Chapter 15 Lecture Outline
M&M Kinetics
Introduce Kinetics with M&M's and spreadsheet M&M Kinetics
- Zero Order Example (start with 100 on overhead, eat 10 every 5 seconds)
- First Order Example (start with 100 on overhead, eat 1/5 every 5 seconds)
- Second Order Example (start with 100 on overhead, eat 0.006 of number squared)
- Experiment with General Graph to show effect of rate.Reaction Rate and Order
Measuring Reaction Rates
- Rates as [delta]C/[delta]t
- Overhead, and Figure 15.2 from Textbook
- Rate for specific product or reactant
- For the reaction 2 NO2 <-> 2 NO + O2 At 300 C
- Data from Zumdahl
| time | NO2 | NO | O2 |
| 0 | 0.0100 | 0 | 0 |
| 50 | 0.0079 | 0.0021 | 0.0011 |
| 100 | 0.0065 | 0.0035 | 0.0018 |
| 150 | 0.0055 | 0.0045 | 0.0023 |
| 200 | 0.0048 | 0.0052 | 0.0026 |
| 250 | 0.0043 | 0.0057 | 0.0029 |
| 300 | 0.0038 | 0.0062 | 0.0031 |
| 350 | 0.0034 | 0.0066 | 0.0033 |
| 400 | 0.0031 | 0.0069 | 0.0035 |
- Graph concentration vs time for NO2
- Calculate concentration of other species and graph
- Calculate average rate
- from 0 to 50
- from 200 to 250
- Completed Spreadsheet
- Compare rates for each species
Reaction Rates
- CD-ROM 15-2 side bar (Bleach and dye)
- Surface Area. (
internet © Saunders, 1997)
- Concentration. CD-ROM, 15-4 (
internet © Saunders, 1997)
- Mg (s) + 2 HCl (aq) -> H2 (g) + Mg2+ + 2 Cl1- (aq)
- 2 N2O5 -> 4 NO2 + O2 (Initial Rate vs concentration)
Rate Equations
- Calculating Reaction Rates: Completed Spreadsheet
Calculate rate of NO2 loss, NO gain, O2 gain, and reaction
for the reaction:
2 NO2 <-> 2 NO + O2 At 300 C
Data from Zumdahl
| time (s) | NO2(atm) | NO (atm) | O2 (atm) | Rate rxn (atm/s) |
| 0 | 0.0100 | 0 | 0
|
| 50 | 0.0079 | 0.0021 | 0.0011 | 2.1*10-5 |
| 100 | 0.0065 | 0.0035 | 0.0018 | 1.40*10-5 |
| 150 | 0.0055 | 0.0045 | 0.0023 | 1.00*10-5 |
| 200 | 0.0048 | 0.0052 | 0.0026 | 7.00*10-6 |
| 250 | 0.0043 | 0.0057 | 0.0029 | 5.00*10-6 |
| 300 | 0.0038 | 0.0062 | 0.0031 | 5.00*10-6 |
| 350 | 0.0034 | 0.0066 | 0.0033 | 4.00*10-6 |
| 400 | 0.0031 | 0.0069 | 0.0035 | 3.00*10-6 |
- The rate law
- Expression:
for a reaction: aA + bB -> cC + dD
rate is rate = k [A]x [B]y
- For reaction of NO2, calculate 1st order rate constant for each rate.
- For reaction of NO2, calculate 2nd order rate constant for each rate.
- Fits 2ndorder better so write rate law
rate = k [NO2]2
Calculate average value of k, from above:
rate = 0.277 atm-1 sec-1 [NO2]2
- Discuss units for kinetics problems (Handout). Note: Table 15.1 page 706, correct units for k all sec-1
Method of Initial Rates
Determine rate law from initial rates or rates at different concentrations. (Lecture Problems).
- Derive Relationship
ln(Rate1/Rate2) = x ln(A1/A2)
Where:
- Rate1 is the rate under the first conditions.
- Rate2 is the rate under the second conditions.
- A1 is the concentration of A for the first conditions.
- A2 is the concentration of A for the second conditions.
- x is the reaction order for A.
- For the reaction:
NH4+(aq) + NO2-(aq) -> N2(g) + 2 H2O (l)
With the following experimental data.
