1. Fred Omega Garces Chemistry 201 Miramar College
13.1 Expressing the Reaction Rate Differential Rate and Integrated Rate Laws
Fred Omega Garces
Chemistry 201
Miramar College

2. Outline Kinetic Molecular Theory Kinetics: Intro
PreExponential Factor
    Collision Frequency
    Steric Factor
Activation Energy
    Arrhenius Equation
    Reaction Coordinate Diagram
Catalyst factor
Maxwell-Boltzmann Diagram
Reaction Mechanism & Chapman cycle
Elementary Step
Rate Determining Step
Kinetics: Intro
Factors influencing Reaction Rates
Rates & Stoic Relations
Expressing reaction Rates
Rate Law and components
Differential Rate Law
Initial Rate Method
Rate Laws and Half-Lives
    0th Order
    1st Order
    2nd Order

3. Thermodynamics Vs. Kinetics
ThermoChem: Will a reaction occur ?
Kinetics: If so, how fast ?
• Iron rusting; G -,
but rate of rusting depends on reaction conditions.
4 Fe (s) O2(g) + xH2O(g) D 2Fe2O3•XH2O (s)
Whether it occurs overnight or over many years, the reaction condition influence how fast it occurs.
• Conversion of diamond to graphite is
thermodynamically favorable (G -).
C (diamond) D C (graphite) G = kJ/mol
Kinetics makes this reaction nearly impossible.

4. Factors Influencing Reaction Rates
Factors Affecting Rates
Concentration of Reactant - Molecules must collide in order to react. More reactants means the faster rxn. Note: rate µ concentration
Physical state - Molecules must mix to form product. The more finely a reactant chemical is divided, the greater its surface area per unit mass, the more contact it makes with other reactants the greater the speed of the reaction. Stirring and mixing helps in this process.
Note: rate µ collision frequency
Temperature- Molecules must collide with enough energy to react. Note: rate µ collision energy (energy is temperature dependent)
Catalyst- The pathway in which the reactant convert to product influence the reactivity.

5. Rates ; Another word for Speed
Rate - Change / time (A positive quantity)
Speed -  distance /  time =  x / time
Wages -  money /  time =  $ / time
Reaction -  conc. /  time =  [A] / time
Differential Rate Law; An expression that relates the rate to the concentration,
  [A] with respect to time.
General Form:  [A]
 t

6. Instantaneous Vs. Average rates (Integrated Form)
Consider: Reactant  Product
Experiment will monitor concentration of reactant (product) as a function of time.
rate = - [ c] = =
 t
rate (Avg.) = M s-1
rate = - [ c] = =
 t
rate (Avg.) = M s-1
rate = - dc = Tangent at dt

7. Stoichiometry Relationship: Example
H2 (g) + I2 (g)  2 HI (g)
Rate = -  [H2] = -  [I2] ? +  [HI]
Rxn  t t  t
-  [H2] = +  [HI] ? - 1 mol H2
t t mol HI
Rate of reaction does not depend on which specie is monitored during the reaction.
- [H2] = +  [HI]
 t t
Disappearance
Appearance
Rate =  [H2] = -  [I2] = +  [HI]
Rxn  t  t t

8. Stoichiometry Rate Relationship
aA + bB  cC+ dD
Rate = -  [A] = -  [B ] = +  [C ] = +  [D ]
Rxn a t b t c t d t
10 mol/min
2H2 + O2  2 H2O
Rate = -  [H2] = -  [O2 ] = +  [H2O ]
Rxn  t t  t
-5 mol/min mol/min mol/min
2O3  3 O2
Rate = -  [O3] = +  [O2 ]
Rxn  t t

9. Methods of Initial Rates
Methods of initial rates is used to determine the rxn rates:
2 NO2(g) 2 NO(g) + O2(g)
Rate Law c * Rate Reaction = d[NO2] / dt
Rate = -  [NO2 ] = - [NO2] t= [NO2] t=50
Rxn  t s s
Rate = -  [NO2 ] = M
Rxn  t s s
Rate = -  [NO2 ] = - (-2.7•10-5) M
Rxn  t s
Rate = -  [NO ] = •10-5 M
Rxn  t s
= Rate Reaction
The rate of a reaction does not depend on which specie is measured.
* Integrated Rate Law; how the concentrations depend on time.

