Download 3 Chemical Equilibrium

South Pasadena • AP Chemistry
Name
3 ▪ Chemical Equilibrium
Period
3.1
PROBLEMS
1. Write the equilibrium constant expressions for the
following reactions. How are they related to one
another?
(a) 2 N2O (g) + 3 O2 (g)  4 NO2 (g)
Keq =
[NO2]4eq
[N2O]2eq[O2]3eq
(b) N2O (g) + 3/2 O2 (g)  2 NO2 (g)
[NO2]2eq
K′eq =
= (Keq)1/2
[N2O] eq[O2]3/2eq
(c) 4 NO2 (g)  2 N2O (g) + 3 O2 (g)
K″eq =
[N2O]2eq[O2]3eq
= (Keq)−1
[NO2]4eq
2. Calculate the value of the equilibrium constant for
the following system, given the data shown:
H2 (g) + CO2 (g)  H2O (g) + CO (g)
Concentrations at equilibrium:
[H2]eq = 1.5 M, [CO2]eq = 2.5 M
[H2O]eq = 0.5 M, [CO]eq = 3.0 M
Keq =
[H2O]eq[CO]eq (0.5)(3.0)
=
= 0.40
[H2]eq[CO2]eq (1.5)(2.5)
3. Chlorine molecules will dissociate at high
temperatures into chlorine atoms. At 3000C, for
example, Kc for the equilibrium shown is 0.55. If
the partial pressure of chlorine molecules is 1.5
atm, calculate the partial pressure of the chlorine
atoms:
Cl2 (g)  2 Cl (g)
Kp = Kc(R·T)∆n = (0.55)((0.08206)(3273 K))2−1
= 148
Kp =
(PCl)2
(PCl2)
PCl = 15 atm
148 =
(PCl)2
(1.5)
–
Date
EQUILIBRIUM
4. Suppose that 0.50 moles of hydrogen gas, 0.50
moles of iodine gas, and 0.75 moles of hydrogen
iodide gas are introduced into a 2.0 liter vessel and
the system is allowed to reach equilibrium.
H2 (g) + I2 (g)  2 HI (g)
Calculate the concentrations of all three substances
at equilibrium. At the temperature of the
experiment, Kc equals 2.0 × 10−2.
Q=
[HI]2
(0.375)2
=
= 2.25 > Keq
[H2][I2] (0.25)(0.25)
H2 (g) + I2 (g)  2 HI (g)
M
Initial
0.25
0.25
0.375
Change
+x
+x
−2x
Equil.
0.25+x
0.25+x
0.375−2x
Keq =
[HI]2eq
(0.375−2x)2
=
= 2.0 × 10−2
[H2]eq[I2]eq (0.25+x)(0.25+x)
x = 0.16
[H2]eq = [I2]eq = 0.41 M; [HI]eq = 0.06 M
5. If the mechanism of a chemical equilibrium
consists of two reversible elementary steps, each
with its own equilibrium constant Kc1 and Kc2,
what expression relates the equilibrium constant
Kc for the overall equilibrium to the two constants
Kc1 and Kc2?
[B]
[A]
[D]
Step 2: C  D
Kc2 =
[C]
Overall: A + C  B + D
[B][D]
Kc =
= Kc1 · Kc2
[A][C]
Step 1: A  B
Kc1 =
6. When 2.0 mol of carbon disulfide and 4.0 mol of
chlorine are placed in a 1.0 liter flask, the
following equilibrium system results. At
equilibrium, the flask is found to contain 0.30 mol
of carbon tetrachloride. What quantities of the
other components are present in this equilibrium
mixture?
CS2 (g) + 3 Cl2 (g)  S2Cl2 (g) + CCl4 (g)
CS2 (g) + 3 Cl2 (g)  S2Cl2 (g) + CCl4 (g)
M
b. Is the mixture at equilibrium?
No.
c. In which direction will the system move to
reach equilibrium?
The reaction will shift to the left, because
Qp > Kp.
d. When the system reaches equilibrium, what
will be the partial pressures of the components
in the system?
Initial
2.0
4.0
0.0
0.0
Change
−0.30
−0.90
+0.30
+0.30
torr
Equil.
1.7
3.1
0.30
0.30
Initial
675
43
23
Change
+2x
−2x
−x
Equil.
675+2x
43−2x
23−x
[CS2]eq = 1.7 M, [Cl2]eq = 3.1 M,
[S2Cl2]eq = 0.30 M
7. 3.0 moles each of carbon monoxide, hydrogen,
and carbon are placed in a 2.0 L vessel and
allowed to come to equilibrium according to the
equation: CO (g) + H2 (g)  C (s) + H2O (g)
If the equilibrium constant at the temperature of
the experiment is 4.0, what is the equilibrium
concentration of water vapor?
