Chemical Kinetics

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Chemical Kinetics

• The area of chemistry that concerns reaction
  rates and reaction mechanisms.

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Reaction Rate
The change in concentration of a reactant or
product per unit of time
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2NO2(g) ? 2NO(g) O2(g)
Reaction Rates
1. Can measure disappearance of reactants
2. Can measure appearance of products
3. Are proportional stoichiometrically
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2NO2(g) ? 2NO(g) O2(g)
Reaction Rates
4. Are equal to the slope tangent to that
point
5. Change as the reaction proceeds, if
the rate is dependent upon concentration
?NO2
?t
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Rate Laws
Differential rate laws express (reveal) the
relationship between the concentration of
reactants and the rate of the reaction.
The differential rate law is usually just called
the rate law.
Integrated rate laws express (reveal) the
relationship between concentration of reactants
and time
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Writing a (differential) Rate Law
Problem - Write the rate law, determine the value
of the rate constant, k, and the overall order
for the following reaction
2 NO(g) Cl2(g) ? 2 NOCl(g)
Experiment NO (mol/L) Cl2 (mol/L) Rate Mol/Ls
1 0.250 0.250 1.43 x 10-6
2 0.500 0.250 5.72 x 10-6
3 0.250 0.500 2.86 x 10-6
4 0.500 0.500 11.4 x 10-6
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Writing a Rate Law
Part 1 Determine the values for the exponents
in the rate law
R kNOxCl2y
Experiment NO (mol/L) Cl2 (mol/L) Rate Mol/Ls
1 0.250 0.250 1.43 x 10-6
2 0.500 0.250 5.72 x 10-6
3 0.250 0.500 2.86 x 10-6
4 0.500 0.500 1.14 x 10-5
In experiment 1 and 2, Cl2 is constant while
NO doubles.
The rate quadruples, so
the reaction is second order with respect to NO
? R kNO2Cl2y
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Writing a Rate Law
Part 1 Determine the values for the exponents
in the rate law
R kNO2Cl2y
Experiment NO (mol/L) Cl2 (mol/L) Rate Mol/Ls
1 0.250 0.250 1.43 x 10-6
2 0.500 0.250 5.72 x 10-6
3 0.250 0.500 2.86 x 10-6
4 0.500 0.500 1.14 x 10-5
In experiment 2 and 4, NO is constant while
Cl2 doubles.
The rate doubles, so the
reaction is first order with respect to Cl2
? R kNO2Cl2
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Writing a Rate Law
Part 2 Determine the value for k, the rate
constant, by using any set of experimental data
R kNO2Cl2
Experiment NO (mol/L) Cl2 (mol/L) Rate Mol/Ls
1 0.250 0.250 1.43 x 10-6
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Writing a Rate Law
Part 3 Determine the overall order for the
reaction.
R kNO2Cl2
3
2

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? The reaction is 3rd order
Overall order is the sum of the exponents, or
orders, of the reactants
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Determining Order withConcentration vs. Time data
(the Integrated Rate Law)
Zero Order
First Order
Second Order
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Solving an Integrated Rate Law
Problem Find the integrated rate law and the
value for the rate constant, k
Time (s) H2O2 (mol/L)
0 1.00
120 0.91
300 0.78
600 0.59
1200 0.37
1800 0.22
2400 0.13
3000 0.082
3600 0.050
A graphing calculator with linear regression
analysis greatly simplifies this process!!
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Time vs. H2O2
Time (s) H2O2
0 1.00
120 0.91
300 0.78
600 0.59
1200 0.37
1800 0.22
2400 0.13
3000 0.082
3600 0.050
Regression results
y ax b a -2.64 x 10-4 b 0.841 r2
0.8891 r -0.9429
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Time vs. lnH2O2
Time (s) lnH2O2
0 0
120 -0.0943
300 -0.2485
600 -0.5276
1200 -0.9943
1800 -1.514
2400 -2.04
3000 -2.501
3600 -2.996
Regression results
y ax b a -8.35 x 10-4 b -.005 r2
0.99978 r -0.9999
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Time vs. 1/H2O2
Time (s) 1/H2O2
0 1.00
120 1.0989
300 1.2821
600 1.6949
1200 2.7027
1800 4.5455
2400 7.6923
3000 12.195
3600 20.000
Regression results
y ax b a 0.00460 b -0.847 r2 0.8723 r
0.9340
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And the winner is Time vs. lnH2O2
1. As a result, the reaction is 1st order
2. The (differential) rate law is
3. The integrated rate law is
4. Butwhat is the rate constant, k ?
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Finding the Rate Constant, k
Method 1 Calculate the slope from the Time vs.
lnH2O2 table.
Time (s) lnH2O2
0 0
120 -0.0943
300 -0.2485
600 -0.5276
1200 -0.9943
1800 -1.514
2400 -2.04
3000 -2.501
3600 -2.996
Now remember
? k -slope
k 8.32 x 10-4s-1
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Finding the Rate Constant, k
Method 2 Obtain k from the linear regresssion
analysis.
Regression results
y ax b a -8.35 x 10-4 b -.005 r2
0.99978 r -0.9999
Now remember
? k -slope
k 8.35 x 10-4s-1
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Rate Laws Summary
Zero Order First Order Second Order
Rate Law Rate k Rate kA Rate kA2
Integrated Rate Law A -kt A0 lnA -kt lnA0
Plot that produces a straight line A versus t lnA versus t
Relationship of rate constant to slope of straight line Slope -k Slope -k Slope k
Half-Life
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Reaction Mechanism
The reaction mechanism is the series of
elementary steps by which a chemical reaction
occurs.

