Question

Using the given data, determine the rate constant of this reaction.

A+2B---> C+D

Trial	(A)	(B)(M)	Rate (M/s)
1	0.380	0.330	0.0153
2	0.380	0.660	0.153
3	0.760	0.330	0.0612

k=

(inculde units)

Answer 1

A + 2B ----> C + D

Let the rate be r = k[A]^m[B]ⁿ ---(1)

Apply the tabulated data to (1) we have

1st values ---> 0.0153 M/s = k[0.380]^m[0.330]ⁿ ---(2)

2nd values ---> 0.153 M/s = k[0.380]^m[0.660]ⁿ ---(3)

3rd values ---> 0.0612 M/s = k[0.760]^m[0.330]ⁿ ---(4)

(3)/(2) gives 2ⁿ = 10 ===> n log 2 = log10 = 1 ---> n = 1 / log2 = 3.32

(4)/(2) gives 2^m = 4 ===> m = 2

Plug the valuesof m & n in (2) gives

0.0153 M/s = k[0.380M]²[0.330M]^3.32

k = 4.20 M^-4.32s^-1