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NAME: KEY CHEMISTRY 418, Fall, 2010(10F) Section Number: 10 Midterm Examination #1, October 7, 2010 Answer each question in the space provided; use back of page if extra space is needed. Answer questions so the grader can READILY understand your work; only work on the exam sheet will be considered. Write answers, where appropriate, with reasonable numbers of significant figures. You may use only the handbook of “Essential Data and Equations for a Course in Physical Chemistry”, a calculator, and a straight edge.
1. (20 points) An experimental setup has two compartments (V 1 = 1 L, V 2 = 3 L), connected by a valve. (^) DO NOT WRITE IN THIS SPACE p. 1________/ p. 2________/ p. 3________/ p. 4________/ p. 5________/ ============= p. 6_______/ (Extra credit) ============= TOTAL PTS
/
Initially, the valve is closed and 50 g of Ar gas is placed at room temperature into compartment one, 100 g of Xe gas is placed into compartment two. Use ideal gas approximation to determine the total pressure in the setup once the valve is open and the system attains equilibrium at room temperature.
Answer: For each gas: PV = nRT
mol g mol
g M Xe
mol n Xe g g mol
g M Ar
n Ar g 0. 762
- 293 /
1. 252 ; ( )^100
( )^50
V = V 1 + V 2 = 4 L
At room temperature:
Pa m
J
L
L^ m
mol J mol K K V
P Ar n ArRT 775907. 7 775907. 7 4 0. 001
( ) ( )^1.^2528.^3144 / /^298.^15
3 3
×
= = × ×
Pa m
J
L
L^ m
mol J mol K K V
P Xe n XeRT 472237. 8 472237. 8 4 0. 001
( ) ( )^0.^7628.^3144 / /^298.^15
3 3
×
= = × ×
Ptotal = P 1 ( Ar )+ P 2 ( Xe )= 775907. 7 Pa + 472237. 8 Pa = 1248145. 5 Pa = 12. 318 atm
Although this is a pretty high pressure, the gases used are very close to being perfect so ideal gas law is readily applicable. Also, since the ideal gas approximation is used, there is no need to account for possible interactions between the molecules of different kind.
2. (20 points) a) (4 points). State the Zeroth Law of thermodynamics
Answer: Two systems that are separately in thermal equilibrium with a third system are also in thermal equilibrium with one another
b) (4 points). Explain in a sentence or two the improvements of the van der Waals equation of state compared to the ideal gas equation of state
Answer: van der Waals equation of state introduces the size of the molecules, which accounts for repulsive interactions, and describes the attractive forces (constant a). This combination allows for a better prediction of gas behavior at high pressures and low temperatures and theoretically predicts the existence of a liquid phase (albeit more qualitatively than quantitatively describes it).
c) (4 points). Explain in one or two sentences the difference between reversible and irreversible processes
Answer: If the direction of a process can be changed by an infinitesimal force, this process is reversible (also a process that can be described as moving between equilibrium states). If an infinitesimal force does not change the direction of a process, such process is called irreversible (also spontaneous).
d) (4 points). State the Third Law of thermodynamics
Answer: The entropy of a pure, perfectly crystalline substance is zero at 0 K.
e) (4 points). Explain why reaction enthalpies depend on temperature. You can use formulas or graphics if it would help your explanation.
Answer: According to the equation describing temperature-dependence of enthalpy of a reaction on page 5-10 of the Blue Book, the value of enthalpy depends on the heat capacities of reactants and products. These values can obviously have different temperature dependences, as illustrated by Shomate equation, which means that the enthalpy change for a reaction generally depends on temperature.
4. (20 points). Calculate absolute molar entropy of diamond at 500°C and 1 bar. Use any information available in the Blue Book and assume that the heat capacity of diamond can be approximated as a constant within the temperature interval studied.
At constant pressure:
mol K
J
K
K
mol K
J
mol K
dT J T
C Cdiamond S S
K
K
pm
m Θ^ K =^ m Θ K + ∫ = × + × 298. 15 =^8.^2 ×
(^15) ( , ) 2. 4 6. 1 ln 773. 15
15
, , 773. 15 , 298. 15
Although the number is very small, you can see that it more than triples at this high temperature. This means that the solid diamond is very well ordered at room temperature but its disorder should affect its properties significantly at 500°C.
5. (20 points). Use Hess’ Law and any enthalpies of formation available in Table 5.8 to calculate the standard enthalpy ( Δ H (^) rzn^ Θ) for a reaction of partial oxidation of sucrose: C 12 (^) H 22 O 11 ( s )+ 6 O 2 ( g )→ 2 C 2 H 5 OH ( l )+ 8 CO 2 ( aq )+ 5 H 2 O ( l )
where CO 2 ( aq ) is CO 2 dissolved in water (see below) The following reactions can be used:
- C 12 (^) H 22 O 11 ( s )+ 12 O 2 ( g )→ 12 CO 2 ( g )+ 11 H 2 O ( l ),Δ HI^ =− 5639. 7 kJ ( 1 )This is a combustion reaction
- (^) CO 2 (^) ( g )→ CO 2 ( aq ),Δ HII^ =− 19. 4 kJ ( 2 )Dissolution of CO 2 in water.
Answer:
Δ H (^) rzn Θ^ =Δ HI + 8 ×Δ HII − 4 Δ Hf Θ( CO 2 ( g ))+ 2 Δ Hf Θ( C 2 O 5 OH ( l ))− 6 Δ H^ Θ f ( H 2 O ( l ))
mol
kJ mol
kJ mol
H kJ mol kJ mol kJ Δ (^) rzn =− + ×− − × − + × − − ×− Θ
The stoichiometric coefficients have the units of mol but as long as the reaction used is balanced, the units of the enthalpy of the reaction will be acceptable as either kJ or as kJ/mol for a full credit.
mol
kJ mol
kJ mol
kJ mol
H kJ mol kJ mol kJ Δ (^) rzn Θ =− 5639. 7 / − 155. 2 / + 1574 − 555. 2 + 1714. 8 =− 3061. 3