• The sum of the elementary steps must give the
  overall balanced equation for the reaction
• The mechanism must agree with the
  experimentally determined rate law

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Rate-Determining Step

• In a multi-step reaction, the slowest step is the
  rate-determining step. It therefore determines
  the rate of the reaction.

The experimental rate law must agree with the
rate-determining step
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Identifying the Rate-Determining Step
For the reaction 2H2(g) 2NO(g) ? N2(g)
2H2O(g) The experimental rate law is R
kNO2H2
Which step in the reaction mechanism is the
rate-determining (slowest) step?
Step 1 H2(g) 2NO(g) ? N2O(g) H2O(g)
Step 2 N2O(g) H2(g) ? N2(g) H2O(g)
Step 1 agrees with the experimental rate law
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Identifying Intermediates
For the reaction 2H2(g) 2NO(g) ? N2(g)
2H2O(g)
Which species in the reaction mechanism are
intermediates (do not show up in the final,
balanced equation?)
Step 1 H2(g) 2NO(g) ? N2O(g) H2O(g)
Step 2 N2O(g) H2(g) ? N2(g) H2O(g)
2H2(g) 2NO(g) ? N2(g) 2H2O(g)
? N2O(g) is an intermediate
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Collision Model
Key Idea Molecules must collide to react.
However, only a small fraction of collisions
produces a reaction. Why?
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Collision Model
Collisions must have sufficient energy to produce
the reaction (must equal or exceed the activation
energy).
1.
Colliding particles must be correctly oriented to
one another in order to produce a reaction.
2.
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Factors Affecting Rate
Increasing temperature always increases the rate
of a reaction.

• Particles collide more frequently

• Particles collide more energetically

Increasing surface area increases the rate of a
reaction
Increasing Concentration USUALLY increases the
rate of a reaction

Presence of Catalysts, which lower the activation
energy by providing alternate pathways
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Endothermic Reactions
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Exothermic Reactions
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The Arrhenius Equation

• k rate constant at temperature T
• A frequency factor
• Ea activation energy
• R Gas constant, 8.314 J/Kmol

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The Arrhenius Equation, Rearranged

• Simplifies solving for Ea
• -Ea / R is the slope when (1/T) is plotted
  against ln(k)
• ln(A) is the y-intercept
• Linear regression analysis of a table of (1/T)
  vs. ln(k) can quickly yield a slope
• Ea -R(slope)

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Catalysis

• Catalyst A substance that speeds up a reaction
  without being consumed
• Enzyme A large molecule (usually a protein)
  that catalyzes biological reactions.
• Homogeneous catalyst Present in the same phase
  as the reacting molecules.
• Heterogeneous catalyst Present in a different
  phase than the reacting molecules.

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Lowering of Activation Energy by a Catalyst
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Catalysts Increase the Number of Effective
Collisions
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Heterogeneous Catalysis
Carbon monoxide and nitrogen monoxide adsorbed on
a platinum surface
Step 1 Adsorption and activation of the
reactants.
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Heterogeneous Catalysis
Carbon monoxide and nitrogen monoxide arranged
prior to reacting
Step 2 Migration of the adsorbed reactants on
the surface.
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Heterogeneous Catalysis
Carbon dioxide and nitrogen form from previous
molecules
Step 3 Reaction of the adsorbed substances.
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Heterogeneous Catalysis
Carbon dioxide and nitrogen gases escape (desorb)
from the platinum surface
Step 4 Escape, or desorption, of the products.