Experimental Data
| [NH41+] (M) |
[NO21-] (M) |
Rate (M s-1 |
| 0.100 |
0.005 |
1.35*10-7 |
| 0.100 |
0.010 |
2.75*10-7 |
| 0.200 |
0.010 |
5.4*10-7 |
- Determine the order of a reaction
- Determine k for reaction:
- Write the rate law
- Use the rate law to calculate the rate for new concentrations.
Integrated Rate Equation:
(Lecture Problems)
- Plot of concentration vs time, and rate vs time
- Units for axis
- Units for area under curve
- Integration of rate equation
- Derive 1st order integrated rate equation
- Use the integrated rate equation to find the concentration at time t for:
C2H5Cl -> C2H4 + HCl at 700 K, k = 2.50*10-3 min-1
- If the initial concentration of C2H5Cl is 3.45 atm, how long does it take for the partial pressure of C2H5Cl to drop to 1.00 atm?
- What is the partial pressure of C2H5Cl after six hours?
- How long does it take for the C2H5Cl concentration to drop to 1/2 the initial concentration?
- How long does it take for 99% of the C2H5Cl to react?
- Graph Concentration vs time from integrated rate equation (overhead).
Rate laws, integrated rate equation, and units
| Reaction Order |
Rate |
Rate Law |
Integrated Rate Equation |
Linear Plot |
Slope |
Units for k |
| 0 |
[delta] concentration
-----------------------
[delta] time |
rate = k |
[A]0 - [A]t = kt |
[A] vs time |
-k |
concentration time-1
mol liter-1 sec-1
pressure time-1 |
| 1 |
[delta] concentration
-----------------------
[delta] time |
rate = k [A] |
ln ([A]t/[A]0) = - kt |
ln [A] vs time |
-k |
sec-1 |
| 2 |
[delta] concentration
-----------------------
[delta] time |
rate = k [A]2 rate = k [A] [B] |
(1/[A]t) - (1/[A]0) = kt |
(1/[A]) vs time |
k |
concentration-1 time-1
liter mole-1 sec-1
pressure-1 time-1 |
Graphing Kinetics Data
Graphing Kinetics Data (pdf, mathcad)
Collision Theory
- Collision Energy NO + O3 -> NO2 + O2
(
internet© Saunders, 1997)
- Orentation NO + O3 -> NO2 + O2
( internet© Saunders, 1997)
- Successful Reaction NO + O3 -> NO2 + O2
( internet© Saunders, 1997)
- Energy Diagram (Overhead)
- Reaction Energy Diagram F1- + CH3Cl -> CH3F + Cl1- (
internet© Saunders, 1997)
Temperature and the rate constant
- Temperature and reaction rate, bleach demo (
internet © Saunders, 1997)
- Concentration and reaction (collision) rate).
- The Arrhenius Equation
- Graphing ln(k) vs 1/T, idea of linear relationships
- Derive relationship
- Show graph on Lotus
- Calculations with the Arrhenius equiaton
- Given A and Ea , determine k at T
- Find change in rate at two T's from Ea
- Determine Ea from k at two T's
- Arrhenius Plot to find Ea
- (Lecture Problems)
Catalysis
- H2O2 Decomposition (
internet© Saunders, 1997)
- Catalysis reaction steps
- Given the reaction steps:
- NO + O3 -> NO3 + O (slow)
NO3 + O -> NO2 + O2 (fast)
NO + O3 -> NO2 + O2 (overall)
- And an additional step:
NO2 + O -> NO + O2
- Combines as:
NO + O3 -> NO2 + O2
NO2 + O -> NO + O2
O3 + O -> 2 O2
- Which species is a catalyst?
- Which species is an intermediate?
- How does this catalyst effect the rate of the reaction
- Catalysis Energetics and Reaction Rates. (Lecture Problems)
For the ovrall reaction: O3 + O -> 2 O2
- For the catalyzed reaction Ea = 11.9 kJ
- For the uncatalyzed reaction Ea = 14.0 kJ
- Draw an energy level diagram
- What is the rate constant at 298 K for each mechanism?
- For the catalytic destruction of O3 by Cl Ea = 2.1 kJ, what would this rate constant be at 298 K?
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