaction? Express the rate law symbolically in terms of kkk, [A][A], and [B][B]. View Available Hint(s)

Answer 1

Ans :-

Given reaction is :

A + 2B <--------------> C

Let rate of this reaction is :

Rate = K.[A]^p.[B]^q ..................(1)

Where,

K = rate constant

p = Order with respect to A

q = Order with respect to B

Put the results of first experiment in equation (1) :

1.5 x 10^-2 = K(0.50)^p(0.050)^q .....................(2)

Put the results of second experiment in equation (1) :

3.0 x 10^-2 = K(0.50)^p(0.100)^q .....................(3)

Put the results of third experiment in equation (1) :

6.0 x 10^-2 = K(1.00)^p(0.050)^q .....................(4)

Divide equation (2) by (3) :

1.5 x 10^-2 / 3.0 x 10^-2 = K(0.50)^p(0.050)^q / K(0.50)^p(0.100)^q

(0.5)¹ = (0.5)^q

On comparing

q = 1

Divide equation (2) by (4) :

1.5 x 10^-2 / 6.0 x 10^-2 = K(0.50)^p(0.050)^q/ K(1.00)^p(0.050)^q

0.25 = (0.5)^p

(0.5)² = (0.5)^p

On comparing

p = 2

Put the values of p and q in equation (1) :

So,

Rate law expression becomes :

Rate = K.[A]2.[B]