Determine the rate law, including the values of the orders and rate law constant, for the following reaction using the experimental data provided.

Question

A+B yields products

Trial 1 A-0.10M B-0.20M rate- 1.2 x 10^-2 m/min

trial 2 A-0.10M B- 0.40M rate-4.8 x 10^-2 m/min

trial 3 A- 0.20M B-0.40M rate-9.6 x 10^-2 M/min

Answer 1

Trial 1 and 2, A constant, so order for B can be found.

B doubles (0.2 to 0.4), rate increases 4x (4.8/1.2)

so, for B:

(rate/rate)=(B/B)^x

4=2^x

x=2

B is second order

For trials 2 and 3, B is constant, so the order for A can be found.

A doubles (0.2/0.1) from trial 2 to 3 and the rate doubles (9.6/4.8)

(rate/rate)=(A/A)^x

2=2^x

x=1

A is first order

rate=k[A]¹[B]²

Rate law:

rate=k[A][B]²

Rate constant k:

0.012 M/min=k[0.10M][0.20M]²

0.012 M/min=k[0.10M][0.04M²]

0.012M/min=k 0.004M³

(0.012M/min)÷(0.004M³)=k

0.012/0.004=3

M/min / M³ = 1/min*M²

k= 3 1/min*M²

Unit also may be given as M^2-·min^-1, or equivalent