# Question cef19

May 27, 2017

Here's what I got.

#### Explanation:

The integrated rate law for a first-order reaction

$\text{A " -> " products}$

looks like this

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\ln \left(\left({\left[\text{A"]_t)/(["A}\right]}_{0}\right)\right) = - k \cdot t}}}$

Here

• ${\left[A\right]}_{t}$ is the concentration of the reactant after a period of time $t$
• ${\left[A\right]}_{0}$ is the initial concentration of the reactant
• $k$ is the rate constant

In your case, you know that

$k = {\text{200 s}}^{- 1}$

Now, in order for 75% of the reaction to be complete, you need the concentration of the reactant to decrease by 75%.

In other words, you need the concentration of the reactant to be down to 25% of its initial value.

Therefore, you can say that

${\left[A\right]}_{t} = \frac{25}{100} \cdot {\left[A\right]}_{0}$

Plug this into the integrated rate law and solve for $t$

$\ln \left(\frac{\frac{25}{100} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{{\left[A\right]}_{0}}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{{\left[A\right]}_{0}}}}}\right) = - {\text{200 s}}^{- 1}$

This will be equivalent to

$\ln \left(\frac{25}{100}\right) = - {\text{200 s}}^{- 1}$

which will get you

t = (ln(25/100))/(-"200 s"^(-1)) = color(darkgreen)(ul(color(black)("0.007 s")))#

The answer is rounded to one significant figure, the number of sig figs you have for the rate constant.

If you want, you can convert this to milliseconds by using the fact that

$\text{1 s} = {10}^{3}$ $\text{ms}$

You should end up with

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{time needed = 7 ms}}}}$