Unformatted text preview: Part 1: Multiple Choice.
(4 pts each, 44 pts total)
Instructions: Bubble in the correct answer on your Scantron™ form AND circle the answer on your exam.
Each question has one correct answer.
1.) The answer to question 1 is A. Bubble in A on your Scantron™ form. 2.) Consider the sublimation of dry ice:
If K1 is the equilibrium constant at 300 K, and K2 is the equilibrium constant at 400 K, which of the
following inequalities must be true?
A.) K1 = K2 3.) B.) K1 = K2-1 C.) K1K2 = 0 D.) K1 > K2 E.) K1 < K2 For the vaporization of methanol
-1 and ∆Sû = 112.9 J K-1 mol-1. What is the boiling point of methanol at sea level?
∆Hû = 38.0 kJ mol
Assume ∆Hû and ∆Sû are independent of T.
A.) 64 K 4.) B.) 237 K C.) 273 K D.) 337 K E.) 373 K Consider the reaction:
N2 (g) + 3H2 (g)
at equilibrium. What would be the reaction quotient immediately following the reduction of volume by
two at constant temperature before any reaction occurs?
A.) Q = 5.) 1
4 B.) Q = 1
2 C.) Q = K D.) Q = 2K E.) Q = 4K One mole of an ideal gas expands isothermally against a constant pressure of 1 atmosphere. Which of
the following inequalities is true?
A.) ∆P > 0 B.) q > 0 C.) ∆S < 0 D.) ∆V < 0 E.) ∆T < 0 Page 3 of 6 Name:_________________________________________ 6.)
One mole of an ideal gas expands adiabatically against a constant pressure of 1 atmosphere. Which of
the following inequalities is true?
A.) ∆P > 0 C.) ∆S < 0 B.) q > 0 D.) ∆V < 0 At what temperature does K = 1, ∆Gû = 0 for the reaction H2O (l) 7.) A.) –273 °C 8.) B.) 0 °C C.) 100 °C E.) ∆T < 0 H2O (g)?
D.) 273 °C E.) 373 °C How many different ways can you distribute six distinguishable stones between two boxes with five in
the first box and one in the second box?
A.) 1 9.) B.) 3 C.) 6 D.) 9 E.) 15 The caloric content of 10 little cookies can heat up 10 kg of water by 10 °C. What would be the change
in temperature if 1 little cookie was used to heat up 1 kg of water?
A.) 0.1 °C B.) 1.0 °C C.) 10 °C E.) 1000 °C D.) 100 °C For each of the problems 10-12, select the graph that best describes the behavior listed. A.) B.) 0 10.) C.)
0 PN2O4 as a function of (PNO2)2 for N2O4 (g) E.)
0 0 2NO2 (g), at constant T.
C 11.) ln(K) as a function of 1
for a the combustion of liquid methanol (CH3OH).
A or B 12.) ∆Gû as a function of T for the vaporization of water, H2O (l)
E H2O (g). Page 4 of 6 Name:_________________________________________ Part 2: Short Answer Problems (101 pts total)
Instructions: Enter answers in the boxes provided. Show your work. Explain your answer when requested in
15 words or less.
SO2Cl2 (g) SO2 (g) + Cl2 (g) is endothermic with ∆H = 4.6 kJ/mol. SO2Cl2 (g) and SO2 (g) are placed in a bulb at a fixed temperature
with partial pressures of 3.0 atm each.
a) Write the expression for reaction quotient (Q) and calculate its value before any reaction occurs.
Q = PSO2 × PCl2 / PSO2Cl2
Q=0 b) After equilibrium is reached in the bulb at the same fixed temperature, the partial pressure of Cl2 (PCl2) is
found to be 1.0 atm. What are the partial pressures of SO2Cl2 and SO2 (PSO2Cl2 and PSO2)?
Final c) SO2Cl2
+x C l2
1 (x=1) Answers:
PSO2Cl2 = 2 atm
PSO2 = 4 atm Calculate the value of the equilibrium constant for the reaction in part b).
