Name:_____________
(4 points)
Chemistry 114
Fourth Hour Exam
Remember- Show all work for partial credit
1. (10 points) Coal (C(s)) can be converted to methane in the following reaction:
C(s) + 2H2(g) 6 CH4(g)
ÄH rxn = -75 kJ
If I have a system with C(s), H2(g) and CH4(g) that is at equilibrium, what will happen
under the following conditions?
More C(s) is added to the system
More CH4 will form -or- CH4 will disappear -or- No change will be observed
More CH4 (g) is added to the system
More CH4 will form -or- CH4 will disappear -or- No change will be observed
The temperature of the system is increased
More CH4 will form -or- CH4 will disappear -or- No change will be observed
The volume of the system is decreased
More CH4 will form -or- CH4 will disappear -or- No change will be observed
Argon gas is added to the system
More CH4 will form -or- CH4 will disappear -or- No change will be observed
2. (16 points) Fill in the following table
[H+]
pH
[OH-]
pOH
.0056
2.25
1.79x10-2
11.75
1x10-7M
NaOH
5x10-8
7.3
2x10-7
(include water)
6.70
.00189
M
H2SO4
.00378
2.42
2.65x10-12
11.58
.0056M
HCl
__10_M
HNO3
10
-1
1x10-15
15
2
3.(12 points) What is the pH of 4.5 mM Ammonia (KB = 1.8x10-5)
B + H2O WBH+ + OHX = [OH-] = [BH+]
[B] = .0045 -X ..0045
KB = 1.8x10-5 = X2/.0045
1.8x10-5 x .0045 = X2
X=sqrt(1.8x10-5 x .0045)
= 2.8x10-4
pOH = -log (2.8x10-4) = 3.55
pH= 14-3.55 = 10.45
4. (12 points) In the following pairs of compounds, circle the compound that is most
acidic.
A.
HCl
or
CH3COOH
B.
NaOH
or
NH3
C.
HClO2
or
HClO4
D.
Sodium Acetate
E.
Chromium(VI) oxide
F.
Sulfur dioxide
or
Ammonium chloride
or
or
Chromium(VI) nitrate
Strontium oxide
3
5. (12 points) I am going to mix 25 mls of a .15M solution of benzoic acid (a weak acid,
KA =6.4x10-5) with 20 mls of a .10M solution of sodium benzoate (the sodium salt of this
weak acid). What is the pH of the resulting solution?
Mixture of acid and conjugate base = buffer, use Henderson-Hasselbach Equation
pH = pKa + log [A-]/[HA]
pKa = -log(6.4x10-5) = 4.18
[HA] =(.025 l x .15M) / .045 l = .0833
[A-] = (.02 l x .10M) /.045 l = .0444
pH = 4.18 + log (.0444/.0833)
= 3.91
6. (12 points) Define the following terms
A. Entropy- (S) a measure of randomness.
B. Spontaneous process - A process that will occur without outside intervention.
C. Second law of Thermodynamics - The entropy of the universe is always increasing.
D. ÄSsystem
The change in entropy of a system, S final - S Initial
E. Buffer - A solution that resists changes in pH when either acid or base is added.
F. Lewis acid - An electron acceptor or electrophile.
4
7. The chemical reaction for the burning of ethane is:
2 C2H6 (g) + 7 O2(g) 64 CO2 (g) + 6 H2O(g)
A. (4 points) Should this process have a + or - ÄS? Explain:
10 gas molecules as products, 9 gas molecules as reactants, the number of gas
molecules increase so the randomness of the system increases, so ÄS should be +.
B.(8 points) From the data table below, calculate the actual ÄS for this reaction.
S
(J/K@mol)
C2H6(g)
266.9
O2(g)
205
CO2(g)
214
H2O(l)
70
H2O(g)
189
ÄS = 3n Product Sproduct - 3n Reactant S Reactant
=[4(214)+6(189)] - [(2(266.9) + 7(205)]
= +21.2 J/K@mol
8. (12 points) Tin has two solid forms, called white tin and gray tin.
ÄH for the transition Tin(white) 6Tin(gray) is -2 kJ/mole
ÄS for the transition Tin(white) 6Tin(gray) is -8 J/mole
At what temperature are white tin and gray tin in equilibrium?
At equilibrium ÄG = 0 = ÄH -TÄS
0 = -2000 J/mol - X(-8) J/mol
2000 = 8X
X = 2000/8
X = 250 K = -23oC = -9.4o F
Note: White tin is the normal metallic tin seen at room temperature. If tin is exposed to
temperatures below -9.4 F it starts turning into gray tin that crumbles away. This is
called ‘tin disease’