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

Dec 19, 2015

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

Explanation:

The rate law for a second-order reaction is

$\frac{1}{\left[A\right]} = \frac{1}{{\left[A\right]}_{0}} + k t$

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

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

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

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

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