# The half-life of a first-order reaction is 1.5 hours. How much time is needed for 94% of the reactant to change to product?

Dec 27, 2017

$\text{7.6 h}$

#### Explanation:

The thing to remember about first-order reactions is that their half-life is independent of the initial concentration of the reactant.

In other words, it doesn't matter how much reactant you have, the half-life of the reaction, i.e. the time needed for half of the initial concentration of the reactant to be consumed, will be constant.

Mathematically, you can express this as

${\left[\text{A"]_t = ["A}\right]}_{0} \cdot {e}^{- k t}$

Here

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

Now, start by isolating $t$ on one side of the equation. To do that, take the natural log of both sides

$\ln \left({\left[\text{A"]_t) = ln(["A}\right]}_{0} \cdot {e}^{- k t}\right)$

This will get you

$\ln \left({\left[\text{A"]) - ln(["A}\right]}_{0}\right) = - k t \cdot \ln \left(e\right)$

$\ln \left(\left[\text{A"]_0) - ln(["A}\right]\right) = k t$

Finally, you have

t = ln (["A"]_0/["A"]_t)/k " " " "color(darkorange)("(*)")

Now, you know that after the passing of one half-life, ${t}_{\text{1/2}}$, the concentration of the reactant is reduced to half of its initial value.

["A"]_ t = ["A"]_ 0/2 " "->" " "after t"_"1/2"

This means that you have

ln(["A"]_0/["A"]_t) = ln(color(red)(cancel(color(black)(["A"]_0)))/(color(red)(cancel(color(black)(["A"]_0)))/2)) = ln(2)

and so

${t}_{\text{1/2}} = \ln \frac{2}{k}$

This means that the rate constant of the reaction is equal to

$k = \ln \frac{2}{t} _ \text{1/2}$

$k = \ln \frac{2}{\text{1.5 h" = "0.462 h}} ^ \left(- 1\right)$

Finally, you know that in order for 94% of the reactant to be consumed, you need to have

${\left[\text{A"]_t = (1 - 94/100) * ["A}\right]}_{0}$

This basically means that when 94% of the reactant is consumed, you are left with 6% of what you started with.

Plug this into equation $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\text{(*)}}$ to find

t = ln( (color(red)(cancel(color(black)(["A"]_0))))/(3/100 * color(red)(cancel(color(black)(["A"]_0)))))/"0.462 h"^(-1) = color(darkgreen)(ul(color(black)("7.6 h")))

The answer is rounded to two sig figs, the number of sig figs you have for the half-life of the reaction.