K = PSO2 × PCl2 / PSO2Cl2
= 4×1/2 = 2
K=2 d) If some Cl2 is added to the equilibrium mixture, will the pressure of SO2 increase, decrease, or stay
constant as the system approaches the new equilibrium state? Circle the answer and explain.
Decrease Same Increase Explanation: When more product is
added, reaction goes backwards,
producing reactants and eliminating
products like SO2. Page 5 of 6 Name:_________________________________________ (18 pts)
Consider the reaction of silver chloride (AgCl) dissolving in water.
+ a) - AgCl (s)
Ag (aq) + Cl (aq)
Dissolving 14.3 g of AgCl (s) consumes 6.5 kJ of heat. What is the temperature change if 14.3 g of
AgCl(s) totally dissolves in 1.00 L of water initially at 20 °C?
q = m cp ∆T
∆T = q / m cp
= (–6500 J) / (1000 g) (4.184 J g-1 K-1)
= –1.6 K b) Answer:
–1.6 K In actuality the equilibrium constant (K) for this reaction is very small (1.6 x 10-10 at 25.0 °C). How will
this affect the temperature change predicted in part (a)? Explain.
Answer: K is very small, less product is
formaed, the temperature change is
smaller than in (a) (23 pts)
A study of the geology of the earth shows that rocks older than 2 billion years contain iron in the form of
FeS2. In rocks less than 2 billion years old, iron appears mostly as the oxide Fe2O3 (hematite).
4 FeS2 (s) + 8 H2O (l) + 15 O2 (g)
a) Calculate ∆H° for the above reaction.
∆Hº = [2 HºFe2O2 + 8 HºH2SO4]
– [4 HºFeS2 + 8 HºH2O + 15 HºO2]
= 2(–824.2) + 8(–814.0)
– [4(–178.2) + 8(–285.8) + 15(0)] kJ
= –5161.2 kJ b) Answer:
∆Hºreaction = –5161.2 kJ Calculate ∆S° for the above reaction.
∆Sº = [2 SºFe2O2 + 8 SºH2SO4]
– [4 SºFeS2 + 8 SºH2O + 15 SºO2]
= 2(87.4) + 8(156.9)
– [4(52.93) + 8(69.91) + 15(205.03)] J/K
= –2416.45 J/K c) 2 Fe2O3 (s) + 8 H2SO4 (l) Answer:
∆Sºreaction = –2416.45 J/K Over what temperature range is this reaction spontaneous? As always, show your calculations.
0 > ∆Gº = ∆Hº – T∆Sº < 0
T < ∆Hº / ∆Sº (for ∆Sº < 0)
T < (-5161.2 × 103 J) / (–2416 J/K) = 2136 K Answer:
T < 2136 K Page 6 of 6 Name:_________________________________________ (23 pts)
Consider the hydrogenation of formaldehyde (H2C=O) to form methanol (CH3OH).
CH2O + H2 H-H
O=O Average Bond Energy
C ≡C H–C=O
H a) CH3OH + H-H ---> ∆Hº = 743 + 436 – (413 + 360 + 463) = –57 kJ Answer:
∆Hº = –57 kJ The formation of which species, formaldehyde or methanol, is more exothermic (i.e. has the lower
∆Hºf (CH3OH) = 2 C–C + ½ O=O + 2 H–H
– [ 3 C–H + 1 C–O + 1 O–H ] = –246 kJ
∆Hºf (CH2O) = 2 C–C + ½ O=O + H-H
– [ 2 C–H + 1 C=O ] = –188 kJ c) H
H Estimate ∆Ho for this reaction.
bonds broken: 2 C–H + 1 C=O + 1 H–H
bonds made: 3 C–H + 1 C–O + 1 O–H b) 360
methanol (CH3OH) The combustion of which species, formaldehyde or methanol, produces more heat per mole?
CH2O + O2 ! CO2 + H2O
∆H = 2 C–H + 1 C=O + 1 O=O
– [ 2 C=O + 2 O-H ] = –346 kJ
CH3OH + 3/2 O2 ! CO2 + 2 H2O
∆H = 3 C–H + 1 C–O + 1 O–H + 3/2 O=O
– [ 2 C=O + 4 O-H ] = –530 kJ Answer:
methanol (CH3OH) ...
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