Using Rate Laws to Predict
Concentrations at Time = t
If our experimental study of a reaction shows
that its rate is first order, we can calculate the
concentrations of reactant and products at a
time t.
Process: A  products
rate = k[A]
Using the form for the instantaneous rate −
rearranging
𝒅[𝑨]
[𝑨]
= −𝒌 𝒅𝒕
integrating from 0 to t
rearranging
𝒅𝑨
𝒅𝒕
𝒍𝒏 [𝑨]𝒕 −𝒍𝒏 [𝑨]𝟎 = −𝒌𝒕
𝒍𝒏 [𝑨]𝒕 = 𝒍𝒏 [𝑨]𝟎 − 𝒌𝒕
= 𝒌[𝑨]
Using Rate Laws to Predict
Concentrations at Time = t
For any first order reaction
A  products
the concentration of A at time t is given by
𝒍𝒏 [𝑨]𝒕 = − 𝒌𝒕 + 𝒍𝒏 [𝑨]𝟎
where [A]o is the initial concentration and k is the
rate constant.
ln[A] is the natural logarithm of A and that is the power to
which e (2.71828) must be raised to get A.
Plots of First Order Reactions
𝒍𝒏 [𝑨]𝒕 = − 𝒌𝒕 + 𝒍𝒏 [𝑨]𝟎
y
=
mx + b
A plot of 𝒍𝒏 [𝑨]𝒕 versus time will be linear
with slope = -k and y-intercept = 𝒍𝒏 [𝑨]𝟎 .
ln [A]0
slope = -k
Order of a Reaction - Example
0
1.000
2500
0.947
5000
0.895
7500
0.848
10000
0.803
pressure of SO2Cl2 (atm)
Analyze rate data for the reaction SO2Cl2(g)  SO2(g) + Cl2(g) at
320°C to prove the reaction is first order and to find the rate
constant.
time (s)
pressure of
1
SO2Cl2 (atm)
0.95
0.9
0.85
0.8
0
2000
4000
6000
8000
10000
time (s)
A common mistake is to plot
the data given without
thought as to “why.”
A plot of P vs. time is not linear. I used
the “add trendline” feature of Excel.
Plots of First Order Reactions
If the reaction is 1st order, should a plot of P
versus t be linear? (rate = k[A])
𝒍𝒏 [𝑨]𝒕 = − 𝒌𝒕 + 𝒍𝒏 [𝑨]𝟎
[𝑨]𝒕 = [𝑨]𝟎 𝒆−𝒌𝒕
A plot of [𝑨]𝒕 versus time will be a decreasing
exponential.
Order of a Reaction - Example
When we plotted P versus t, we did not get a straight line. So
we calculate ln P and plot ln P vs t.
time
(s)
pressure
of SO2Cl2
(atm)
ln P
0
1.000
0
2500
0.947
-0.0545
5000
0.895
-0.1109
7500
0.848
-0.1649
0.803
-0.2194
ln(pressure of SO2Cl2)
10000
0
-0.05
-0.1
-0.15
-0.2
-0.25
0
Why is it okay to use
pressure instead of molarity
when analyzing the data?
2000
4000
6000
8000
10000
time (s)
A plot of ln P vs. time is linear. This proves
that the reaction is first order in SO2Cl2.
Order of a Reaction - Example
Now we can determine the first order rate constant k:
time
(s)
ln P
k (s-1)
0
0
0
2500
-0.0545
2.18 x 10-5
5000
-0.1109
2.26 x 10-5
7500
-0.1649
2.16 x 10-5
10000
-0.2194
2.18 x 10-5
average k is
k = 2.20 x 10-5 s-1
at T = 320°C.
General equation for concentration
in a first order reaction:
𝒍𝒏 [𝑨]𝒕 = − 𝒌𝒕 + 𝒍𝒏 [𝑨]𝟎
For this case:
𝒍𝒏 𝑷𝒕 = − 𝒌𝒕 + 𝒍𝒏 𝑷𝟎
since Po = 1, ln Po = 0
𝒍𝒏 𝑷𝒕 = − 𝒌𝒕
𝒍𝒏 𝑷𝒕
𝒌=−
𝒕
Plot of a Zero Order Reaction
First, write the rate law: rate = k
Integrating leads to: [𝑨]𝒕 = − 𝒌𝒕 + [𝑨]𝟎
y =
mx + b
A plot of [𝑨]𝒕 versus time will be linear with
slope = -k and y-intercept = [𝑨]𝟎 .
slope = -k
Plot of a Second Order Reaction
First, write the rate law: rate = k[A]2
Integrating leads to:
𝟏
[𝑨]𝒕
= 𝒌𝒕 +
y =
A plot of
𝟏
[𝑨]𝒕
𝟏
[𝑨]𝟎
mx + b
versus time will be linear with
slope = k and y-intercept =
𝟏
[𝑨]𝟎
You mark the slope and y-intercept for this plot.
