Arnoldi
Honors Chemistry
Kinetics and Equilibrium
Practice Test
COMPLETION
1. ______A heterogeneous reaction is a reaction that includes
different __.
2. ______In equilibrium, the __ of the forward and reverse
reactions are equal.
3. ______The reaction rate describes the __ with which the reaction
takes place.
4. ______An intermediate species (will / will not) appear as a reactant
or product.
5. ______An inhibitor causes the reaction rate to __.
6. ______The actual path the reactants take as they change into products is called the __.
7. ______The rate-determining step is the __ step in the reaction
mechanism.
8. ______Kinetics is the branch of chemistry concerned with __ and
__.
______
9. ______Collision theory is used to describe the collisions between
particles, and the __ that result from those collisions.
10. ______Increasing temperature __ reaction rate because there will be __ collisions, and those collisions will be __ effective.
______
______
11. ______Increasing concentration (must / won’t / may) increase
reaction rate because…
12. ______In a heterogeneous reaction, increasing surface area will
__ reaction rate.
13. ______Effective collisions require correct particle __ and sufficient
__.
______
14. ______The higher the K value, the ( more / less / can’t tell) products are made.
15. ______The higher the Ksp value, the (more / less / can’t tell) a
substance dissolves in water.
16. ______compounds tend to react faster simply because of the
way they are bonded.
17. ______make a reaction proceed faster by lowering Activation
Energy.
18. ______What kind of substances tend to react faster: ionic or covalent?
19. ______work by getting in the way of a reaction, thus making it
proceed more slowly.
20. ______Increasing temperature about __ ‘C effectively doubles reaction rate.
TRUE and FALSE (use + for true and 0 for false)
21. _____ At equilibrium, the amounts of products and reactants are equal.
22. _____ At equilibrium, all changes have ended.
23. _____ If a reaction is at equilibrium, the reactants or products can never be
totally used up.
24. _____ KCl dissolving in water will reach equilibrium very readily.
SHORT ANSWER
25. Given the following reaction: 2 A + 3 B à A2B3
a. Write the Mass Action Expression
b. Write the Equilibrium Expression
c. Why can’t you write the rate law (for sure) for this reaction given the information above?
26. The following reaction occurs in a single-step mechanism.
2 A(aq) + B (g) + 1 C(g) à A2BC(aq)
(a) Write the rate law expression assuming that it is a one step reaction.
(b) What would happen to the rate of reaction in [A] was halved?
(c) What would happen to the rate of reaction if [B] was tripled?
(d) What would happen to the rate of reaction if [C] was quartered?
(e) What would happen if both (b) and (c) occurred?
(f) How would you know if the above reaction was not a one-step mechanism?
27. Answer the following questions using the mechanism provided: 2X(aq) + Y2(g) à 2 XY (aq)
Answer Options to Rate Questions: Increase, Decrease, No Change
(1) X + Y2 à XY + Y (slow)
(2) Y + X à XY (fast)
2X + Y2 à 2 XY
a. If [X] is increased, what will happen to the rate of the reaction? ______
Why?
b. What is the intermediate product? ______
c. If surface area decreases, what will happen to the rate? ______
Why?
d. Which step, (1) or (2), is the rate determining step? ______
e. If pressure decreases, what will happen to the rate? ______
Why?
f. If a catalyst is added, what will happen to the rate? ______
28. For the reaction: 2 A(aq) + B2(g) à 2AB (aq) several experiments were performed and the following data was obtained:
Experiment [A] [B] rate, mole/sec
1 0.500 0.250 0.850
2 0.500 0.500 3.40
3 1.00 0.500 6.80
a. Determine the rate law for the reaction.
b. Determine the rate of reaction when [A] = 0.750 and [B2] = 0.850
29. Answer the following questions given this equilibrium:
energy + AQ(aq) + R2(g) <----> AR2(aq) + Q(g)
a. What happens to the concentration of Q if the system is heated? _____
b. Which way will equilibrium shift (or doesn’t it) if pressure is increased? _____
c. What happens to the value of K if more R2(g) is added? _____
d. Which way will equilibrium shift (or doesn’t it) if we remove Q? ______
e. Which way will equilibrium shift (or doesn’t it) if we add an inhibitor? _____
f. What happens to the K value if the volume of the container is increased? _____
g. How can we make more Q? (multiple answers)
PROBLEMS Show all work. Circle final answers. Use significant figures.
1. The Ksp for Lead (II) Bromide at 25’C is 6.6 X 10-6.
a. Write the dissolving equation for Lead (II) Bromide in water.
b. Determine the concentration of Lead and Bromide ions in this situation.
c. If these ions are dissolved in 1500.0 mL of water, what mass of lead (II) bromide dissolved?
d. Write one chemical that you could add to the above solution to make the lead (II) bromide precipitate out of solution. Explain your answer.
2. What is the mass action expression for the following reaction:
2 AX(g) ó A2 (g) + X2 (g)
3. If the KEQ for the synthesis of AX is 25.0, what will the concentration of AX be at equilibrium, if the concentrations of A2 (g) and X2 (g) at equilibrium are 0.500 M and 0.400 M respectively.