10. Reaction Rate Relationship
Rates of a reaction in terms of any specie in the reaction: aA +bB g pP + rR
Coefficient- The coefficients from the balance equation must be taken into account when calculating rates (speed of rxn).
Stoichiometry- Although mole ratio from overall stoichiometry does not relate to the order of the rate or the Rate Law directly (to be discussed later), it does provide the relationship of the rate of reactant consumption and product generation.
In our previous example: 2 NO2(g) . 2 NO(g) + O2(g)
Reactant NO2 , and product NO, have coefficient of 2, but product O2 has coefficient of 1. This leads to the production of NO having the same rate as NO2 consumption, but the production of O2 being only half as fast.
t = 1
t = 2
t = 3
t = 4
Reaction Progress

11. In Class Exercise
In Class activity: Number heads - Break in to groups of 4 and answer the following question. I will call one of the members to explain answer in 10 min.
Acrylonitrile is produced from propene, ammonia and oxygen by the following reaction: 2C3H6 + 2NH3 + 3O2  2CH2CHCN + 6H2O
Relate the rates of reaction of starting materials and products.
Draw a Concentration Vs. Time diagram for the reaction.
If 120 moles of H2O is produced in 1-min. How many moles of propene and ammonia are consumed in 1-hr. Assume 1-L vessel.
If grams of Oxygen is consumed in 1-min, how long (min) will it take to produce 100 g of acrylonitrile. Assume 1-L vessel.
Strategy: The rates for different species participating in a chemical reaction are related by the stoichiometric coefficients of the balanced chemical equation.

12. Summary of reaction rate
The average reaction rate is the change in reactant (or product) concentration in a certain time increment. The rate slows down however as reactants are used up. The instantaneous rate at time t is obtained from the slope of the tangent on a concentration-time curve at time t. The method of initial rates is generally used to avoid complications of back reaction.
The next step is to solve the differential rate law which leads to the integrated rate law. (This requires Calculus)

13. Solutions to the Differential Rate Law*
Differential Form of the Rate Law -
Expression that provides rate dependence on time.
Also referred to as simply the Rate Law
aA + bB  mM+ nN
rate = K [A]x [B]y [C]w
C = catalyst
x, y and w - order of reaction
x-th order in A: y-th order in B: w-th order in C, (the catalyst)
x, & y is not related to the coefficient of the balance equation; it is experimentally determined
K - rate constant
* Zumdahl

14. Methods of Initial Rates
In our discussion so far, the reverse reaction has not been considered: 2 NO2(g) 2 NO(g) + O2(g)
Suppose 2 NO(g) + O2(g)  2 NO2(g also occurred?
In determining the concentration of species in a reaction, the formation of products may interfere with the measurements.
This can be avoided if the reaction is studied under conditions in which the reverse reaction makes negligible contribution. This is the Initial Rates Methods
Methods of Initial Rates: Rates of the reaction depends only on the concentration of the reactants.

15. Initial Rate Methodology Determining the Order of a Reaction
• Several experiments are carried out in which the initial concentration of all substances are known but are varied in the different trials
• The reaction is allowed to proceed for a short period (~3% of completion) such that the average rate during this time of the reaction is close to the instantaneous rate.
• The result of the analysis yields the Differential Rate Law (or Rate Law)
• The experiment is such that the rate is a function of only the reactants in the reaction.

16. Rate Law Determination
Consider the combination reaction of NO and O2 to produce NO2 :
2 NO(g) + O2(g)  2 NO2 (g)
Determination of the Rate Law (via Methods of Initial Rates)
Initial Concentrations
(mol/ L) Initial Rate
Experiment [NO] [O2] (mol/L•s)
Based on these data, what is the rate law? What is the value of the rate constant, K?

