16. II. CHEMICAL KINETICS / A. Reaction Rate _
II. Chemical Kinetics
A. Reaction Rate
1. Rate is how quickly or slowly anything is taking place: events per unit time.
e.g. steps per minute, miles per hour, exams per week, moles per second.
2. Factors affecting reaction rates
a) chemical nature of the reactants
b) concentration of the reactants
c) temperature
d) ability of reactants to make contact
e) presence of a catalyst
3. Rate in terms of moles per unit time
a) Consider the hypothetical reaction A(g) → B(g)
Average Rate
Time (min) Moles of B Δt Δmol Δmol/Δt
0 / 010 / 0.26 / 10 / 0.26 / 0.026
20 / 0.45 / 10 / 0.19 / 0.019
30 / 0.59 / 10 / 0.14 / 0.014
40 / 0.70 / 10 / 0.11 / 0.011
50 / 0.78 / 10 / 0.08 / 0.008
60 / 0.84 / 10 / 0.06 / 0.006
Example(1): How many moles of B are produced during the first 10 minutes? What is the average rate during the first 10 minutes?
Example(2): How many moles of B are produced between 50 and 60 minutes? What is the average rate during this time interval?
Example(3): On the graph below plot the number of moles of B vs time.
1.0-
0.9-
0.8-
0.7-
0.6-
0.5-
0.4-
0.3-
0.2-
0.1-
│ │ │ │ │ │
10 20 30 40 50 60
Example(4): On the graph below plot the rate of the reaction vs. time.
0.025-
-
0.020-
-
0.015-
-
0.010-
-
0.005-
-
- │ │ │ │ │ │
10 20 30 40 50 60
Example(5): On the graph below plot the number of moles of A vs. time, and the number of moles of B vs. time.
Time Moles of B Moles of A
0 / 0 / 1.010 / 0.26 / 0.74
20 / 0.45 / 0.55
30 / 0.59 / 0.41
40 / 0.70 / 0.30
50 / 0.78 / 0.22
1.0-
0.9-
0.8-
0.7-
0.6-
0.5-
0.4-
0.3-
0.2-
0.1-
│ │ │ │ │ │
10 20 30 40 50 60
4. Rate in terms of moles/liter/sec
The problem with measuring in terms of moles/sec is you cannot compare two experiments, unless everything is the same.
Example(6): Two students are asked to go outside and measure the rate of rain fall. One student catches rain in a cup for one hour, the other catches rain in a large pan of one hour. Each then measures the volume of the rain collected in one hour. Student one reports that it is raining at a rate of 0.5 quarts per hour, whereas the other student reports 10 quarts per hour.
To avoid this problem in kinetics, we measure the number of moles per liter per unit time,
mol/L/unit time. Or since mol/L is the same as Molarity, rate is measured in M/unit time (Usually M/sec.).
Example(7): If the reaction in example of A(g) → B(g) were taking place in a 2 liter container,
What is the average rate during the first 10 minutes?
Example(8): Sketch the concentration of B versus time, and the concentration of A versus time
[ ]
Time
5 Instantaneous rate: the slope of the tangent to the curve at any given time.
6. Rate and stoichiometry
Example(8): For the reaction A → B, how does the rate of B compare to the rate of A?
Example(9): For the reaction 2HI → H2 + I2, how does the rate of H2 compare to the rate of I2? How does the rate of HI compare to the rate of H2?
Example(10): If the rate of formation of H2 is 3.0x10–2 M/s, what are the rates of I2 and HI?
In general for aA + bB → cC + dD
Reaction rate = –1 rate A = –1 rate B = 1 rate C = 1 rate D
a b c d
Example(11): For the reaction 2N2O5 → 4NO2 + O2, write the reaction rate in terms of each substance.
Example(12): If the rate for N2O5 is –4.2x10-7 M/s, a) what is the reaction rate? b) What is the rate of formation of NO2? c) What is the rate of formation of O2?
19. II. CHEMICAL KINETICS / B. Dependence of Rate on Concentration _
B. Dependence of Rate on Concentration
1. Collision Model of the Rate Law
2. Reaction Order
In general rate = k[A]x[B]y[C]z, where x, y and z are determined by experiment.
a) Reaction Order: the sum of the exponents in the rate equation.
Example(1): Below are 3 reactions and their rate equations. Determine the order of each reaction.
2HI → H2 + I2 Rate = k[HI]2
2N2O5 → 4NO2 + O2 Rate = k[N2O5]1
2NO + Cl2 → 2NOCl Rate = k[NO]2[Cl2]1
3. Rate Constant Units
a) First order
b) Second Order
c) Third Order
4. Finding the Rate Law From Initial Rate
Example (2): The data below were collected for the reaction 2NO + 2H2 → N2 + 2H2O.
a)Find the powers on the concentrations in the rate law equation for the reaction.
b) Find the value of k for the rate equation for the reaction.
c) What is the rate of the reaction when [NO] = 0.10 M, and [H2] = 0.20 M?