A2 (g) + X2 (g) ó 2 AX(g)
In this reaction, are the products or reactants favored? How do you know?
What is the equilibrium constant for the reverse reaction?
4. For the reaction:
2 AX (s) + Q2(aq) ó 2 AQ (aq) + X2 (g)
a. What is the equilibrium expression?
b. 2.00 moles of each reactant were placed in 5.00 L of water inside of a closed container. 1.50 moles of AQ were in solution at equilibrium. What is the value of Keq?
5. If about 0.003819 grams of AgCl(s) can dissolve in 2.00 L of water at room temperature, what is the Ksp of silver chloride?
Practice Test Answer Key
1. phases 21. 0
2. rates 22. 0
3. speed 23. +
4. will not 24. 0
5. slow down
6. reaction mechanism
7. slowest
8. reaction rates / mechanisms
9. reactions
10. increases / more / more
11. may… it only does so if that reactant is in the rate determining step
12. increase
13. orientation / energy
14. more
15. more
16. ionic
17. catalysts
18. Ionic
19. Inhibitors
20. 10
25. a. Mass Action Expression = [A2B3]
[A]2[B]3
b. Same as (a)
c. You don’t know the actual mechanism by which the reaction proceeds.
26. a. R = k [A]2 [B] [C]
b. Rate would decrease by ¼.
c. Rate would increase by 3X.
d. Rate would decrease by ¼.
e. down by ¼ and up by 3 x = up by ¾
f. the exponents for the rate law would not be 2, 1, 1
27. a. The rate will increase. X is in the rate determining step.
b. Y
c. The rate will decrease. (it is a heterogeneous reaction)
d. The first step.
e. That means the concentration of Y2 decreases. Since Y2 is in the rate- determining step, this will slow down the reaction,.
f. It will increase.
28. a. When [A] doubled, the rate doubled. When [B] doubled, rate quadrupled.
R = k [A]1[B]2
b. For Experiment 1… 0.0850 = k (0.500)1(0.250)2
2.72 = k
For these conditions… R = (2.72)(0.750)1(0.850)2
R = 1.4739 à 1.47 mole / sec
29. a. increases g. remove the products as they are made
b. no change (causing a shift right to fill in hole)
c. it increases
d. right increase the [ ] of either reactant
e. it doesn’t (causing a shift right to use them up)
f. no change
increase energy
(causing a shift right to use it up)
pressure has no effect… 1 on each side
PROBLEMS Show all work. Circle final answers. Use significant figures.
1. a. PbBr2(s) in in H2O(l) Pb+2(aq) + 2Br-1(aq)
if Pb+2 = x, Br-1 = (2) (x)
b. Ksp = [Pb+2]1[Br-1]2
6.6 X 10-6 = [x]1[2x]2
6.6 X 10-6 = 4x3
0.012 M = x = Pb+2 Br-1 = 0.24 M
c. Pb+2 and PbBr2 are in a 1:1 ratio, so, [Pb+2] = [PbBr2]
[PbBr2] = M = mol
L
[PbBr2] (L) = mol
[3.0 X 10-3 mol / L] (1.5000 L) = 0.0045 mol PbBr2 X 367 g PbBr2 = 1.65 g PbBr2
d. Any soluble chemical containing Pb+2 or Br-1… for example KBr.
The extra ions (in this example Br-1) cause the equilibrium to shift left, resulting in precipitation of the lead (II) bromide.
2. MAE = [A2][X2]
[AX]2
3. Keq = [AX]2
[A2]1[X2]1
(Keq) ([A2]1) ([X2]1) = [AX]2
(25.0) (0.500) (0.400) = [AX]2
2.24 M = [AX]
The products are favored, as the Keq is > 1.
The equilibrium for the reverse reaction is the inverse value… 1/25.0 = 0.0400
4. a. K = [AQ]2 [X2]1
[Q2]1
b. At equilibrium…
1.50 mols AQ made X 1 mol X2 made = 0.750 moles X2 made at equilibrium
2 moles AQ made
1.50 mols AQ made X 1 mol Q2 used = 0.750 moles Q2 used
2 moles AQ made
So, 2.00 – 0.750 = 1.25 moles Q2 remain at equilibrium
1.50 mols AQ made X 2 mol AX used = 1.50 moles AX used
2 moles AQ made
So, 2.00 – 1.50 = 0.50 moles AX remain at equilibrium, but not dissolved (s)
All moles are divided by 5.00 L, so concentrations are…
0.300 M AQ 0.150 M X2 0.250 M Q2 N/A AX
K = [0.300]2 [0.150]1 = 0.0540
[0.250]1
5. If about 0.003819 grams of AgCl(s) can dissolve in 2.00 L of water at room temperature, what is the Ksp of silver chloride?
AgCl(s) + H2O(l) ó Ag+1(aq) + Cl-1(aq)
Ksp = [Ag+1][Cl-1]
[AgCl] = [Ag+1] = [Cl-1] (because all in 1:1:1 ratio)
So… [for all], find moles first… 0.003819 g AgCl X 1 mol AgCl = 2.65 X 10-5 mol
144 g AgCl
[all] = 2.65 X 10-5 mol = 1.33 M
2.00 L
Ksp = [1.33][1.33] = 2.66