Question

Using the given data, determine the rate constant of this reaction.
A + 2B --> C + D

Given Data:
Trial [A](M) [B](M) RATE (M/s)
1 .230 .330 0.0231
2 .230 .660 0.0231
3 .460 .330 0.0924
k=? the units are M-1 s-1

Answer 1

Answer- Given, concentration of reactant and rate and need to calculate the rate constant. For the calculating the rate constant we must know the rate law.

Assume the rate law-

Rate = k [A]^m [B]ⁿ

There is m and n order with respect to A and B,

so rate law are as follow -

Rate1 = k [A]₁^m [B]₁ⁿ

Rate2 = k [A]₂^m [B]₂ⁿ

Rate 3 = k [A]₃^m [B]₃ⁿ

Order with respect to A -

Rate3/ Rate1 = k [A]₃^m [B]₃ⁿ / k [A]₁^m [B]₁ⁿ

0.0924 / 0.0231 = (0.460)^m /(0.230)^m * (0.330)ⁿ / (0.330)ⁿ

4 = (2)^m

m = 2 , so order with respect to A is second order

No order of B-

Rate2/Rate1 = k [A]₂^m [B]₂ⁿ / k [A]₁^m [B]₁ⁿ

0.0231/ 0.0231 = (0.230)^m /(0.230)^m *(0.660)ⁿ /(0.330)ⁿ

1 = (2)ⁿ

So, n = 0, so order with respect B is zero

So overall order = 2+0 = 2

So rate law, rate = k [A]²

So overall order is second order

Now rate constant-

Rate = k [A]²

0.0231 M/s = k *(0.230 M)²

k = 0.0231 M.s^-1/ 0.0529 M²

= 0.437 M^-1 s^-1

So, rate constant is 0.437 M^-1 s^-1