Chemistry - solve for second order constant in k(M^-1min^-1)

Question

The following reaction 2A → B + C is second order, following the rate law Rate = k[A]². If the starting concentration [A₀] = 0.250 M, and after 215 min, [A(t)] = 0.0150 M, what is the second order rate constant k (in M⁻¹min⁻¹)?

Answer 1

Hi Paw K.!

Because we are given two concentrations, one at the initial point in the reaction and one at a later time, and the time at which the concentration of A was measured later on during the course of the reaction, we can use the second-order integrated rate law to find the rate constant, k.

We start with the integrated rate law for a second order reaction:

1/[A]_t = kt + 1/[A]₀

From here, we just rearrange the equation and solve for k:

1/[A]_t = kt + 1/[A]₀

k = [(1/[A]_t) - (1/[A]₀)]/t

Now, we can just plug-and-chug

<=> [(1/0.0150M) - (1/0.250M)]/215min

k = 0.291M^-1min^-1

I hope this helps!

Cheers