Question

Starting with `"1.20 g"` of `A`, what is the mass of `A` remaining undecomposed after `"1.00 h"`?

The reaction `A\rarr` products is first order in `A`.
If `1.20 g A` is allowed to decompose for `38 min` , the mass of `A` remaining undecomposed is found to be `0.30 g`. What is the half-life, `t_(1/2)`, of this reaction?
Answer to the half-life question: `(\color(indianred)(\text{19 minutes}))`

Answer 1

Here's what I got.

Explanation:

As you know, the half-life of a reaction describes the time needed for half of the initial amount of a reactant to be consumed.

Keep in mind that for a first-order reaction, the half-life does not depend on the initial amount of the reactant.

So, let's say that you have `A_0` as the initial amount of the reactant `"A"`. You can say that you have

• `1/2 * A_0 = A_0/2 ->` after one half-life
• `1/2 * A_0/2 = A_0/4 ->` after two half-lives
• `1/2 * A_0/4 = A_0/8 ->` after three half-lives
• `vdots`

This means that with every passing half-life, the amount of reactant `"A"` will be halved.

Mathematically, this can be expressed as

`color(blue)(ul(color(black)(A_t = A_0 * (1/2)^n)))`

Here

• `A_t` is the amount that remains undecayed after in `t` time interval
• `A_0` is the initial mass of the sample
• `n` is the number of half-lives that pass in the `t` time interval

Now, you know that you start with `"1.20 g"` of reactant `"A"` and the reaction proceeds for `"38 min"`, at which point you are left with `"0.30 g"` of `"A"`.

The numbers actually allow for a very quick calculation here. Notice that the amount of `"A"` decreased by a factor of `4`, since

`(1.20 color(red)(cancel(color(black)("g"))))/(0.30color(red)(cancel(color(black)("g")))) = color(blue)(4)`

Since you know that the amount of `"A"` decreases by a factor of `2`, i.e. it gets halved, with every passing half-life, you can say that `2` half-lives must pass in order for the amount of `"A"` to decrease by a factor of `color(blue)(4)`.

Therefore, the half-life of the reaction is

`"38 min"/2 = color(darkgreen)(ul(color(black)("19 min")))`

You can get the same result by using the half-life equation

`0.30 color(red)(cancel(color(black)("g"))) = 1.20 color(red)(cancel(color(black)("g"))) * (1/2)^n`

Rearrange to get

`(1/2)^n = 0.30/1.20`

`(1/2)^n = 1/4`

`(1/2)^n = (1/2)^2 implies n= 2`

Since `n` tells you how many half-lives passed in the given time period, i.e. in `"38 min"`, you can say that

`n = "total time"/t_"1/2" implies t_"1/2" = "total time"/n`

Therefore, you have

`t_"1/2" = "38 min"/2 = color(darkgreen)(ul(color(black)("19 min")))`

Now that you know the half-life, you can determine the amount of reactant `"A"` that remains undecayed after

`1.00 color(red)(cancel(color(black)("h"))) * "60 min"/(1color(red)(cancel(color(black)("h")))) = "60 min"`

by using the half-life equation

`A_"60 min" = "1.20 g" * (1/2)^( (60 color(red)(cancel(color(black)("min"))))/(19color(red)(cancel(color(black)("min"))))`

`A_"60 min" = "1.20 g" * (1/2)^(60/19) = color(darkgreen)(ul(color(black)("0.13 g")))`

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