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>>
View the other Key Equations and Concepts in this
chapter
Integrated Rate Laws
>> Parts of this equation/concept include:
The integrated rate laws used in this chapter assume that there
is only one reactant that affects the rate of reaction. The integrated
rate law has a very different form for first-, second-, and zero-order
reactions. Integrated rate laws for other types of reactions are
beyond the scope of this chapter.
| A. Determining Order
and Rate Constants from Concentration-Versus-Time Data |
Each integrated rate law can be arranged as a linear equation.
The order is determined by graphing each of the three possibilities
and determining which of the three is linear. The process can be
quite tedious; consequently, it is normally done using a computer
spreadsheet. With a spreadsheet, a regression analysis tests the
linearity of the plot. The closer the correlation coefficient is
to unity (1), the more linear the line. If a spreadsheet is unavailable,
graph the data and trust your eyes. They are normally better than
you give them credit for.
In each of the three graphs, time is graphed on the x-axis.
To test for a zero-order reaction, graph concentration of reactant
on the y-axis. To test for a first-order reaction, graph
the natural log of reactant concentration on the y-axis.
To test for a second-order reaction, graph the reciprocal of reactant
concentration on the y-axis. Remember that the linear graph
is the correct one.
Once the correct (linear) graph is determined, the rate constant
is determined from the slope of that line. If you remember that
the rate constant is always an absolute value, you need not worry
whether the slope is k (as with second-order reactions) or
–k (with zero- and first-order reactions).
>> Example 1
Determine the rate law, including the value of the rate constant,
from the following rate-versus-time data.
| Time (s) |
Concentration (M) |
| 0 |
0.500 |
| 10 |
0.197 |
| 20 |
0.078 |
| 30 |
0.031 |
| 40 |
0.012 |
| 50 |
0.005 |
Solution:
To determine the order of the reaction, you need values for the
natural log of the concentration and the reciprocal of the concentration,
so that you can make the appropriate graphs. You can use a spreadsheet
to do this if one is available.
| Time (s) |
Concentration (M) |
ln (concentration) |
1/concentration |
| 0 |
0.500 |
–0.69315 |
2 |
| 10 |
0.197 |
–1.62455 |
5.076142 |
| 20 |
0.078 |
–2.55105 |
12.82051 |
| 30 |
0.031 |
–3.47377 |
32.25806 |
| 40 |
0.012 |
–4.42285 |
83.33333 |
| 50 |
0.005 |
–5.29832 |
200 |
Graph time versus concentration and check for linearity. If you
use a spreadsheet, check the correlation coefficient (R2).
The closer the value is to 1, the more like a line the data are.
It is better to compare all three graphs than to make a decision
based on just one.
Figure 14.2A is the graph of concentration versus time. A best-fit
line is shown as a dash, a best curve through all points is shown
as a solid line. The correlation coefficient of the line is 0.7386.
Figure 14.2B is the graph of ln[concentration] versus time. A
best-fit line is shown as a dash, a best curve through all points
is shown as a solid line. The correlation coefficient of the line
is 0.9999.
Figure 14.2C is a graph of 1/concentration versus time. A best-fit
line is shown as a dash, a best curve through all points is shown
as a solid line. The correlation coefficient of the line is 0.7517.
Since Figure B, graphing ln[A], has the best-fit line, this is
a first-order reaction. Therefore the rate law is rate = k[A].
From the graph the equation of the line is y = (–0.0924)x
+ 0.7004. The absolute value of the slope is the rate constant.
Since the rate law is first order and the time units are seconds,
k = 0.0924 s–1.
>> Example 2
Determine the rate law, including the value of the rate constant,
from the following rate-versus-time data.
| Time (min) |
Concentration (M) |
| 0 |
0.467 |
| 1 |
0.267 |
| 3 |
0.144 |
| 5 |
0.099 |
| 6 |
0.085 |
| 7 |
0.075 |
Solution:
Determine the natural logarithm and reciprocal for each reactant
concentration
| Time (min) |
Concentration (M) |
ln (concentration) |
1/concentration |
| 0 |
0.467 |
–0.76143 |
2.141328 |
| 1 |
0.267 |
–1.32051 |
3.745318 |
| 3 |
0.144 |
–1.93794 |
6.944444 |
| 5 |
0.099 |
–2.31264 |
10.10101 |
| 6 |
0.085 |
–2.4651 |
11.76471 |
| 7 |
0.075 |
–2.59027 |
13.33333 |
Make the three graphs (determine the correlation coefficient
if using a spreadsheet) to determine which is most linear. Here
the graphs are Figure 14.3A for concentration versus time, Figure
14.3B for ln (concentration) versus time, and Figure 14.3C for
1/concentration versus time. The trend (straight) line is dashed;
the best-fit curve is solid.
The correlation coefficient of Figure 14.3A = 0.8157; that for
Figure 14.3B = 0.9516; and that for Figure 14.3C = 1.0000. Therefore
the best-fit line is the one with 1/[A] on the y-axis.
So this is a second-order reaction. The rate law is rate = k[A]2.
The value of the rate constant is the absolute value of the slope
of the line in Figure 14.3C, which is 1.599 M–1•min–1.
(The units are typical of a second-order reaction. Time is in
minutes, since that is how it is given as data.)
