Chapter 13 Kinetics - Gordon State College

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CHEMISTRY
Chapter 13
Chemical Kinetics
Factors that Affect Reaction
Rates
• Kinetics is the study of how fast chemical reactions
occur.
• There are 4 important factors which affect rates of
reactions:
–
–
–
–
reactant concentration,
temperature,
action of catalysts, and
surface area.
• Goal: to understand chemical reactions at the molecular
level.
Reaction Rates
Change of Rate with Time
• Consider:
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
Reaction Rates
• For the reaction A  B there are two ways of measuring
rate:
– the speed at which the products appear (i.e. change in moles of B per unit time), or
– the speed at which the reactants disappear (i.e. the change in moles of A per unit
time).
moles of A 
Average rate with respect to A  
t
Reaction Rates
Reaction Rate and Stoichiometry
• In general for:
aA + bB  cC + dD
1 A 
1 B 1 C 1 D
Rate  



a t
b t
c t
d t
• Rate is (-) if reagent is consumed.
• Rate is (+) if compound is produced.
• Rate will ultimately be (+) because change in
concentration will be negative. Two (-)’s
become (+).
Take Note:
• Rate must ALWAYS be a positive value!
Take Note!
• Since the direction of equilibrium
changes as more product is produced,
rates have to be determined as soon as
the experiment has begun.
2N2O5
4NO2 + O2
[N2O5] (mol/L)
Time (sec)
0.100
0
0.0707
50
0.0500
100
0.0250
200
0.0125
300
0.00625
400
Sample Problem
• A. How is the rate at which ozone (O3)
disappears related to the rate at which O2
appears in the reaction: 2 O3 (g)  3 O2 (g)?
• B. If the rate at which O2 appears, [O2]/t, is
6.0 x 10-5 M/s at a particular instant, at what
rate is O3 disappearing at this same time,
-[O3]/t?
Answers
• A. “Related to” means compare, so write the
rate expression comparing the compounds.
• B. 4.0 x 10-5 M/s
Sample Problem
• The decomposition of N2O5 proceeds
according to the following equation:
2 N2O5 (g) 4 NO2 (g) + O2 (g)
If the rate of the decomposition of N2O5 at a
particular instant in a reaction vessel is 4.2 x 10-7
M/s, what is the rate of appearance of: (a) NO2,
(b) O2 ?
Differential Rate Law
- is a rate law that expresses how
rate is dependent on
concentration
Example:
Rate = k[A]n
Differential First Order Rate Law
• First Order Reaction
– Rate dependent on concentration
– If concentration of starting reagent was
doubled, rate of production of compounds
would also double
Concentration and Rate
•
•
•
•
Using Initial Rates to Determines Rate
Laws
A reaction is zero order in a reactant if the change in
concentration of that reactant produces no effect.
A reaction is first order if doubling the concentration
causes the rate to double.
A reacting is nth order if doubling the concentration
causes an 2n increase in rate.
Note that the rate constant does not depend on
concentration.
Differential Rate Law
• For single reactants: A  C
Rate = k[A]n
• For 2 or more reactants: A + B  C
Rate = k[A]n[B]m
Rate = k[A]n[B]m[C]p
Problem
• NH4+
•
•
•
•
+
NO2-
N2
+
2H2O
Give the general rate law equation for rxn.
Derive rate order.
Derive general rate order.
Solve for the rate constant k.
To Determine the Orders of the
Reaction (n, m, p, etc….)
• 1. Write Rate law equation.
• 2. Get ratio of 2 rate laws from successive
experiments.
• Ratio = rate Expt.2 = k2[NH4+]n[NO2-]m
rate Expt.1 k1[NH4+]n[NO2-]m
• 3. Derive reaction order.
• 4. Derive overall reaction order.
Experimental Data
[NO2-]initial Initial Rate
Expt.
[NH4]initial
1
0.100 M
0.0050 M
1.35 x 10-7
2
0.100 M
0.010 M
2.70 x 10-7
3
0.200 M
0.010 M
5.40 x 10-7
A+BC
Experiment
Number
[A] (mol·L-1)
[B] (mol·L-1)
Initial Rate
(mol·L-1·s-1)
1
0.100
0.100
4.0 x 10-5
2
0.100
0.200
4.0 x 10-5
3
0.200
0.100
1.6 x 10-4
(a) Determine the differential rate law
(b) Calculate the rate constant
(c) Calculate the rate when [A]=0.050 mol·L-1 and [B]=0.100 mol·L-1
Use the data in table 12.5 to determine
1) The orders for all three reactants
2) The overall reaction order
3) The value of the rate constant
2NO(g) + 2H2(g)  N2(g) + 2H2O(g)
Experiment
Number
[NO] (mol·L-1)
[H2] (mol·L-1)
Initial Rate
(mol·L-1·s-1)
1
0.10
0.10
1.23 x 10-3
2
0.10
0.20
2.46 x 10-3
3
0.20
0.10
4.92 x 10-3
(a) Determine the differential rate law
(b) Calculate the rate constant
(c) Calculate the rate when [NO]=0.050 mol·L-1 and [H2]=0.150 mol·L-1
Sample Problem:.
Consider the general reaction aA + bB cC and the
following average rate data over some time period Δt:

