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

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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}))#

The reaction

If

Answer to the half-life question:

##### 1 Answer

#### Answer:

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

#1/2 * A_0 = A_0/2 -># afterone half-life#1/2 * A_0/2 = A_0/4 -># aftertwo half-lives#1/2 * A_0/4 = A_0/8 -># afterthree half-lives#vdots#

This means that with **every passing half-life**, the amount of reactant **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 thatremains undecayedafter in#t# time interval#A_0# is the initial mass of the sample#n# is thenumber of half-livesthat pass in the#t# time interval

Now, you know that you start with

The numbers actually allow for a very quick calculation here. Notice that the amount of

#(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 *halved*, with every passing half-life, you can say that **half-lives** must pass in order for the amount of

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 *how many half-lives* passed in the given time period, i.e. in

#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

#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.