CO (g) + H2 (g)  C (s) + H2O (g)
M
Initial
1.5
1.5
0.0
Change
−x
−x
+x
Equil.
1.5−x
1.5−x
x
Kc =
[H2O]eq
x
=
= 4.0
[CO]eq[H2]eq (1.5−x)(1.5−x)
x = 9 – 12x + 4x2
(PNO)2eq(PCl2)eq
(PNOCl)2eq
(43 − 2x)2(23 − x)
= 0.060
(675 + 2x)2
x=3
PNOCl = 681 torr, PNO = 37 torr,
PCl2 = 20 torr
9. Sulfuryl chloride decomposes at high temperatures
to produce sulfur dioxide and chlorine gases:
SO2Cl2 (g)  SO2 (g) + Cl2 (g)
At 375C, the equilibrium constant Kc is 0.045. If
there are 2.0 grams of sulfuryl chloride, 0.17 gram
of sulfur dioxide, and 0.19 gram of chlorine
present in a 1.0 L flask,
2.0 g  1 mol 
= 0.015 M
1.0 L 134.97 g
0.17 g  1 mol 
[SO2] =
= 0.0027 M
1.0 L 64.07 g
0.19 g  1 mol 
[Cl2] =
= 0.0027 M
1.0 L 70.90 g
[SO2Cl2] =
[H2O]eq = 1.0 M
8. Nitrosyl chloride NOCl decomposes to nitric oxide
and chlorine when heated:
2 NOCl (g)  2 NO (g) + Cl2 (g)
At 600K, the equilibrium constant Kp is 0.060. In
a vessel at 600K, there is a mixture of all three
gases. The partial pressure of NOCl is 675 torr,
the partial pressure of NO is 43 torr and the partial
pressure of chlorine is 23 torr.
a. What is the value of the reaction quotient?
2
(PNO) (PCl2) (43 torr) (23 torr)
=
(PNOCl)2
(675 torr)2
= 0.093
Qp =
Kp =
a. What is the value of the reaction quotient?
x = 1, or 2.25
2
2 NOCl (g)  2 NO (g) + Cl2 (g)
Q=
[SO2][Cl2] (0.0020)(0.0027)
=
[SO2Cl2]
(0.015)
= 4.8 × 10−4
b. Is the system at equilibrium?
No.
c. In which direction will the system move to
reach equilibrium?
Q < Kc, so reaction will shift to the right.
10. Ammonium chloride is placed inside a closed
vessel where it comes into equilibrium at 400C
according to the equation shown. Only these three
substances are present inside the vessel. If Kp for
the system at 400C is 0.640, what is the pressure
inside the vessel?
NH4Cl (s)  NH3 (g) + HCl (g)
12. When ammonia is dissolved in water, the
following equilibrium is established. If the
equilibrium constant is 1.8 × 10−5, calculate the
hydroxide ion concentration in the solution if
0.100 mol of ammonia is dissolved in sufficient
water to make 500 mL of solution.
NH3 (aq) + H2O (l)  NH4+ (aq) + OH− (aq)
atm
NH4Cl (s)  NH3 (g) + HCl (g)
M
Initial
0
0
Initial
0.200
0
0
Change
+x
+x
Change
−x
+x
+x
Equil.
x
x
Equil.
0.200−x
x
x
Kp = (PNH3)eq(PHCl)eq = (x)(x) = 0.640
Kc =
x = 0.80
Ptotal = PNH3 + PHCl = 0.80 atm + 0.80 atm
= 1.60 atm
11. Bromine and chlorine react to produce bromine
monochloride according to the equation. Kc =
36.0 under the conditions of the experiment.
Br2 (g) + Cl2 (g)  2 BrCl (g)
If 0.180 moles of bromine gas and 0.180 moles of
chlorine gas are introduced into a 3.0 Liter flask
and allowed to come to equilibrium, what is the
equilibrium concentration of the bromine
monochloride? How much BrCl is produced?
Br2 (g) + Cl2 (g)  2 BrCl (g)
Initial
0.060
0.060
0
Change
−x
−x
+2x
Equil.
0.060−x
0.060−x
2
Kc =
[NH4+]eq[OH−]eq
(x)(x)
=
= 1.8 × 10−5
[NH3]eq
0.200−x
x = 0.0019 M
PNH3 = PHCl = 0.80 atm
M
NH3 (aq) + H2O (l)  NH4+ (aq) + OH− (aq)
2
2x
[BrCl] eq
(2x)
=
= 36.0
[Br2]eq[Cl2]eq (0.060−x)(0.060−x)
2x
= 6.0
0.060 − x
x = 0.045 M
[BrCl]eq = 0.090 M, nBrCl = 0.27 mol
[OH−]eq = 0.0019 M
13. The following reaction is exothermic:
Ti (s) + 2 Cl2 (g)  TiCl4 (g)
List all the ways the yield of the product TiCl4
could be increased.

