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

## The reaction $A \setminus \rightarrow$ 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, ${t}_{\frac{1}{2}}$, of this reaction? Answer to the half-life question: $\left(\setminus \textcolor{\in \mathrm{di} a n red}{\setminus \textrm{19 \min u t e s}}\right)$

Feb 23, 2017

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 $\text{A}$. You can say that you have

• $\frac{1}{2} \cdot {A}_{0} = {A}_{0} / 2 \to$ after one half-life
• $\frac{1}{2} \cdot {A}_{0} / 2 = {A}_{0} / 4 \to$ after two half-lives
• $\frac{1}{2} \cdot {A}_{0} / 4 = {A}_{0} / 8 \to$ after three half-lives
• $\vdots$

This means that with every passing half-life, the amount of reactant $\text{A}$ will be halved.

Mathematically, this can be expressed as

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{{A}_{t} = {A}_{0} \cdot {\left(\frac{1}{2}\right)}^{n}}}}$

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 $\text{1.20 g}$ of reactant $\text{A}$ and the reaction proceeds for $\text{38 min}$, at which point you are left with $\text{0.30 g}$ of $\text{A}$.

The numbers actually allow for a very quick calculation here. Notice that the amount of $\text{A}$ decreased by a factor of $4$, since

$\left(1.20 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))))/(0.30color(red)(cancel(color(black)("g}}}}\right) = \textcolor{b l u e}{4}$

Since you know that the amount of $\text{A}$ 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 $\text{A}$ to decrease by a factor of $\textcolor{b l u e}{4}$.

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

$0.30 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))) = 1.20 color(red)(cancel(color(black)("g}}}} \cdot {\left(\frac{1}{2}\right)}^{n}$

Rearrange to get

${\left(\frac{1}{2}\right)}^{n} = \frac{0.30}{1.20}$

${\left(\frac{1}{2}\right)}^{n} = \frac{1}{4}$

${\left(\frac{1}{2}\right)}^{n} = {\left(\frac{1}{2}\right)}^{2} \implies n = 2$

Since $n$ tells you how many half-lives passed in the given time period, i.e. in $\text{38 min}$, you can say that

$n = \frac{\text{total time"/t_"1/2" implies t_"1/2" = "total time}}{n}$

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 $\text{A}$ 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.