contestada

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

Respuesta :

The chemiacl reaction given:

[tex]2A\rightarrow A_2[/tex]

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).

[tex]\frac{1}{\lbrack A\rbrack_t}=\frac{1}{\lbrack A\rbrack_o}+kt[/tex]

[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.

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

Otras preguntas