Question

it has been found that ddt in the soil decomposes by a first - order reaction with a half - life of 12.0 years.
what amount of time would it take ddt in a soil sample to decompose from a concentration of 375 ppbm (parts per billion by mass) to 15 ppbm?

Explanation:

Step1: Find the rate - constant k

For a first - order reaction, the half - life formula is $t_{1/2}=\frac{\ln2}{k}$. Given $t_{1/2} = 12.0$ years, we can solve for $k$.
$k=\frac{\ln2}{t_{1/2}}=\frac{\ln2}{12.0\ years}\approx0.0578\ years^{-1}$

Step2: Use the first - order integrated rate law

The first - order integrated rate law is $\ln\frac{[A]_0}{[A]_t}=kt$, where $[A]_0$ is the initial concentration, $[A]_t$ is the final concentration, $k$ is the rate constant, and $t$ is the time.
We have $[A]_0 = 375$ ppbm, $[A]_t=15$ ppbm, and $k = 0.0578\ years^{-1}$.
First, calculate $\ln\frac{[A]_0}{[A]_t}=\ln\frac{375}{15}=\ln25\approx3.2189$.
Then, since $\ln\frac{[A]_0}{[A]_t}=kt$, we can solve for $t$.
$t=\frac{\ln\frac{[A]_0}{[A]_t}}{k}=\frac{3.2189}{0.0578\ years^{-1}}\approx55.7$ years

Answer:

55.7 years