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Half-lives
1 half-life
100 % 50 %^ 25 %^ 12.5 %^ 6.25 %
The time it takes for a reactant to reach half of its initial amount
16M 8M 4M 2M 1M
Original concentration
t1/ 2 half-lives
t1/ 3 half-lives
t1/ 4 half-lives
t1/
x
t 0
[A] [A]
General Equation for Half-life
Where “x” = number of half-lives
This equation can be used for any order as long as no time component is involved.
Half-lives
0
1
2
3
4
5
6
7
0 0.5 1 1.5 2 2.5 3
[A] (mol/L)
Time (s)
Half lives are equal
1 st^ half-life
2 nd^ half-life 3 rd^ half-life
0
1
2
3
4
5
6
7
0 0.5 1 1.5 2 2.5 3
[A] (mol/L)
Time (s)
Half-lives increase over time
4 thhalf-life
1 st^ half-life
2 nd^ half-life 3 rd^ half-life (^4) thhalf-life
1 st^ Order 2 nd^ Order
Are there any differences in half-lives over time between zero-, first-, and second order reactions?
Half-lives
0
1
2
3
4
5
6
7
0 0.5 1 1.5 2 2.5 3
[A] (mol/L)
Time (s)
0 th^ Order
Half lives decrease
1 st^ half-life
2 nd^ half-life
3 rd^ half-life 4 thhalf-life
With first-order half-life problems, you can sometimes use short cuts (such as the general rate law equation) because successive half-lives are the same time period.
With second- and zero- order problems that involve time, you must use the integrated half-life equations because successive half-lives have different time periods.
Half-life Equations
kt [A]
[A]
ln 0
0
0 [A] 0
1 kt
2
[A]
1
[A] kt 2
[A]
0
2 nd^ Order
1 st^ Order
0 th^ Order
k
t
ln(2) 1/2
0
1/ [A]
1
k
t
k
t 2
[A] 0 1/2
Half-lives are related to the rate constant, k , by the following equations
x
t 0
[A] [A]
General Half-life Equation
Where “x” = number of half-lives
Integrated Half-life Equations
Half-Life Examples
A certain reaction has a rate constant of 1.25 hr-^1. What is the half life of this reaction in hours? Minutes?
How many half lives will it take for the same reaction above to go from 6.0M to 0.020M?
If a 2 nd^ order reaction has a half life of 3.1 days, how many hours will it take for 10.0% of the reactant to disappear if the concentration starts at 1.0M?
k
t
ln(2) 1/2
0
1/ [A]
1
k
t
k
t 2
[A] 0 1/2
0.555hr
8.2 half-lives
Graphing Integrated Rate Laws
ln[A] t kt ln[A] 0
[A] 0
[A]
kt t
[A] t kt [A] 0
y = m x + b
1 st-order rate law
2 nd-order rate law
0 th-order rate law
Equation of a straight line
ln[A] t vs. t
1/[A] (^) t vs. t
[A] (^) t vs. t
ln
[A]
t
time
1/[A]
t
time
[A]
t
time
ln[A] 0
1/[A] 0
[A] 0
slope y-intercept
Time (s) [A] (M) 0 1 1 0. 2 0. 3 0. 4 0. 5 0. 6 0. 7 0. 8 0.
Time (s) ln[A] 0 0 1 -0. 2 -1. 3 -1. 4 -1. 5 -1. 6 -1. 7 -2. 8 -2.
Time (s) 1/[A] 1/M 0 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9
Time vs. Conc. Time vs. ln(Conc.) Time vs. 1/(Conc.)
Determining order by graphing
To determine the “order” of the reactant, you must graph the data three different ways: Time vs. Conc.; Time vs. ln(Conc.); and Time vs. 1/Conc.
Multistep Reactions – Reaction
Mechanisms
NO 2 + CO → NO + CO 2
NO 2 + CO NO + CO 2
This reaction could happen in just one step…or in multiple steps. How can we figure it out?
First we need to learn about these steps.
