Question

the second - order reaction a → products takes 13.5 s for the concentration of a to decrease from 0.740 m to 0.303 m. what is the value of k for this reaction?

Explanation:

Step1: Identify the second - order integrated rate law

The second - order integrated rate law is $\frac{1}{[A]_t}-\frac{1}{[A]_0}=kt$, where $[A]_t$ is the concentration at time $t$, $[A]_0$ is the initial concentration, $k$ is the rate constant, and $t$ is the time.

Step2: Substitute the given values

We are given that $[A]_0 = 0.740\ M$, $[A]_t=0.303\ M$, and $t = 13.5\ s$.
Substitute these values into the equation: $\frac{1}{0.303}-\frac{1}{0.740}=k\times13.5$.
First, calculate $\frac{1}{0.303}\approx3.3003$ and $\frac{1}{0.740}\approx1.3514$.
Then $\frac{1}{0.303}-\frac{1}{0.740}=3.3003 - 1.3514=1.9489$.

Step3: Solve for $k$

We have the equation $1.9489 = k\times13.5$.
To find $k$, divide both sides of the equation by $13.5$: $k=\frac{1.9489}{13.5}\approx0.144\ M^{-1}s^{-1}$.

Answer:

$0.144\ M^{-1}s^{-1}$