>> back
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| B. Determining Reactant Concentration After Some Time |
Use the appropriate integrated rate law to determine the relationship
between reactant concentration and time.
>> Example 3
The reaction of A has a rate constant of 1.4 x 10–4
s–1. If 1.0 M of reactant reacts for 25
minutes, how much is left?
Solution:
Because this is a first-order reaction (the units on rate constant
give it away), the relevant integrated rate law is
ln[A] = –kt + ln[A]0
The a nswer is determined by algebraically solving this equation
for the unknown value, in this case, [A]. The units on k
must match the time units. It is probably easier to change the
units of time than to change those of the rate constant. Therefore
| t |
= |
25 min |
 |
|
= |
1500 s |
ln[A] = –(1.4 x 10–4/s)(1500 s) + ln(1.0)
ln[A] = –0.21 + 0
ln[A] = –0.21
[A] = 0.81 M
>> Example 4
If the reactant concentration for a second-order reaction decreases
from 0.10 M to 0.03 M in 1.00 hour, what is the
rate constant for the reaction?
Solution:
This is a second-order reaction (the problem says so), so the
relevant integrated rate law is
1/[A] = kt + 1/[A]0
Substituting the concentration values into this equation (and
keeping units),
1/(0.03 M) = kt + 1/(0.10 M)
33.3 M–1 = kt + 10 M–1
23.3 M–1 = kt
The units on the rate constant depend on the value used for time.
The easiest thing is to use time as hours. So
23.3 M–1 = k(1.00 hr)
23.3 M–1•hr–1
= k
If you use time in units of minutes, t = 60.0 min
23.3 M–1 = k (60.0 min)
0.39 M–1•min–1
= k
If you use time in units of seconds, t = 3600 s
23.3 M–1 = k(3600 s)
6.5 x 10–3 M–1•s–1
= k
Any of these three answers is correct.
>> Example 5
How long does it take 2.00 M of reactant to decrease in
concentration to 0.75 M if the rate constant is 0.67 M/min?
Solution:
The units of rate constant imply that this is a zero-order reaction
(rate = k). Therefore, the relevant integrated rate law
is
[A] = –kt + [A]0
In this example, the equation is solved for time. The units of
time will be minutes, which is determined from the rate constant.
(0.75 M) = –(0.67 M/min)t + (2.00
M)
–1.25 M = (–0.67 M/min)t
1.9 min = t
>> back
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The relationship between half-life and rate constant depends on
the order of the reaction. Probably the most important relationship
is for a first-order reaction. First-order reactions are the only
type of reaction where the concentration of reactant does not affect
the half-life.
>> Determining Rate
Constant from Half-life
The equations for half-life can be determined by solving the appropriate
integrated rate law for time, assuming that the final reactant concentration
is half the initial concentration. Alternately, these equations
have already been worked out (Equations 14.12 and 14.16).
>> Example 6
What is the rate constant for a first-order reaction with a half-life
of 300.0 seconds?
Solution:
Since this is a first-order reaction, t1/2
= (ln 2)/k. Solving this equation for k,
300k = 0.6931
k = 2.310 x 10–3 s–1
>> Example 7
What is the rate constant for a second-order reaction with a
half-life of 35 seconds when the initial concentration is 0.50
M?
Solution:
Since this is a second-order reaction, the equation for the half-life
is
so
(35 s)(0.50 M)k = 1
(17.5 M•s)k = 1
k = 0.057 M–1•s–1
>> Determining Reactant
Concentration from Half-life
Normally, the integrated rate law is used to determine concentration.
The rate constant is determined from half-life, as in the preceding
examples.
>> Example 8
A first-order reaction has a half-life of 0.500 hour. If the
initial concentration was 1.00 M, how much remains after
1.13 hour?
Solution:
Since the reaction is first order, the equation for the half-life
is
So 1.39 hr–1 = k.
Using this rate constant and the integrated rate law for a first-order
reaction, solving for [A]
ln[A] = –kt + ln[A]0 (Make sure the units on
k and t match!)
ln[A] = –(1.39 hr–1)(1.13 hr) + ln(1.00)
ln[A] = –1.57 + 0
[A] = 0.21 M
If the relevant time is a multiple of the half-life, it might be
easier to think through the logic of the definition rather than
to do all the calculations (although the calculation will still
work).
>> Example 9
A first-order reaction has a half-life of 0.50 hour. If the initial
concentration is 0.80 M, what is the concentration after
1.50 hours?
Solution:
One and one-half hours is three half-lives. A half-life is the
time required for the reactant concentration to decrease by half.
So after 0.50 hour, the reactant concentration is 0.40 M.
Another half-life (1.00 hour) decreases that value by one-half
to 0.20 M. Another half-life (1.50 hour) decreases 0.20
M to 0.10 M.
>> Determining Half-life
from Rate Constant
Determining half-life from rate constant is a matter of solving
for time in the half-life equations rather than k.
>> Example 10
What is the half-life of a reaction with an initial concentration
of 0.300 M and a rate constant of 0.00385 M–1•s–1?
Solution:
The units on the rate constant indicate a second-order reaction.
Therefore the equation for half-life is
| t1/2 |
= |
| 1 |
|
| (0.00385 M–1•s–1)(0.300
M) |
|
t1/2 = 866 s
>> View
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