1. AP Chemistry B Exam Prep Session Kinetics

2. and for gases: Kp = (PR)r(PS)s (PA)a(PB)b aA +bB + …↔ rR +sS + . .
The equilibrium expression is based upon the overall balanced equation:
aA +bB + …↔ rR +sS
Kc = [R]r[S]s
  [A]a[B]b
K > 1 products favored
K < 1 reactants favored
In heterogeneous equilibria only use gases & aqueous! Omitted solids; pure liquids; water (in aqueous solutions) because their [ ]’s do not change.
and for gases:
Kp = (PR)r(PS)s
(PA)a(PB)b

3. Types of equilibrium problems:
Kc general types
Ka acid equilibrium constants, weak acids to moderate acids; strong acids Ka>>>1
Kb base equilibrium constants, weak bases to moderate; strong bases Kb>>>1
Kw ion product constant for water; basis for pH scale @ 25˚C
Ksp solubility product constant, point of saturation for “insoluble” precipitation reactions
Kp using pressures of gases; DO NOT use brackets, must use (P) notation for equilibrium expression.

4. Converting from Kp to Kc
Kp using pressures of gases; DO NOT use brackets, must use (P) notation for equilibrium expression.
Kp = Kc(RT)Δn
Based upon Δn

5. Reaction Rates C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
All reactions slow down over time.
Therefore, the best indicator of the rate of a reaction is the instantaneous rate near the beginning.

6. Progress of Reaction Which reagent is the reactant? How can you tell?
What is the value of ∆G?
Do [A] & [B] look like same ratio in terms of stoichiometry?
Figure: UNE2
Title:
Exercise 14.2
Caption:
Graph of concentration versus time.

7. Which expression correctly states the rate of reaction?
N2(g) H2(g) NH3(g)
Which expression correctly states the rate of reaction?

8. N2(g) H2(g) NH3(g)

9. For the reaction: A 2 B, which of the following is a correct expression for the reaction rate?1. 
[A] 3. 
[A]
Rate
Rate = =  t 2  t 2. 
[B] 4. 
[B]
Rate =
Rate = - 2  t 2  t

10. For the reaction: 2 A  3 B
the rate of appearance of B, D[B]/Dt, is 0.30 M/s. What is the value of the rate of disappearance of A, –D[A]/Dt?
0.05 M/s
0.10 M/s
0.20 M/s
0.40 M/s
0.60 M/s

11. Correct Answer: 0.05 M/s 0.10 M/s 0.20 M/s 0.40 M/s 0.60 M/s  [A]
[B] 
[A]
Rate = =  2  3 t t 
[A] 2 
[B]
Rate = =  t 3  t 
[A]
2(0.30 M
/s)
Rate = = =
0.20 M /s  t 3

12. Factors Affecting Rate:
How to Speed Up a Reaction
[Use Collision Theory, Kinetic Molecular Theory]
increase the concentration of reactants
- increase molarity of solutions
- increase partial pressure of gases
[collision model: more collisions]
more surface area between unlike phases
increase the temperature
[collision model: more & harder collisions]
add a catalyst

13. Maxwell–Boltzmann Distributions
If the dotted line represents when reactions will happen (activation energy), as the temperature increases, so does the fraction of molecules that can overcome the activation energy barrier.
As a result, the reaction rate increases.

14. Catalysts
Catalysts increase the rate of a reaction by decreasing the activation energy of the reaction.
Catalysts change the mechanism by which the process occurs.
2006 #6

15. Rate cannot be determined by overall reaction.
Rate law must match the rate determining step of the reaction mechanism.
H2O2  2 OH- slow
H2O2 + OH-  H2O HO2 fast
HO2 + OH-  H2O O2 very fast
Overall reaction?
RDS?
Rate law?
2009 #3

16. a) Which species is the catalyst? b) Which is an intermediate?
A proposed mechanism for the decomposition of N2O is given below:
NO + N2O  N2 + NO2
2 NO2  2 NO + O2
a) Which species is the catalyst?
b) Which is an intermediate? NO
NO2
N2O N2 O2