Experiment # / [NO] / [H2] / Initial Rate, M/s1 / 0.010 / 0.010 / 0.06
2 / 0.010 / 0.020 / 0.12
3 / 0.020 / 0.010 / 0.24
24. II. CHEMICAL KINETICS / C. Theory of Reaction Rates _
C. Theory of Reaction Rates
1. Postulates
a) Reactant molecules must collide in order to react.
b) Molecules must collide with enough force or energy in order to react.
c) Molecules must have the correct orientation upon collision in order to react.
2. Effect of Concentration
3. Activation Energy (Ea): the minimum amount of kinetic energy that the molecules must
possess for a collision to be effective.
a) As molecules approach one another on a collision course, they slow down as they get closer to one another. This is the result of the repulsion that their electron clouds have for one another.
b) Their kinetic energy is converted into potential energy.
c) If they do not have enough initial kinetic energy, they will not be able to overcome the repulsion, and will veer off.
d) If they do have sufficient initial kinetic energy, they will collide and make new bonds and/or break existing bonds
4. ΔE or ΔH of the Reaction
a) Exothermic reaction
b) Endothermic reacton
c) Relation of the rate constant to Ea: The greater the Ea the smaller is the rate law constant.
Example (1): Consider two first order reactions, one with a high Ea and one with a low Ea. If the concentrations are the same which reaction would be slower? What difference is there between the two rate laws?
5. Effect of Temperature
a) Relation of the rate constant to temperature: The greater the temperature the large is the rate constant.
Example (2): Consider two identical reactions, one taking place at a low temperature, the other at a higher temperature. If the concentrations are the same, which reaction would be faster? What difference is there between the two rate laws?
b) Temperature measures the average kinetic energy of the molecules in a sample, but not all molecules have the same amount of kinetic energy.
c) Distribution of kinetic energy.
d) Change in kinetic energy distribution with temperature.
e) Arrhenius Equation: relates the rate constant to the Ea and the temperature.
k = Ae-Ea/RT or k= A . For a given reaction A is a constant.
eEa/RT
6. Molecular Orientation
a) Even if molecules collide with the require activation energy they may not react. They must have the appropriate alignment.
Example(3) H2(g) + I2(g) → 2HI(g)
Example(4) Cl2(g) + NO(g) → Cl(g) + NOCl(g)
7. Multiple Step Reactions
a) Rate Determining Step: In a multiple step reaction, the slowest step determines the
the rate of the reaction, and thus the rate equation.
Example(5): The following reaction A → C occurs in 2 steps: A → B, and B → P.
a) If the first step is the slower step, what is the rate equation for the reaction?
b) If the second step is the slower step, what is the rate equation for the reaction?
b) Intermediate: a species that is produced and consumed during a reaction.
Example(6): The reaction 2NO2(g) + F2(g) → 2NO2F(g) occurs in two steps:
Step 1: NO2(g) + F2(g) → NO2F(g) + F(g)
Step 2: F(g) + NO2(g) → NO2F(g)
a) If the rate equation for the overall reaction is: rate=k[NO2][F2], which step is the rate determining step?
b) What is the intermediate during the reaction?
Example(7): The reaction : 2H2(g) + 2NO(g) → N2(g) + 2H2O(g) occurs in two steps.
If the slower step is: H2(g) + 2NO(g) → N2O(g) + H2O(g)
a) What is the rate equation,
b) What is the equation for the faster step?
c) What is (are) the intermediate(s)?
27 II. CHEMICAL KINETICS / D. Catalysts _
D. Catalysis
1. Catalyst: Increases the rate of a reaction, but is not used up during the reaction.
2. A catalyst gives an alternate pathway to the products that has a lower activation energy than the pathway without the catalyst.
E
Reaction Coordinate
Fraction
of
Molecules
Kinetic Energy
3. Relation of the rate constant to the presence of a catalyst: If a reaction is catalyzed the rate constant increases.
Example (1): Consider two identical reactions, one taking place without a catalyst, the other with a catalyst. If the concentrations and the temperatures are the same, which reaction would be faster? What difference is there between the two rate laws?
3. Types of catalyst
a) Homogeneous Catalyst: the catalyst exist in the same phase as the reactants.
Example(2): The reaction H2O2(aq) → H2O(l) + O2(g) can be catalyzed by the addition of HBr.
2H+(aq) + 2Br–(aq) + H2O2(aq) → Br2(aq) 2H2O(l)
Br2(aq) + H2O2(aq) → 2Br–(aq) + 2H+(aq) + O2(g)
b) Heterogeneous Catalyst: the catalyst exist in a different phase than the reactants.
Example(3): Platinum as a surface catalyst.
Example(3): Enzymes are biochemical catalyst.
Substrate + Enzyme ⇄ ES + Product(s)
6. Summary of what a catalyst does and does not do.
a) The catalyst enters into the reaction.
b) At the end of the reaction it is regenerated, so it is not used up.
c) It increases the rate of the reaction by providing an alternative pathway with a lower
activation energy.
d) Since the activation energy is lower the reaction can take place rapidly without
increasing the temperature.
e) It does not increase the amount of product formed.
f) The ΔH of the reaction does not change.