Chemistry 202/212
Worksheet 10
Winter 2005
February 3
A(g) ↔ 2 B(g)
1. The following data are for the system:
Time (s)
PA (atm)
PB (atm
0
1.00
0.00
20
0.83
0.34
Oregon State University
40
0.72
0.56
60
0.65
0.70
80
0.62
0.76
100
0.62
0.76
(a) How long does it take the system to reach equilibrium?
About 80 s
(b) How does the rate of the forward reaction compare with the rate of the reverse
reaction after 30 s? after 90 s?
at 30 s: rate forward > rate reverse
at 90 s: rate forward = rate reverse
3 A(g) + 2 B(g) ↔ C(g)
2. Complete the table below for the reaction:
Time(s)
PA (atm)
PB (atm)
PC (atm)
0
2.450
1.500
0.000
10
2.000
1.200
0.150
20
1.625
0.986
0.275
30
1.325
0.750
0.375
40
1.100
0.600
0.450
50
0.950
0.500
0.500
3. Write equilibrium constant (K) expressions for the following reactions:
PIF2 5
K
(a) I2(g) + 5 F2(g) ↔ 2 IF5(g)
PI 2  PF52
(b) CO(g) + 2 H2(g) ↔ CH3OH(l)
K
(c) 2 H2S(g) + 3 O2(g) ↔ 2 H2O(l) + 2 SO2(g)
K
(d) SnO2(s) + 2 H2(g) ↔ Sn(s) + 2 H2O(l)
K
1
PCO  PH2 2
2
PSO
2
PH2 2S  PO3 2
1
PH2 2
60
0.950
0.500
0.500
4. Given the following data at a certain temperature,
2 N2(g) + O2(g) ↔ 2 N2O(g)
N2O4(g) ↔ 2 NO2(g)
½ N2(g) + O2(g) ↔ NO2(g)
K = 1.2∙10-35
K = 4.6∙10-3
K = 4.1∙10-9
calculate K for the reaction between one mole of dinitrogen oxide gas and oxygen gas to
give dinitrogen tetroxide gas.
½[2 N2O(g) ↔ 2 N2(g) + O2(g)]
2 NO2(g) ↔ N2O4(g)
2[½ N2(g) + O2(g) ↔ NO2(g)]
K = (1.2∙10-35) -½
K = (4.6∙10-3) -1
K = (4.1∙10-9)2
N2O(g) ↔ 1½ O2(g) + N2O4(g)
K = (1.2∙10-35) -½∙(4.6∙10-3) -1∙(4.1∙10-9)2
K = 1.1∙103
5. The reversible reaction between hydrogen chloride gas and one mole of oxygen gas
produces steam and chloride gas:
4 HCl(g) + O2(g) ↔ 2 Cl2(g) + 2 H2O(g)
K = 0.79
Predict the direction in which the system will move to reach equilibrium if one starts with
(a) PH2O = PHCl = PO2 = 0.20 atm
(b) PHCl = 0.30 atm, PH2O = 0.35 atm, PCl2 = 0.2 atm, PO2 = 0.15 atm
K
PH2 2O  PCl2 2
4
PHCl
 PO2
(a) Q 
 0.79
(0.2) 2 (0.0) 2
 0.0  0.79
(0.2) 4 (0.2) 2
Therefore reaction moves in forward direction.
(0.35) 2 (0.2) 2
 27  0.79
(0.3) 4 (0.15) 2
Therefore reaction moves in reverse direction.
(b) Q 
6. Solid ammonium carbamate, NH4CO2NH2, decomposes at 25ºC to ammonia and
carbon dioxide according to the following reaction and K value.
NH4CO2NH2(s) ↔ 2 NH3(g) + CO2(g)
K = 2.3∙10-4
In a sealed 10.0 L flask, 7.50 g of NH4CO2NH2(s) is allowed to decompose at 25ºC.
(a) What is the total pressure in the flask when equilibrium is established?
Assume that x moles of NH4CO2NH2(s) decomposes. That means 2x moles of
NH3(g) and x moles of CO2(g) are produced.
NH4CO2NH2(s) ↔ 2 NH3(g) + CO2(g)
-x
2x
2
K  PCO 2  PNH
= 2.3∙10-4
3
x
Since the partial pressure of a gas is proportional to the number of moles…
Px = pressure exerted by x moles of gas
K = (2Px)2∙Px = 4Px3 = 2.3∙10-4
Px = 0.039 atm
So, the total pressure of gas is 3 Px (2 Px + Px) or 3(0.039 atm)=0.116 atm.
(b) What percentage of NH4CO2NH2(s) decomposed?
Using PV=nRT or Px V=xRT we find that
x = 0.016 mol
Using a MW = 78 g/mol we find that 1.258 g or decomposed or 17%.
(c) Did this reaction take place relatively quickly or slowly?
The rate of the reaction cannot be determined from the equilibrium constant.
7. Consider the system
SO3(g) ↔ SO2(g) + ½ O2(g)
∆H = 98.9 kJ
(a) Predict whether the forward or reverse reaction will occur when the equilibrium is
disturbed by
i. adding oxygen gas
reverse
ii. decreasing the pressure of the system
forward
iii. adding argon gas
reverse (constant volume)
nothing (constant pressure)
iv. removing SO2(g)
forward
v. decreasing the temperature
reverse
(b) Which of the above factors will increase the value of K? Which will decrease it?
Decreasing the temperature will increase the value of K.
None of the factors will decrease the value of K.
8. For each of the following reactions, indicate the Brønsted-Lowry acids and bases.
What are the conjugate acid/base pairs?
(a) H3O+(aq) + CN-(aq) ↔ HCN(aq) + H2O
(b) HNO2(aq) + OH-(aq) ↔ NO2-(aq) + H20
(c) HCHO2(aq) + H2O ↔ CHO2-(aq) + H3O+(aq)
acid/base
base/acid
(a)
H3O+(aq)/HCN(aq)
CN-(aq)/H2O
(b)
HNO2(aq)/ NO2-(aq)
OH-(aq)/ H20
(c) HCHO2(aq)/ CHO2 (aq) H2O/ H3O+(aq)
9. Find the pH and pOH of solutions with the following [H+]. Classify each as acidic or
basic.
[H+]
pH pOH
(a) 6.0 M
-0.78 14.78 acidic
(b) 0.33 M
0.48
13.52 acidic
(c) 4.6∙10-8 M
7.34
6.66
basic
(d) 2.3∙10-14 M
13.64
0.36
basic
10. Calculate [H+] and [OH-] in solutions with the following pH.
pH
[H+]
[OH-]
-4
(a) 4.0
1∙10 M
1∙10-10 M
(b) 8.52
3.0∙10-9 M
3.3∙10-6 M
(c) 0.00
1.0 M
1.0∙10-14 M
(d) 12.60
2.5∙10-13 M
4.0∙10-2 M
11. What is the pH of a solution obtained by adding 13.0 g of NaOH to 795 mL of a
0.200 M solution of Sr(OH)2? Assume no volume change after NaOH is added.
mol OH - in original solution 
0.200 mol Sr(OH) 2
2 mol OH 0.795 L


 0.318 mol OH L
1 mol Sr(OH) 2
mol OH - added from NaOH 
13.0 g NaOH mol NaOH 1 mol OH 

 0.325 mol OH 40 g NaOH 1 mole NaOH
(0.318 mol OH -  0.325 mol OH - )
[OH] 
 0.809 M
0.795 L
-
 110 -14 
 110 -14 
  - log 
  13.91
pH  - log[H ]  - log 
- 
 [OH ] 
 [0.809] 
