Question

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?

Answer 1

`"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.

![https://chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/Kinetics/Reaction_Rates/Half-lives_and_Pharmacokinetics#First_Order_Kinetics]()

Mathematically, you can express this as

`["A"]_t = ["A"]_0 * e^(-kt)`

Here

• `["A"]_t` represents the concentration of the reactant after a time `t`
• `["A"]_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(["A"]_t) = ln(["A"]_0 * e^(-kt))`

This will get you

`ln(["A"]) - ln(["A"]_0) = - kt * ln(e)`

`ln(["A"]_0) - ln(["A"]) = kt`

Finally, you have

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

Now, you know that after the passing of one half-life, `t_"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_"1/2" = ln(2)/k`

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

`k = ln(2)/t_"1/2"`

In your case, you have

`k = ln(2)/"1.5 h" = "0.462 h"^(-1)`

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

`["A"]_t = (1 - 94/100) * ["A"]_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 `color(darkorange)("(*)")` 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.