17. Correct Answer: NO
NO2
N2O N2 O2
Note in the mechanism below that NO is regenerated. By definition a catalyst does not undergo permanent chemical change.
NO + N2O  N2 + NO2
2 NO2  2 NO + O2
Intermediates are produced and consumed within the process of the reaction = NO2

18. Sample AP Multiple Choice
Factors that affect the rate of a chemical reaction include which of the following?
Frequency of collisions of reactant particles
Kinetic energy of collisions of reactant particles
Orientation of reactant particles during collisions
II only
I and II only
I and III only
II and III only
I, II, and III

19. CAUTION! Requires memorization & must be determined experimentally.
Rate Laws
CAUTION! Requires memorization & must be determined experimentally.

20. Rate Laws Rate constant: same for respective reaction @ constant temp
Use reactions where reverse reaction condition can be neglected (negligible)
Rate = k [NO]2 [Br2]
Rate constant: same for respective constant temp
Order of reaction
2 + 1 = 3rd order

21. Types of Rate Laws Integrated Rate Law
Differential Rate Law
Integrated Rate Law
Shows how the rate of a reaction depends on concentrations (if I change [ ] how does rate change?
Uses [initial] & initial rate data
No memorizing, solve
Concentrations of species in the reaction depend upon time
At intervals of time, what are the concentrations
Use of graphs to determine order
Have to memorize!

22. Sample Problem 16.3
Determining Reaction Orders from Initial Rate Data
PROBLEM:
Many gaseous reactions occur in a car engine and exhaust system. One of these is
rate = k[NO2]m[CO]n
Use the following data to determine the individual and overall reaction orders.
Experiment
Initial Rate (mol/L*s)
Initial [NO2] (mol/L)
Initial [CO] (mol/L) 1 2 3
0.0050
0.080
0.10
0.40
0.20
PLAN:
Solve for each reactant using the general rate law using the method described previously.
SOLUTION:
rate = k [NO2]m[CO]n
First, choose two experiments in which [CO] remains constant and the [NO2] varies.

23. Determining Reaction Orders from Initial Rate Data
Sample Problem 16.3
Determining Reaction Orders from Initial Rate Data
[NO2] 2
[NO2] 1 m =
rate 2
rate 1
k [NO2]m2[CO]n2
k [NO2]m1 [CO]n1 =
The reaction is 2nd order in NO2.
0.40
0.10 = m
0.080
0.0050 ;
16 = 4m and m = 2
[CO] 3
[CO] 1 n =
rate 3
rate 1 =
k [NO2]m3[CO]n3
k [NO2]m1 [CO]n1
The reaction is zero order in CO.
0.20
0.10 n
0.0050 = ;
1 = 2n and n = 0
rate = k [NO2]2[CO]0 = k [NO2]2

24. Sample AP Multiple Choice
The table shows the results from a rate study of the reaction X + Y  Z. Starting with known concentration of X and Y in experiment 1, the rate of formation of Z was measured. If the reaction was first order with respect to X and second order with respect to Y, the initial rate of formation of Z in experiment 2 would be
R/4
R/2 R 2R 4R
Experiment
[X]o
[Y] o
Initial rate of Formation of Z (mol L-1sec-1) 1
0.40
0.10 R 2
0.20 ?

25. Integrated Rate Law
Order is determined by plotting data such that a straight line is obtained (1st order is most common followed by 2nd, lastly 0 order)
Once order is determined, can use appropriate equation to solve for rate constant (k) or k = slope
Data dictates which strategy must be used. For integrated rate law only given time vs. concentration no initial rate!

26. Integrated rate laws and reaction orders.
1/[A]t = kt + 1/[A]0
ln[A]t = -kt + ln[A]0
[A]t = -kt + [A]0

27. Figure 16.8
Graphical determination of the reaction order for the decomposition of N2O5.

28. Figure 16.8
Graphical determination of the reaction order for the decomposition of N2O5.

29. Figure 16.8
Graphical determination of the reaction order for the decomposition of N2O5.

30. A Sample AP Free Response Question
Topic: Kinetics 2010B #6
sitory/ap10_frq_chemistry_formb.pdf