.
Determining the Order of a Reaction from
Concentration vs Time Data
On the Unit 3 test, I typically give concentration vs
time data and ask a series of questions about the
reaction.
Give some thought as to how you would use conc vs
time data to get: order of reaction, rate law, value of k,
etc.
If your thought is to plot the data, be sure you know
HOW to plot data correctly, how to interpret the plot,
and how to give written support to your conclusion.
Half-Life of a Reaction
The half-life of a reaction is a convenient way
to describe the speed of a reaction.
The half-life is the time it takes for the
concentration of a reactant to drop to onehalf of its initial value. If a reaction has a
short half-life, it is a fast reaction.
The equation that gives the half-life depends
on the rate law for the reaction. We will focus
on the half-lives of first order reactions.
Half-Life of a First Order Reaction
The half-life (t½) is the time it takes for the
concentration of a reactant A to drop to
one-half of its initial value.
For a first order reaction, ln[A]t = - kt + ln[A]o
Rearranging
[𝑨]𝒕
𝒍𝒏
[𝑨]𝟎
= −𝒌𝒕
[𝑨]𝒕𝟏
𝟐
At the end of one half-life, 𝒍𝒏
= −𝒌𝒕𝟏 𝟐
[𝑨]𝟎
𝟏
[𝑨]𝟎
𝟐
𝒍𝒏
= −𝒌𝒕𝟏
𝟐
[𝑨]𝟎
Half-Life of a First Order Reaction
𝒍𝒏
At the end of one half-life,
𝟏
[𝑨]𝟎
𝟐
[𝑨]𝟎
= −𝒌𝒕𝟏 𝟐
ln ½ = - kt½
- ln 2 = - kt½
ln 2 = kt½
0.693 = kt½
𝒕𝟏
or
𝟐
𝟎. 𝟔𝟗𝟑
=
𝒌
𝟎. 𝟔𝟗𝟑
𝒌=
𝒕𝟏
𝟐
t½ of a First Order Reaction - Example
SO2Cl2(g)  SO2(g) + Cl2(g) T = 320°C
Determine the time required for the pressure of SO2Cl2 to drop
from 1.000 atm to 0.500 atm. How long does it take for the
pressure to drop from 0.050 atm to 0.025 atm?
time (s)
pressure of
SO2Cl2 (atm)
0
1.000
2500
0.947
5000
0.895
7500
0.848
10000
0.803
We have already determined
that the reaction is first order
and that the average k is
k = 2.20 x 10-5 s-1
at T = 320°C.
t½ of a First Order Reaction - Example
SO2Cl2(g)  SO2(g) + Cl2(g) T = 320°C
Determine the time required for the pressure of SO2Cl2 to drop
from 1.000 atm to 0.500 atm. How long does it take for the
pressure to drop from 0.050 atm to 0.025 atm?
𝒕𝟏
𝟐
𝟎. 𝟔𝟗𝟑
𝟎. 𝟔𝟗𝟑
𝟒 𝒔 = 𝟖. 𝟕𝟓 𝒉𝒓
=
=
=
𝟑.
𝟏𝟓𝒙𝟏𝟎
𝒌
𝟐. 𝟐𝟎𝒙𝟏𝟎−𝟓 𝒔−𝟏
The answer to BOTH questions is t½, because in both
cases the final pressure is one-half the initial
pressure.
We did not have to convert pressure to molarity because halflives apply to the ratio of two pressures: P2 = M2RT = M2
P1 M1RT M1
Half-Life of a First Order Reaction
The half-life applies to ANY
drop in concentration (or
pressure) of one-half.
[𝑨]𝒕 = [𝑨]𝟎 𝒆−𝒌𝒕
t½ of a First Order Reaction - Example
How much time is required for a 5.75 mg sample of
51Cr to decay to 1.50 mg if it has a half-life of 27.8
days?
Radioactive decay is a first order reaction process.
Since the reaction is first order, we know 𝒍𝒏
[𝑨]𝒕
[𝑨]𝟎
= −𝒌𝒕
Knowing the half-life gives us k (not t):
𝟎. 𝟔𝟗𝟑
𝟎. 𝟔𝟗𝟑
𝒌=
=
= 𝟎. 𝟎𝟐𝟒𝟗𝟑 𝒅𝒂𝒚𝒔−𝟏
𝒕𝟏
𝟐𝟕. 𝟖 𝒅𝒂𝒚𝒔
𝟐
𝟏. 𝟓𝟎 𝒎𝒈
𝒍𝒏
= −𝟎. 𝟎𝟐𝟒𝟗𝟑𝒅𝒂𝒚𝒔−𝟏 𝒕
𝟓. 𝟕𝟓 𝒎𝒈
t = 53.9 days