Δ[A]
 0.0080 mol  L1  s1
Δt

Δ[B]
 0.0120 mol  L1  s 1
Δt
Δ[C]
 0.0160 mol  L1  s1
Δt
Determine a set of possible coefficients to balance this
general reaction.
Problem
• Reaction: A + B  C obeys the rate law:
Rate = k[A]2[B].
• A. If [A] is doubled (keeping B constant), how
will rate change?
• B. Will rate constant k change? Explain.
• C. What are the reaction orders for A & B?
• D. What are the units of the rate constant?
You now know that….
• The rate expression correlates consumption of
reactant to production of product. For a
reaction: 3A  2B
- 1[A] = 1[B]
3 t
2 t
• The differential rate law allows you to correlate
rate with concentration based on the format:
Rate = k [A]n
You also know that…
• 1. Rate of consumption of reactant decreases
over time because the concentration of
reactant decreases. Lower concentration
equates to lower rate.
• 2. If a graph of concentration vs. time were
constructed, the graph is not a straight line
Expt. 25
1400
1200
Time in seconds
1000
800
Series1
600
400
200
0
0
0.002
0.004
0.006
0.008
Concentration (M)
0.01
0.012
0.014
Expt. 25
0.014
Concentration in Molarity
0.012
0.01
0.008
Series1
0.006
0.004
0.002
0
0
200
400
600
800
Time in seconds
1000
1200
1400
How can we make the line straight?
What is the relationship between
concentration and time?
By graphing concentration vs. 1/time?
Concentration vs. Reciprocal of Time
0.006
Reciprocal of Time (1/sec)
0.005
0.004
0.003
Series1
0.002
0.001
0
0
0.002
0.004
0.006
0.008
Concentration (M)
0.01
0.012
0.014
The Integrated Rate Law makes
this possible!
First Order Reaction
-3
0
200
400
600
800
1000
1200
1400
1600
-3.5
ln[concentration]
-4
-4.5
Series1
-5
-5.5
-6
-6.5
Time in seconds
Integrated Rate Law
• Expresses the dependence of
concentration on time
Integrated Rate Laws
• Zero Order:
[A]t = -kt + [A]o
• First Order:
ln[A]t = -kt + ln[A]o
• Second Order:
1
[A]t
= kt + 1
[A]o
where [A]o is the initial concentration and
[A]t is the final concentration.
Integrated First-Order Rate Law
• ln[A]t = -kt + ln[A]0
• Eqn. shows [concn] as a function of time
• Gives straight-line plot since equation is of
the form y = mx + b
The Change of
Concentration with Time
Zero Order Reactions
•A plot of [A]t versus t is a
straight line with slope -k and
intercept [A]0.
The Change of
Concentration with Time
First Order Reactions
•A plot of ln[A]t versus t is a
straight line with slope -k and
intercept ln[A]0.
The Change of
Concentration with Time
Second Order Reactions
•A plot of 1/[A]t versus t is a
straight line with slope k and
intercept 1/[A]0.

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