Chapter 12
Chemical Kinetics
The Rate of a Chemical Reaction
Section 12.1
Reaction Rates
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Reaction Rate
� Change in concentration of a reactant or product
per unit time.
[A] means concentration of A in mol/L; A is the
reactant or product being considered.
[ ]
2 1
2 1
concentration of A at time concentration of A at time Rate =
A=
−
−
∆
∆
t t
t t
t
Section 12.1
Reaction Rates
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The Decomposition of Nitrogen Dioxide
Section 12.1
Reaction Rates
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The Decomposition of Nitrogen Dioxide
Section 12.1
Reaction Rates
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Instantaneous Rate
� Value of the rate at a particular time.
� Can be obtained by computing the slope of a line
tangent to the curve at that point.
Reaction Rate and Stoichiometry
What do coefficients have to do with rate?
• If there are coefficients, then change in the number of molecules of one substance is a multiple of the change in the number of molecules of another.
• To be consistent, the change in the concentration of each substance is multiplied by 1/coefficient.
• So, for the reaction:
aA + bB � cC + dD
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Reaction Rate and Stoichiometry
H2 (g) + I2 (g) → 2 HI(g)
– For the above reaction, for every 1 mole of H2 used, 1 mole of I2 will also be used and 2 moles of HI made.
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Practice Problems:
• Be able to relate the overall reaction rate to the
change in concentration of a particular reactant or
product
• Be able to relate the change concentration of one
reactant or product to anther using their respective
coefficients. Watch out for the sign convention!
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Practice Problems:
Consider the reaction:
4PH3 (g) � P4 (g) + 6H2 (g)
If in a certain experiment, over a specific time
period, 0.0048 moles of PH3 are consumed in a 2.0 L
container each second of reaction, what are the rates
of production of P4 and H2, respectively?
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Practice Problems:
Consider the reaction:
4PH3 (g) � P4 (g) + 6H2 (g)
If in a certain experiment, over a specific time period 0.0048 moles of PH3+
are consumed in a 2.0 L container each second of reaction, what are the rates
of production of P4 and H2, respectively?
Rate = s
L)mol/2.00.048(][3 −−
=∆
∆−
t
PH = 2.4 ×
310
−mol/L•s
t∆
]PH[∆
4
1
t∆
]P[∆ 34 −= = 2.4 × 3
10−
/4 = 6.0 × 4
10−
mol/L•s
t∆
]PH[∆
4
6
t∆
]H[∆ 32 −= = 6(2.4 × 3
10−
)/4 = 3.6 × 3
10−
mol/L•s
The Rate Law
• The rate law of a reaction is the mathematical relationship between the rate of the reaction and the concentrations of the reactants and homogeneous catalysts as well.
• The rate law must be determined experimentally!
• The rate of a reaction is directly proportional to the concentration of each reactant raised to a power.
• For the reaction aA + bB →→→→ products the rate law would have the form given below.
– n and m are called the orders for each reactant.
– k is called the rate constant.
Section 12.2
Rate Laws: An Introduction
The Rate Law (aka differential rate law)
� Relates Rate to Reactant Concentrations.(remember, the rate changes; it depends on concentration!)
� For the decomposition of nitrogen dioxide:
2NO2(g) → 2NO(g) + O2(g)
Rate = k[NO2]n:
� k = rate constant
� n = order of the reactant
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Section 12.2
Rate Laws: An Introduction
Rate Law
Rate = k[NO2]n
� The concentrations of the products do not appear in the
rate law because the reaction rate is being studied
under conditions where the reverse reaction does not
contribute to the overall rate.
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Section 12.2
Rate Laws: An Introduction
Rate Law
Rate = k[NO2]n
� The value of the exponent n must be determined by
experiment; it cannot be written from the balanced
equation.
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Rate = k[A]n
� If a reaction is zero order, the rate of the reaction is
always the same.
� Doubling [A] will have no effect on the reaction rate.
� If a reaction is first order, the rate is directly
proportional to the reactant concentration.
� Doubling [A] will double the rate of the reaction.
� If a reaction is second order, the rate is directly
proportional to the square of the reactant
concentration.
� Doubling [A] will quadruple the rate of the reaction.
Section 12.2
Rate Laws: An Introduction
Types of Rate Laws
�
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Section 12.2
Rate Laws: An Introduction
Rate Laws: Which one is more useful?
� Because the differential and integrated rate laws for a
given reaction are related in a well–defined way, the
experimental determination of either of the rate laws is
sufficient.
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Section 12.2
Rate Laws: An Introduction
Rate Laws: Which one is more useful?
� Experimental convenience usually dictates which type of
rate law is determined experimentally.
� Knowing the rate law for a reaction is important mainly
because we can usually infer the individual steps
involved in the reaction from the specific form of the
rate law.
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Section 12.3
Determining the Form of the Rate Law
� Determine experimentally the power to which each
reactant concentration must be raised in the rate law.