17. GuideLines for Solving Rate Law
Experiments are chosen such that the concentration of all except one of the reactant does not change. If there are two reactants in the reaction then a minimum of three experiments is required. Three reactants require a minimum of four experiments.
The two experiments that are selected are divided between each other in the general formula:
Expt 1 = Rate1 = K [A1]x [B1]y [C1]z …
Expt 2 = Rate2 = K [A2]x [B2]y [C2]z …
The reaction order is determined by solving the ratios.
Once the order for each reactant is determined the rate constant (K) is solved by using one of the experimental conditions.
…. but first,

18. A Quick Math Review Exponential Operations: Solving for exponents:
20 = 1
21 = 2
22 = 4
23 = 8
24 = 16
25 = 32
26 = 64
Solving for exponents:
Ax = B
X • lnA = lnB
X = lnB / lnA
i.e., 26 = 64
Try x = 64, solve for x
x = ln(64) / ln(2)
= 2.77/.693
= 6

19. Rate Law Determination; Ratio of Experiments
Now consider our pervious problem:
Initial Concentrations
(mol/ L) Initial Rate
Experiment [NO] [O2] (mol/L • s)
Select experiments in which concentration of [NO] is constant but the concentration of O2 is doubled: Experiments 1 and 2
Expt Rate k [NO]x [O2]y k [0.020]x [0.020]y
Expt Rate k [NO]x[O2]y k [0.020]x [0.010]y = =

20. Rate Law Determination; Ratio of Experiments
Select experiments in which concentration of [NO] is constant but the concentration of O2 is doubled: Experiments 1 and 2
Expt Rate k [NO]x [O2]y k [0.020]x [0.020]y
Expt Rate k [NO]x[O2]y k [0.020]x [0.010]y
Result:
2 = 2y; y = 1 g 1st order in O2
Similarly, selecting experiments 2 and 5 leads to
Expt Rate k [NO]x [O2]y k [0.020]x [0.020]y
Expt Rate k [NO]x[O2]y k [0.010]x [0.020]y
This leads to: 4 = 2x; x = 2 g 2nd order in NO. = = = =

21. Rate Law; Solving for rate Constant
The general rate law is:
Rate law = k [NO]2 [O2]
the rate constant k is determine by selecting one of the experiments and solving the equation.
Consider experiment #1 :
Rate = = k [0.020]2 [0.010]
k = / (4•10-4) (0.010) = 7.1•103 M-2 s-1
Rate Law: Rate = 7.1•103 M-2s-1 [NO]2 [O2]

22. In Class Exercise The following data were obtained. (Units are mol/L)
Acrylonitrile is produced from propene, ammonia and oxygen by the following reaction at 25°C: 2C3H6 + 2NH3 + 3O2  2CH2CHCN + 6H2O
The following data were obtained. (Units are mol/L)
C3H6 NH3 O2 Initial Rates
1.00 e e e e -7
2.00 e e e e -7
3.00 e e e e -7
1.00 e e e e -7
1.00 e e e e -7
1.00 e e e e -7
1.00 e e e e -7
Rate Law: Rate = 5.18 e5 M-5s-1 [C3H6]1 • [NH3]2 • [O2]3

23. Summary
Reaction rates depend on reactant concentrations, temperature, and the presence of a catalyst. The concentration dependence is given by the rat law, rate = k [A]m •[B]n, where k is the rate constant, m and n specify the reaction order with respect to reactants A and B, and m+n is the overall reaction order. The values of m and n must be determined by experiments; they can’t be deduced from the stoichiometry of the overall reaction.
The integrated rate law is a concentration-time equation that allows us to calculate concentrations at any time or the time required for an initial concentration to reach any particular value.
Answer last problem Rate = 5.25 e5 M-5s-1 [C3H6]1 • [NH3]2 • [O2]3