Add Cl2 (g)

Remove TiCl4 (g)

Increase pressure

Lower the temperature
AP Chemistry 2004B #1
N2 (g) + 3 H2 (g)  2 NH3 (g)
For the reaction represented above, the value of the equilibrium constant, Kp, is 3.1 × 10−4 at 700. K.
(a) Write the expression for the equilibrium constant, Kp, for the reaction.
Kp =
(PNH3)2eq
(PN2)eq(PH2)3eq
(b) Assume the initial partial pressures of the gases are as follows:
PN 2 =0.411 atm, PH 2 =0.903 atm, and PNH 3 =0.224 atm.
(i) Calculate the value of the reaction quotient, Q, at these initial conditions.
Qp =
(PNH3)2
(0.224)2
= 0.166
3=
(PN2)(PH2) (0.411)(0.903)3
(ii) Predict the direction in which the reaction will proceed at 700. K if the initial partial pressures are those
given above. Justify your answer.
The reaction will proceed in the reverse direction (to the left) because Q p < Kp.
(c) Calculate the value of the equilibrium constant, Kc, given that the value of Kp for the reaction at 700. K is
3.1 × 10−4.
Kc =
Kp
(3.1 × 10−4)
= 1.02
∆n =
(R·T)
((0.08206)(700 K))2−4
(d) The value of Kp for the reaction represented below is 8.3 × 10−3 at 700. K.
NH3 (g) + H2S (g)  NH4HS (g)
Calculate the value of Kp at 700. K for each of the reactions represented below.
(i) NH4HS (g)  NH3 (g) + H2S (g)
K′p =
(PNH3)(PH2S)
= (Kp)−1 = 120
(PNH4HS)
(ii) 2 H2S (g) + N2 (g) + 3 H2 (g)  2 NH4HS (g)
(PNH4HS)2
(PNH4HS)2
(PNH3)2
= (K′p)−2·(Kp)
2
3=
2
2·
(PH2S) (PN2)(PH2) (PNH3) (PH2S) (PN2)(PH2)3
= (8.3 × 10−3)−2(3.1 × 10−4) = 4.5
K″p =
AP Chemistry 1995 #1
H2 (g) + CO2 (g)  H2O (g) + CO (g)
When H2(g) is mixed with CO2(g) at 2,000 K, equilibrium is achieved according to the equation above. In one
experiment, the following equilibrium concentrations were measured.
[H2] = 0.20 mol/L
[CO2] = 0.30 mol/L
[H2O] = [CO] = 0.55 mol/L
(a) What is the mole fraction of CO(g) in the equilibrium mixture?
Assume 1 L
XCO =
0.55 mol
0.55 mol
=
= 0.45
(0.20 mol + 0.30 mol + 0.55 mol + 0.55 mol) 1.60 mol
(b) Using the equilibrium concentrations given above, calculate the value of Kc, the equilibrium constant for the
reaction.
Kc =
[H2O][CO] (0.55)(0.55)
=
= 5.0
[H2][CO2] (0.20)(0.30)
(c) Determine Kp, in terms of Kc for this system.
Kp = Kc(R·T)∆n = (5.0)((0.08206)(2000 K))2−2 = 5.0
(d) When the system is cooled from 2,000 K to a lower temperature, 30.0 percent of the CO(g) is converted back
to CO2(g). Calculate the value of Kc at this lower temperature.
H2 (g) + CO2 (g)  H2O (g) + CO (g)
M
Initial
0.20
Change +0.165
Equil.
Kc =
0.365
0.30
0.55
0.55
+0.165
−0.165
−0.165
0.465
0.385
0.385
[H2O][CO] (0.385)(0.385)
=
= 0.87
[H2][CO2] (0.365)(0.465)
(e) In a different experiment, 0.50 mole of H2(g) is mixed with 0.50 mole of CO2(g) in a 3.0-liter reaction vessel
at 2,000 K. Calculate the equilibrium concentration, in moles per liter, of CO(g) at this temperature.
H2 (g) + CO2 (g)  H2O (g) + CO (g)
M
Initial
0.17
0.17
0
0
Change
−x
−x
+x
+x
Equil.
0.17−x
0.17−x
x
x
Kc =
[H2O][CO]
(x)(x)
=
= 0.87
[H2][CO2] (0.17−x)(0.17−x)
x = 0.081
[CO] = 0.081 M