Step 2: NO 3 + CO → NO 2 + CO 2
Multistep Reactions – Reaction
Mechanisms
Reactions that have multiple steps are made of individual steps called elementary steps. Elementary steps are single (concerted) steps.
Elementary steps must add up to the overall reaction equation.
Intermediates are chemical species that are created during the reaction, but not a part of the overall reaction equation. Intermediates are created in an elementary step and consumed in another
Step 1: 2NO 2 → NO + NO 3
NO 2 + CO → NO + CO 2
Step 1: N 2 O 4 →NO 3 + NO
Step 2: NO 3 + CO → NO 2 + CO 2
These are not plausible steps for the reaction because the steps add up to a different overall reaction
N 2 O 4 + CO → NO+ NO 2 + CO 2
Molecularity and Rate Laws of Elementary steps
Unimolecular
Bimolecular
Termolecular
A → Product(s)
A + A +A → Product(s) A + B + B → Product(s) A + B + C → Product(s)
A + A → Product(s) A + B → Product(s)
Rate = k [A]
Rate = k [A]^2 Rate = k [A][B]
Rate = k [A]^3 Rate = k [A][B]^2 Rate = k [A][B][C]
Rate laws for elementary steps can be determined directly from the balanced equation The molecularity of an elementary step is based on the rate law for that step
Molecularity Equation Rate Law
Step 1: 2NO 2 → NO + NO 3 Step 2: NO 3 + CO → NO 2 + CO 2
Both of these elementary steps are bimolecular
Elementary Step Rate = k 1 [NO 2 ]^2 Elementary Step Rate = k 2 [NO 3 ][CO]
Since elementary steps are single steps, rate laws for each step can be determined from the balanced chemical equation
Overall reaction rates of multi-step reactions
Slow Step Fast Step
Two-step reaction
6,000 kisses per hour
How many kisses can be processed per hour overall ???
Elementary Steps
The overall reaction can go no faster than the slow step. This step is called the rate-determining or rate-limiting step.
Just remember that reaction rates are not static; reaction rates are variable based on concentration.
20,000 kisses per hour
Multiple-step Reactions – Elementary Steps – Rate-limiting Step 1st
Rate 1 < Rate 2
Slow-step 1st Overall Chemical Formula
Rate laws for elementary steps can be determined directly from the balanced equation
If the slow, rate-determining, step is first, the overall rate law is the same as the slow step
The elementary-step formulas must add up to the overall chemical formula
Step 1 Rate Law Overall Rate Law
Notice that the concentrations of “NO 3 ” and “CO” do not affect the rate of reaction
2NO 2 → NO + NO 3 NO 3 + CO → NO 2 + CO 2 NO 3 + CO → NO 2 + CO 2
2NO 2 → NO + NO 3
NO 2 + CO → NO + CO 2
2NO 2 → NO + NO 3
k 1 [NO 2 ]^2
NO 3 + CO → NO 2 + CO 2
k 2 [NO 3 ][CO]
Rate = k 1 [NO 2 ]^2 Rate =^ k^1 [NO^2 ]
2
Arrhenius Equation
T
a R
E
k Ae
-
Svante Arrhenius determined that activation energy (Ea) of a reaction is related to the reaction constant.
That is…at any temperature the lower the activation energy, the faster a reaction will proceed.
T
ln ln
a
R
E k A-
ln T
1 ln
a A R
E k -
y = m x + b
Arrhenius Equation
2 1
a
2
1
T
1
T
1
R
E
k
k ln
We can also do calculations without graphing, but combining two equations into one.
If we know both k 1 and k 2 , along with T 1 and T 2 we calculate the energy of activation
1
1 a RT
ln k ln A- E
2
2 a RT
ln k ln A- E
In a certain equation, the rate constant at 701K was measured as 2.57M-^1 s-^1 and that at 895 K was measured as 556M-^1 s-^1. Find the activation energy of the reaction.
895K
1 701K
1
E
556 ln mol K
J
a Ms
1
Ms
1 -^4 K^1 molK
J
a (^) 3.09x
E 5.38 1.45x10 (^5) J/mol
molK R 8.314J
