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CHM 15 EXAM
PRACTICE
Fill in the blanks or circle one answer as appropriate (2.5 points per question).
1. The equation which relates expansion work (w) done by a system to the change in the number of moles of gas in a reaction is:
w = - ∆ ngRT
2. For a reaction which is endothermic, the final enthalpy of the system (Hf) is > the initial enthalpy (Hi) of the system.
a. > b. < c. = d. ≥ e. ≤
3. The sign of q (positive or negative) for heat lost by the system is negative.
The sign of w (positive or negative) for work done on the system is positive
4. What is the change in the energy of a system that has 89 J of work done on it while losing 100 J of heat.
a. - 189 J b. + 189 J c. - 11 J d. + 11 J e. none of the above
5. What is the formation reaction for C8H18 (l)? [C(s) is the stable form of carbon.]
a. 8 C (s) + 9 H 2 (g) Æ C 8 H 18 (g) b. 2 C 4 H 8 (s) + H 2 (g) Æ C 8 H 18 (l) c. 2 C 4 (s) + 9 H 2 (g) Æ C 8 H 18 (l) d. 8 C (g) + 9 H 2 (g) Æ C 8 H 18 (l) e. none of the above
6. Under which of the following conditions will a gas behave least ideally?.
a. low pressure and high temperature b. high pressure and high temperature c. low pressure and low temperature d. high pressure and low temperature
7. A gas is compressed to 1/3 of its initial volume while the temperature doubles. If moles of gas remains constant (& the gas behaves ideally), then the pressure will:
a. increase by a factor of 2 b. increase by a factor of 3 c. increase by a factor of 6 d. decrease by a factor of 2 e. decrease by a factor of 3 f. decrease by a factor of 6
8. A gas is heated to four times its initial temperature in a sealed container. The average velocity of the gas will:
a. decrease to ¼ of its initial value b. increase to 2 times its initial value c. decrease to ½ of its initial value d. increase to 4 times its initial value e. remain unchanged
Complete the following calculations (Show all of your work!)
(12 points)
9. 0.6549 gram sample of liquid octane, C8H18 (l), a component of gasoline, undergoes complete combustion in a bomb calorimeter that has a heat capacity of 11.93 kJ/ºC. If the initial temperature of the calorimeter is 20.25ºC and the final temperature of the calorimeter is 22.87ºC, what is the heat of combustion at constant volume (qv or ∆E) of octane in the units kJ /mol? [the molar mass of octane = 114.23 g/mole]
∆T = 22.87ºC – 20.25ºC = +2.62ºC
0 = qcal + qcomb... so... ∆E = qcomb = - qcal = - (11.93 kJ/ºC)( +2.62ºC) = - 31.2 6 kJ
moles consumed = 0.6549 g /(114.23 gmol-1) = 5.733x10-3^ mole
∆ E = -31.2 6 kJ / 5.733x10-3^ mole = -5,45 2 kJ/mole
(15 points)
10. (a) Given the values for ∆Hº f for the reactants and products, calculate ∆Hºrxn
for the following reaction in the units specified:
2 C8H18 (l) + 25 O2 (g) Æ 16 CO2 (g) + 18 H2O (l) ∆Hºrxn = ?? kJ /mole C8H
∆Hº f [C8H18 (l)] = -250.1 kJ / mol ∆Hº f [CO2 (g)] = - 393.5 kJ / mol ∆Hº f [H2O (l)] = - 285.8 kJ / mol
∆H = Σ m ∆Hf (prod) - Σ n ∆H (^) f (react) Simply divide the reaction two to find the energy change per mole of octane ∆H = 8mole(-383.5 kJ/mol) + 9mole(- 285.8 kJ/mol) – 1mole*(-250.1 kJ/mol) ∆ H = -5,390.1 kJ / mole octane
Alternatively, do the calculation for the moles as shown and then divide by two.
(b) Explain why the answers to questions 9 and 10(a) do not, and should not, match one another.
The reaction at constant volume releases different energy per mole of octane due to the absence of the work term present at constant pressure.
(25 points)
13. (a) Balance the following reaction ( DO NOT add any other species to the reaction).
4 H 2 O (g) + 3 Fe (s) → Fe 3 O 4 (s) + 4 H 2 (g)
If 40.00 g of Fe is reacted with 75.0 g of H 2 O:
(b) What is the limiting reactant? (40.00 g Fe) / (55.85 g/mole) = 0.7162 mole Fe
(75.0 g H 2 O) / (18.02 g/mole) = 4.162 mole H 2 O
Not even a close call... 0.7162 / 3 < 4.162 / 4... so Fe is limiting!
(c) What is the theoretical yield of H 2 in BOTH moles AND in grams?
mole H 2 = (4 mole H 2 / 3 mole Fe)(0.7162 mole Fe) = 0.9549 mole H 2*
mass H 2 = (2.016 g H 2 / 1 mole)(0.9549 mole H 2 ) = 1.925 g H 2*
(d) What volume would the DRY H 2 occupy at 298 K and 0.932 atm?
V = nRT/P = (0.9549 mole H 2 )(R)(298 K)/(0.932 atm)
V = 25.0 7 L
(e) If the actual amount of DRY H 2 collected at 298 K and 0.932 atm is 3.01 L, what is the percent yield?
% Yield = 100(actual / theoretical) = 100(3.01L/25.07 L) = 12.0 %
(f) Is this reaction ENDOthermi or EXOthermic? Provide an answer and discuss why you believe you are correct.
EXOthermic. Heat is required to convert iron ore (which is much like rust) into the
metallic iron used to produce steel. Hydrogen gas can be used as a reducing agent in the
process. Therefore, the REVERSE of the above reaction is ENDOthermic. As we learned
in this chapter, if we reverse a reaction, the sign of the enthalpy change must also change.
The actual process of making steel from iron ore is more involved than simply adding
heat, but that needn’t be addressed here.
BONUS !!!!!! (for up to 5 points complete the following).
Avogadro’s number is huge. To try to understand how big this number really is consider
the following:
A basketball has a diameter of 9.4 inches. The Earth has a surface area of 196,935,000 sq
miles. How many basketballs are required to cover the Earth with a single layer of
basketballs?
(note: if your calculator does not have a key for π, try using 22 / 7 )
The projected area taken up by a sphere is equal to the cross-section of the sphere at its
largest point.
r = 9.4 inches / 2 = (1 mi / 1,760 yd)(1 yd / 36 inch)(9.4 in / 2) = 7.42x10-5**^ mi
Area of that circle = π r^2 = π (7.42x10-5^ mi)^2 = 1.73x10-8^ mi^2 /BB
BB needed to cover the Earth = area of Earth/Area per BB =
# BB = 196,935,000 sq miles / 1.73x10-8^ mi^2 /BB = 1.14x10^16 BB
[Note:] this is actually a larger # than we would need because we have not accounted for
the less than 100% packing efficiency of the BB (i.e. there will be some empty spaces
between parts of the BB)
What amount in mole(s) does the number of basketballs represent?
mole BB = (1 mole BB / 6.022x10^23 BB)(1.14x10*^16 BB) = 1.89x10-8^ mole
What mass of aluminum contains a number of aluminum atoms equal to the number of
basketballs found above?
mass Al = (27.0 g Al / 1 mole)( 1.89x10-8*^ mole) = 5.11x10-7^ g Al or 0.511 μ g Al
WOW!
Avogadro’s number is REALLY enormous... and atoms are REALLY tiny!!!!
