Question

The reaction 2A → A2 was experimentally determined to be second order with a rate constant, k, equal to 0.0265 M–1min–1. If the initial concentration of A was 2.50 M, what was the concentration of A (in M) after 180.0 min? _____ M

Answer 1

The chemiacl reaction given:

`2A\rightarrow A_2`

We are given that the reaction:

Is second order reaction

Have a rate constant = 0.0265M-1min-1

Initial concentration of A = 2.50 M

time = 180.0 mins

We want the concentration of A after 180.0 minutes.

For the second order reaction, we use the following equation to calculate conentrattion (integrated rate law).

`(1)/(\lbrack A\rbrack_t)=(1)/(\lbrack A\rbrack_o)+kt`

[A]t is the concentration of A at time t

[A]o is the initial concentration (at time = 0)

k is the rate constant for the reaction

Now lets plug in the values.

`\begin{gathered} (1)/(\lbrack A\rbrack_t)=\frac{1}{2.50\text{ M}}+0.0265M^(-1)\min ^(-1)\text{ x 180.0 min} \\ (1)/(\lbrack A\rbrack_t)=5.17 \\ \lbrack A\rbrack_t=\text{ 0.193 M} \end{gathered}`