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Section 12.3
Determining the Form of the Rate Law
Method of Initial Rates
� The value of the initial rate is determined for each
experiment at the same value of t as close to t = 0 as
possible.
� Several experiments are carried out using different
initial concentrations of each of the reactants, and the
initial rate is determined for each run.
� The results are then compared to see how the initial
rate depends on the initial concentrations of each of the
reactants.
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Section 12.3
Determining the Form of the Rate Law
Overall Reaction Order
� The sum of the exponents in the reaction rate equation.
Rate = k[A]n[B]m
Overall reaction order = n + m
k = rate constant
[A] = concentration of reactant A
[B] = concentration of reactant B
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Section 12.3
Determining the Form of the Rate Law
How do exponents (orders) in rate laws compare to
coefficients in balanced equations?
Why?
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CONCEPT CHECK!
Section 12.4
The Integrated Rate Law
First-Order
� Differential:
Rate = k[A]
� Integrated:
ln[A] = –kt + ln[A]o
[A] = concentration of A at time t
k = rate constant
t = time
[A]o = initial concentration of ACopyright © Cengage Learning. All rights reserved 24
Section 12.4
The Integrated Rate Law
Plot of ln[N2O5] vs Time: 1st order
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Section 12.4
The Integrated Rate Law
First-Order Reactions and Half-Life
� Rate = k[A]
� Integrated:
ln[A] = –kt + ln[A]o
We can consider how long it would take for half of a
reactant to be consumed.
Rearrange this equation to solve for t when a concentration
[A] is halved. You will find that ln[A] - ln[A0] is equal to
0.693. This is how the half-life equation is derived.
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Section 12.4
The Integrated Rate Law
First-Order
� Time required for a reactant to reach half its original
concentration
� Half–Life:
k = rate constant
� Half–life of a 1st order reaction does not depend on the
concentration of reactants.
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12
0.693 = t
k
Section 12.4
The Integrated Rate Law
A first order reaction is 35% complete at the end of
55 minutes. What is the value of k?
k = 7.8 × 10–3 min–1
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Section 12.4
The Integrated Rate Law
Second-Order
� Rate = k[A]2
� Integrated:
[A] = concentration of A at time t
k = rate constant
t = time
[A]o = initial concentration of ACopyright © Cengage Learning. All rights reserved 29
[ ] [ ]0
1 1= +
A Akt
Section 12.4
The Integrated Rate Law
Plot of ln[C4H6] vs Time and Plot of 1/[C4H6] vs Time
Section 12.4
The Integrated Rate Law
Second-Order
� Half–Life (2nd order) :
k = rate constant
[A]o = initial concentration of A
� For a 2nd order reaction, Half–life gets longer as the reaction
progresses and the concentration of reactants decrease.
� Each successive half–life is double the preceding one.
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[ ]12
0
1 =
At
k
Section 12.4
The Integrated Rate Law
For a reaction aA � Products,
[A]0 = 5.0 M, and the first two half-lives are 25 and 50
minutes, respectively.
a) Write the rate law for this reaction.
rate = k[A]2
b) Calculate k.
k = 8.0 × 10-3 M–1min–1
c) Calculate [A] at t = 525 minutes.
[A] = 0.23 MCopyright © Cengage Learning. All rights reserved 32
EXERCISE!
Section 12.4
The Integrated Rate Law
Zero-Order
� Rate = k[A]0 = k
� Integrated:
[A] = –kt + [A]o
[A] = concentration of A at time t
k = rate constant
t = time
[A]o = initial concentration of A
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Section 12.4
The Integrated Rate Law
Plot of [A] vs Time
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Section 12.4
The Integrated Rate Law
Zero-Order
� Half–Life:
k = rate constant
[A]o = initial concentration of A
� Half–life gets shorter as the reaction progresses and the
concentration of reactants decrease.
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[ ]0
12
A =
2t
k
Section 12.4
The Integrated Rate Law
Check your Royal Purple Papers
� What equations are given?
� Are they labeled?
� Which apply to first order reactions only?
� Which are second order?
� Which are zero order?
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Section 12.4
The Integrated Rate Law
How can you tell the difference among 0th, 1st, and 2nd
order rate laws from their graphs?
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CONCEPT CHECK!
Section 12.4
The Integrated Rate Law
Rate Laws
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Section 12.4
The Integrated Rate Law
Summary of the Rate Laws
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Important: Notice that only in the case of a first order reaction
is the half-life independent of [A].
Therefore, if the ½ life is constant, it’s 1st order.
Section 12.4
The Integrated Rate Law
Consider the reaction aA � Products.
[A]0 = 5.0 M and k = 1.0 × 10–2 (assume the units are appropriate
for each case). Calculate [A] after 30.0 seconds have passed,
assuming the reaction is:
a) Zero order
b) First order
c) Second order
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4.7 M
3.7 M
2.0 M
EXERCISE!