Question

A second order reaction where the starting reactant concentration is 0.0963 M has a measured rate constant of 0.70 M/s at 10 oC. What is the concentration of the reactant after 36.0 seconds?

- report your answer in three significant figures

- do not write your answer in scientific notation

- do not include units

Answer 1

Answer 2

The second order rate law is given by the equation:

1/[A]t - 1/[A]0 = kt

where [A]t and [A]0 are the concentrations of the reactant at time t and time zero, respectively, and k is the rate constant.

Rearranging the equation to solve for [A]t, we get:

[A]t = 1/([A]0 + kt[A]0)

Plugging in the given values, we have:

[A]t = 1/(0.0963 M + (0.70 M/s)(36.0 s)(0.0963 M))

[A]t = 0.042 M

Therefore, the concentration of the reactant after 36.0 seconds is 0.042 M.