A second order reaction is 40% completed after one hour. What is the rate constant for this reaction?

Question

A second order reaction is 40% completed after one hour. What is the rate constant for this reaction?

1 Answer

Impact of this question

12429 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-19T16:51:51

$1.1 \times 10^{- 2}$ $1.1xx10^-2$ $M^{- 1}$ $"M"^-1$ ${min}^{- 1}$ $"min"^-1$

Explanation:

The rate law for a second-order reaction is

$\frac{1}{[A]} = \frac{1}{{[A]}_{0}} + k t$ $1/([A])=1/([A]_0)+kt$

We can say that the current concentration $[A]$ $[A]$ is $0.6$ $0.6$ and the original concentration ${[A]}_{0}$ $[A]_0$ is $1$ $1$ . This is true because the reaction is $40 %$ $40%$ complete, which means that $60 %$ $60%$ still has to be reacted, and $60 %$ $60%$ of an original concentration $1$ $1$ is $0.6$ $0.6$ .

$t = 60$ $t=60$ since $60$ $60$ minutes have elapsed.

$\frac{1}{0.6} = \frac{1}{1} + k (60)$ $1/(0.6)=1/1+k(60)$

$1.1 \times 10^{- 2} = k$ $1.1xx10^-2=k$

The rate constant is $1.1 \times 10^{- 2}$ $1.1xx10^-2$ $M^{- 1}$ $"M"^-1$ ${min}^{- 1}$ $"min"^-1$ .

Science

Math

Humanities

... and beyond

A second order reaction is 40% completed after one hour. What is the rate constant for this reaction?

1 Answer

Explanation:

Related questions

Impact of this question