A reactant A decomposes at constant temperature. Its concentration is measured every 20 minutes. Use integrated rate law plots to find the order, k and the half-life, and predict [A] at 120 minutes.
A→products,rate=k[A]m
1
Transform the data three ways
The first example needed several runs, each started fresh. This one follows a single run over time, so the order must come from the shape of the data. Compute ln[A] and 1/[A] for each time, so all three candidate plots can be compared.
Concentration–time data and transformations
t(min)
[A](M)
ln[A]
1/[A](M−1)
0
0.500
−0.693
2.00
20
0.397
−0.924
2.52
40
0.316
−1.152
3.16
60
0.251
−1.382
3.98
80
0.199
−1.614
5.03
2
Look for the straight plot
ln[A] falls by 0.230±0.002 every 20 minutes. Equal steps make a straight line, with R2=0.99999.[A] falls by 0.103, then 0.081,0.065 and 0.052, and 1/[A] rises by 0.52, then 0.64,0.82 and 1.05. Both bend, even though each still has R2 near 0.98. The reaction is first order.
ln[A]t=−kt+ln[A]0
3
Read k from the slope
A least-squares line through the five points (t,ln[A]) has slope −0.01151 min−1. For a first-order plot the slope is −k, so k is 1.151×10−2 min−1. This k is the given the rest of the work starts from.
k=−slope=1.151×10−2 min−1
k=−slope=1.151×10−2 min−1
4
Write the half-life rule
A first-order half-life depends only on k, because setting [A]=[A]0/2 in the integrated law cancels [A]0.
t1/2=kln2
5
Find the half-life
ln2 is 0.693, and 0.693÷0.01151=60.2. The units are 1÷min−1=min.
t1/2=1.151×10−2 min−10.693=60.2 min
t1/2=1.151×10−2 min−10.693=60.2 min
6
Name the starting concentration
The prediction starts from the first reading, 0.500 M at t=0.
[A]0=0.500 M
7
Name the time
The question asks about 120 minutes, in the same unit as k.
t=120 min
8
Write the first-order law
Undoing the logarithm in ln[A]=ln[A]0−kt gives the exponential form, which gives [A] directly.
[A]=[A]0e−kt
9
Predict [A] at 120 minutes
kt=(0.01151 min−1)(120 min)=1.381, a pure number. e−1.381=0.2513, and 0.500 M×0.2513=0.126 M.
[A]=(0.500 M)e−(0.01151)(120)=0.126 M
[A]=(0.500 M)e−(0.01151)(120)=0.126 M
10
Check with half-lives
120 minutes is almost exactly two half-lives, 2×60.2=120.4 min. Two halvings take 0.500 M to 0.125 M, and the prediction, 0.126 M, is just above it, as a time just short of two half-lives should be. The data agree too, since [A] fell from 0.500 M to 0.251 M in the first 60 minutes.
Result
First order, with k=1.151×10−2 min−1 and t1/2=60.2 min.[A] is about 0.126 M at 120 minutes.
rate=1.151×10−2 min−1[A]
Your turn
1. One like the example
A reactant is measured every 10.0 min:[A]=0.400,0.296,0.219 and 0.162 M at t=0,10.0,20.0 and 30.0 min.ln[A]is −0.916,−1.217,−1.519and −1.820. Find the order and k.
Show the answer and explanation
First order, with k=0.0301 min−1.
ln[A] falls by 0.301 or 0.302 every 10.0 min, equal steps, so the ln[A] plot is straight and the reaction is first order. 1/[A] rises by 0.88, then 1.19, then 1.61, so it bends. k comes from the first and last points, (ln0.400−ln0.162)÷30.0 min=0.904÷30.0=0.0301 min−1.
Another reactant reads [A]=0.500,0.333,0.250 and 0.200 M at t=0,20.0,40.0 and 60.0 min. Which plot is straight, and what is k?
Show the answer and explanation
The 1/[A] plot is straight, so the reaction is second order, with k=0.0500 M−1 min−1.
1/[A] is 2.00,3.00,4.00 and 5.00 M−1, rising by 1.00 M−1 every 20.0 min.ln[A] falls by 0.406, then 0.287, then 0.223, so it bends. For second order the slope of 1/[A] is +k, so k=(5.00−2.00) M−1÷60.0 min=0.0500 M−1 min−1.
A reactant starting at 0.80 M takes 2.5 min to fall to 0.40 M, then 5.0 min more to fall to 0.20 M. What is the order, and what is k?
Show the answer and explanation
Second order, with k=0.50 M−1 min−1.
The second half-life is twice the first, which is the signature of second order. A first-order reaction would take 2.5 min for every halving. For second order t1/2=1/(k[A]0), so k=1/(t1/2[A]0)=1÷(2.5 min×0.80 M)=0.50 M−1 min−1. The second half-life confirms it, 1÷(0.50×0.40)=5.0 min.
Fit the same data in Kinetics & Arrhenius, then switch the order to 2. It still returns a rate constant, 0.0376 M−1 min−1, but it puts the starting concentration at 0.545 M instead of the measured 0.500 M.
