Question

How long, in seconds, would it take for the concentration of A to decrease from 0.860 M to 0.310 M?

Answer 1

Answer:

2.038 seconds.

Explanation:

So, in the question above we are given the following parameters in order to solve this question. We are given a rate constant of 0.500 s^-, initial concentration= 0.860 M and final concentration= 0.310 M,the time,t =??.

Assuming that the equation for the first order of reaction is given below,that is;

A ---------------------------------> products.

Recall the formula below;

B= B° e^-kt.

Therefore, e^-kt = B/B°.

-kt = ln B/B°.

kt= ln B°/B.

Where B° and B are the amount of the initial concentration and the amount of the concentration remaining, k is the rate constant and t = time taken for the concentration to decrease.

So, we have; time taken,t = ln( 0.860/.310)/0.500.

==> ln 2.77/0.500.

==> time taken,t =2.038 seconds.