Consider the reaction 3A + B + C ==D + E?

Question

Consider the reaction

3A + B + C== D + E

where the rate law is defined as

-D[A] / Dt = k[A]2[B][C]

An experiment is carried out where [B]0 = [C]0 = 1.00 M and [A]0 = 1.00 x 10-4 M.

After 3.00 minutes, [A] = 3.26 x 10-5 M. The value of k is

Answer 1

Because B and C are in great excess, its change in concentration can be neglected. Hence
[B]~[B]0
[C]~[C]0

The rate law simplifies to
-d[A]/dt=k[B]0[C]0[A]²

This differential equation can be solved by separation of variables
-1/[A]² d[A]=k[B]0[C]0 dt
Integration with lower boundaries [A]0 and t=0 gives
1/[A]-1[A]0=k[B]0[C]0 t

Solving for k
k= 1/t*(1/[A]-1/ [A]0)/([B]0[C]0)
=6891.6 (l³/(mol³min)
=114.9 (l³/(mol³s)

Answer 2

-D[A]/dt = -2.25 X 10-5 M/minute (took the 3 minute concentration of A, subtracted it from the initial concentration and divided by 3.

Now, to find k, just substitute all of the concentrations, set it equal to the value for D[A]/dt and solve for k