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}))#

1 Answer
Feb 23, 2017

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 #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.