Name:_____________________________ Class: _________________ Date: _________________
CHAPTER 18 REVIEW
Chemical Equilibrium
Teacher Notes and Answers
Chapter 18 6. a. Impose a high pressure so the reaction
SECTION 1 that produces fewer gas molecules will be
SHORT ANSWER favored, which is the forward reaction in this
case.
[ C] 2 [ D ]3 b.Keep the temperature low to favor the
1. K =
[ A ] 3 [ B] forward, exothermic reaction, but not so low
as to slow the rate of the forward reaction
[ NO ] 2 too much.
2. a. (1) 2 = 0.1
[N O ] c.No, a catalyst does not change the
2 2 percentage yield.
[ NH +4 ][OH 2 ] −5 7. a. The forward reaction is favored. The
(2) = 2 × 10
[ NH 4 OH] color becomes less intense because the
equilibrium shifts in the direction that
[HI] 2
(3) = 54.0 produces the colorless gas, N2O4.
[H 2 ][I 2 ]
b. Negative; lowering the temperature
[SO 3 ] 2 favors the forward reaction, which is
(4) = 1.8 × 10 − 2 O
2
[SO 2 ] [O 2 ] exothermic, so ∆H has a negative sign.
b. system 3
SECTION 3
c. system 2
3. a. The rate of the forward reaction equals SHORT ANSWER
the rate of the reverse reaction. 1. a
b.The concentrations of the products and the 2. d
reactants remain constant. 3. b
4. 1.1 4. a
5. 0.0024 5. a. (a) base 1 acid 2 acid 1 base 2
(b) base 1 acid 2 base 2 acid 1
SECTION 2 b. b
SHORT ANSWER c. a
1. d 6. The acid is HNO3(aq) and the base is
2. a. b Ca(OH)2(aq).
b. b [ NH +4 ][OH − ]
7. K =
c. a [ NH 3 ]
3. a. reverse 8. a. acid 1base 2 base 1 acid 2
b. neither direction [X − ][H 3 O + ]
c. forward b. K a =
[HX]
d. reverse −8
4. c c. 1.0 × 10
5. Pure solids are not included in equilibrium
expressions because their concentrations
do not change. Their constant value is
incorporated into K.
Original content Copyright © by Holt, Rinehart and Winston. Additions and changes to the original content are the responsibility of the instructor.
Modern Chemistry 1 Chemical Equilibrium
, Name:_____________________________ Class: _________________ Date: _________________
SECTION 4 4. a. Ag 3 PO 4 ( s ) → + 3−
← 3Ag ( aq ) + PO 4 (aq)
−6
SHORT ANSWER b. 5.0 × 10 M
1. (a) c c. 1.7 × 10− 2 0
(b) a 5. a. Ksp = [Pb2 + ][Cl− ]2
(c) b b. 4.5 × 10− 5
2. a. K sp = [Ag + ] 2 [CO 32 − ]
b. reverse reaction
→ 2+ −
3. a. XCl 2 ( s ) ← X ( aq ) + 2Cl ( aq )
Ksp = [X2 + ][Cl− ]2
b. 3 × 10− 8
c. less soluble
Original content Copyright © by Holt, Rinehart and Winston. Additions and changes to the original content are the responsibility of the instructor.
Modern Chemistry 2 Chemical Equilibrium
CHAPTER 18 REVIEW
Chemical Equilibrium
Teacher Notes and Answers
Chapter 18 6. a. Impose a high pressure so the reaction
SECTION 1 that produces fewer gas molecules will be
SHORT ANSWER favored, which is the forward reaction in this
case.
[ C] 2 [ D ]3 b.Keep the temperature low to favor the
1. K =
[ A ] 3 [ B] forward, exothermic reaction, but not so low
as to slow the rate of the forward reaction
[ NO ] 2 too much.
2. a. (1) 2 = 0.1
[N O ] c.No, a catalyst does not change the
2 2 percentage yield.
[ NH +4 ][OH 2 ] −5 7. a. The forward reaction is favored. The
(2) = 2 × 10
[ NH 4 OH] color becomes less intense because the
equilibrium shifts in the direction that
[HI] 2
(3) = 54.0 produces the colorless gas, N2O4.
[H 2 ][I 2 ]
b. Negative; lowering the temperature
[SO 3 ] 2 favors the forward reaction, which is
(4) = 1.8 × 10 − 2 O
2
[SO 2 ] [O 2 ] exothermic, so ∆H has a negative sign.
b. system 3
SECTION 3
c. system 2
3. a. The rate of the forward reaction equals SHORT ANSWER
the rate of the reverse reaction. 1. a
b.The concentrations of the products and the 2. d
reactants remain constant. 3. b
4. 1.1 4. a
5. 0.0024 5. a. (a) base 1 acid 2 acid 1 base 2
(b) base 1 acid 2 base 2 acid 1
SECTION 2 b. b
SHORT ANSWER c. a
1. d 6. The acid is HNO3(aq) and the base is
2. a. b Ca(OH)2(aq).
b. b [ NH +4 ][OH − ]
7. K =
c. a [ NH 3 ]
3. a. reverse 8. a. acid 1base 2 base 1 acid 2
b. neither direction [X − ][H 3 O + ]
c. forward b. K a =
[HX]
d. reverse −8
4. c c. 1.0 × 10
5. Pure solids are not included in equilibrium
expressions because their concentrations
do not change. Their constant value is
incorporated into K.
Original content Copyright © by Holt, Rinehart and Winston. Additions and changes to the original content are the responsibility of the instructor.
Modern Chemistry 1 Chemical Equilibrium
, Name:_____________________________ Class: _________________ Date: _________________
SECTION 4 4. a. Ag 3 PO 4 ( s ) → + 3−
← 3Ag ( aq ) + PO 4 (aq)
−6
SHORT ANSWER b. 5.0 × 10 M
1. (a) c c. 1.7 × 10− 2 0
(b) a 5. a. Ksp = [Pb2 + ][Cl− ]2
(c) b b. 4.5 × 10− 5
2. a. K sp = [Ag + ] 2 [CO 32 − ]
b. reverse reaction
→ 2+ −
3. a. XCl 2 ( s ) ← X ( aq ) + 2Cl ( aq )
Ksp = [X2 + ][Cl− ]2
b. 3 × 10− 8
c. less soluble
Original content Copyright © by Holt, Rinehart and Winston. Additions and changes to the original content are the responsibility of the instructor.
Modern Chemistry 2 Chemical Equilibrium
