2. 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.
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3. The Decomposition of Nitrogen Dioxide
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4. The Decomposition of Nitrogen Dioxide
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5. 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.
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6. Rate Law
Shows how the rate depends on the concentrations of reactants.
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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7. 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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8. 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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9. Types of Rate Laws
Differential Rate Law (rate law) – shows how the rate of a reaction depends on concentrations.
Integrated Rate Law – shows how the concentrations of species in the reaction depend on time.
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10. Rate Laws: A Summary
Because we typically consider reactions only under conditions where the reverse reaction is unimportant, our rate laws will involve only concentrations of reactants.
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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11. Rate Laws: A Summary
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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12. Determine experimentally the power to which each reactant concentration must be raised in the rate law.
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13. 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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14. 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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15. First-Order 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 A
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16. Plot of ln[N2O5] vs Time
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17. k = rate constant First-Order
Time required for a reactant to reach half its original concentration
Half–Life:
k = rate constant
Half–life does not depend on the concentration of reactants.
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18. A first order reaction is 35% complete at the end of 55 minutes
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
ln(0.65) = –k(55) + ln(1)
k = 7.8 x 10-3 min-1. If students use [A] = 35 in the integrated rate law (instead of 65), they will get k = 1.9 x 10-2 min-1.
Note: Use the red box animation to assist in explaining how to solve the problem.
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19. Second-Order Rate = k[A]2 Integrated:
[A] = concentration of A at time t
k = rate constant
t = time
[A]o = initial concentration of A
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20. Plot of ln[C4H6] vs Time and Plot of 1/[C4H6] vs Time20

21. Second-Order Half–Life: [A]o = initial concentration of A
k = rate constant
[A]o = initial concentration of A
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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22. EXERCISE!
For a reaction aA  Products, [A]0 = 5.0 M, and the first two half-lives are 25 and 50 minutes, respectively.
Write the rate law for this reaction.
rate = k[A]2
b) Calculate k.
k = 8.0 × 10-3 M–1min–1
Calculate [A] at t = 525 minutes.
[A] = 0.23 M
a) rate = k[A]2
We know this is second order because the second half–life is double the preceding one.
b) k = 8.0 x 10-3 M–1min–1
25 min = 1 / k(5.0 M)
c) [A] = 0.23 M
(1 / [A]) = (8.0 x 10-3 M–1min–1)(525 min) + (1 / 5.0 M)
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23. 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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24. Plot of [A] vs Time
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25. 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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26. CONCEPT CHECK!
How can you tell the difference among 0th, 1st, and 2nd order rate laws from their graphs?
For the zero-order reaction, the graph of concentration versus time is a straight line with a negative slope. For a first-order graph, the graph is a natural log function. The second-order graph looks similar to the first-order, but with a greater initial slope. Students should be able to write a conceptual explanation of how the half-life is dependent on concentration (or in the case of first-order reactions, not dependent).
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27. Summary of the Rate Laws
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28. Zero order First order Second order 4.7 M 3.7 M 2.0 M EXERCISE!
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:
Zero order
First order
Second order
4.7 M
3.7 M
2.0 M
a) 4.7 M
[A] = –(1.0×10–2)(30.0) + 5.0
b) 3.7 M
ln[A] = –(1.0×10–2)(30.0) + ln(5.0)
c) 2.0 M
(1 / [A]) = (1.0×10–2)(30.0) + (1 / 5.0)
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29. Reaction Mechanism
Most chemical reactions occur by a series of elementary steps.
An intermediate is formed in one step and used up in a subsequent step and thus is never seen as a product in the overall balanced reaction.
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30. NO2(g) + CO(g) → NO(g) + CO2(g)
A Molecular Representation of the Elementary Steps in the Reaction of NO2 and CO
NO2(g) + CO(g) → NO(g) + CO2(g)
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31. Elementary Steps (Molecularity)
Unimolecular – reaction involving one molecule; first order.
Bimolecular – reaction involving the collision of two species; second order.
Termolecular – reaction involving the collision of three species; third order. Very rare.
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32. Rate-Determining Step
A reaction is only as fast as its slowest step.
The rate-determining step (slowest step) determines the rate law and the molecularity of the overall reaction.
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33. Reaction Mechanism Requirements
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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34. 2( ) Decomposition of N2O5 2N2O5(g)  4NO2(g) + O2(g)
Step 1: N2O NO2 + NO (fast)
Step 2: NO2 + NO3 → NO + O2 + NO2 (slow)
Step 3: NO3 + NO → 2NO (fast)
2( )
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35. The reaction A + 2B  C has the following proposed mechanism:
CONCEPT CHECK!
The reaction A + 2B  C has the following proposed mechanism:
A + B D (fast equilibrium)
D + B  C (slow)
Write the rate law for this mechanism.
rate = k[A][B]2
rate = k[A][B]2
The sum of the elementary steps give the overall balanced equation for the reaction.
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36. Collision Model Molecules must collide to react. Main Factors:
Activation energy, Ea
Temperature
Molecular orientations
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37. Activation Energy, Ea
Energy that must be overcome to produce a chemical reaction.
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38. Change in Potential Energy
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39. For Reactants to Form Products
Collision must involve enough energy to produce the reaction (must equal or exceed the activation energy).
Relative orientation of the reactants must allow formation of any new bonds necessary to produce products.
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40. Arrhenius Equation A = frequency factor Ea = activation energy
R = gas constant ( J/K·mol)
T = temperature (in K)
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41. Linear Form of Arrhenius Equation
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42. Linear Form of Arrhenius Equation
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43. EXERCISE!
Chemists commonly use a rule of thumb that an increase of 10 K in temperature doubles the rate of a reaction. What must the activation energy be for this statement to be true for a temperature increase from 25°C to 35°C?
Ea = 53 kJ
Ea = 53 kJ
ln(2) = (Ea / J/K·mol)[(1/298 K) – (1/308 K)]
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44. Catalyst
A substance that speeds up a reaction without being consumed itself.
Provides a new pathway for the reaction with a lower activation energy.
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45. Energy Plots for a Catalyzed and an Uncatalyzed Pathway for a Given Reaction
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46. Effect of a Catalyst on the Number of Reaction-Producing Collisions
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47. Heterogeneous Catalyst
Most often involves gaseous reactants being adsorbed on the surface of a solid catalyst.
Adsorption – collection of one substance on the surface of another substance.
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48. Heterogeneous Catalyst
Adsorption and activation of the reactants.
Migration of the adsorbed reactants on the surface.
Reaction of the adsorbed substances.
Escape, or desorption, of the products.
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49. Homogeneous Catalyst
Exists in the same phase as the reacting molecules.
Enzymes are nature’s catalysts.
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