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The Richard Stockton College of New Jersey
Chemistry Program, School of Natural Sciences and Mathematics PO Box 195, Pomoma, NJ
CHEM 3410: Physical Chemistry I — Fall 2008
Exam 2 — Model Solutions
November 7, 2008
12:45–2 PM
Name: Read all of the following information before starting the exam:
- This is a closed book exam. You are permitted an aid sheet consisting of two sides of a 8.5” x 11” piece of paper. Your aid sheet must be turned in with your exam.
- Show all work, clearly and in order, if you want to get full credit. I reserve the right to take off points if I cannot see how you arrived at your answer (even if your final answer is correct).
- Please keep your written answers brief; be clear and to the point. I will take points off for rambling and for incorrect or irrelevant statements.
- Justify your answers. Clearly state any assumptions you make.
- Circle or otherwise indicate your final answers.
- You have 75 minutes to complete the exam. There are a total of 67 points on the exam, so budget your time accordingly.
- For problems involving calculations, set up your calculations first and then do the computa- tion if time permits.
- Be sure to read all the questions first. You do not have to complete the problems in any particular order.
- Good luck!
Use of wireless communication devices at any time
during the exam is strictly prohibited.
Question Score Total
1 16 2 18 3 17 4 16
Total 67
- ( 16 points) Consider two possible processes for the vaporization of one mole of H 2 O at 95◦C and 1 atm. A. The H 2 O is the only substance in the container which is kept at constant pressure and temperature giving the process:
H 2 O (l, 95 ◦C, 1 atm) → H 2 O (g, 95 ◦C, 1 atm)
B. One mole of He is present in the container along with one mole of H 2 O. The container is kept at constant pressure and temperature. The H 2 O(l) and He(g) are initially not mixed since one is a liquid and the other is a gas. But after vaporizing, the H 2 O(g) then mixes with the He(g) in the container, giving the total process:
H 2 O (l, 95 ◦C, 1 atm) + He (g, 95 ◦C, 1 atm) → (H 2 O + He)mixed (g, 95 ◦C, 1 atm)
You can assume that He(g) and H 2 O(g) are ideal gases. The following data may be useful:
The equilibrium boiling temperature of water is 100◦C at 1 atm. ∆Hvap = 40, 525 J/mol ∆Svap = 108.6 J/molK Assume that both the enthalpy of vaporization and the entropy of vaporization are independent of temperature in the range of 95◦C to 100◦C.
(a) Find ∆G for process A. Is this process spontaneous? ( 4 ) We are told to assume that the enthaply and entropy of vaporization are independent of temperature. Therefore, for process A we can find ∆G in the following manner:
∆GA = ∆Hvap − T ∆Svap = 40525 J − (95 + 273 K)(108. 6 J/K)
∆GA = 560. 2 J and the process is not spontaneous (∆G > 0 ) (b) What are ∆Smixing and ∆Gmixing for mixing one mole of He(g) with one mole H 2 O(g) at a constant pressure of 1 atm and a constant temperature of 95◦C? ( 4 ) We mixing one mole of water gas and one mole of He gas. Therefore the mole fraction of each gas (X) is equal to 0.5.
XHe = 1 mole He 2 total moles gas
The free energy of mixing is:
∆Gmix = RT (XHe ln XHe+XH 2 O ln XH 2 O) = (8. 314 J/molK)(368 K)(.5 ln .5+.5 ln .5)
∆Gmix = − 2121 J/mol Since ∆Hmix = 0 for an ideal gas:
∆Smix = −∆Gmix T
2121 J/mol 368 K = +5. 76 J/molK
Both this quantities make sense since the mixing of two ideal gases is always spontaneous (∆Gmix < 0 and ∆Smix > 0 ). ∆Gmix = − 2121 J/mol and ∆Smix = 5. 76 J/molK
- ( 18 points) The gas phase dissociation of molecule A is given by:
A(g) ⇀↽ B(g) + C(g)
Initially pure A is placed in a container.
(a) At 80◦C and 1 bar, the fraction α of “A” that is dissociated is 0.4. (If n moles of A are initially placed in the container, then αn moles will have dissociated leaving n − αn moles of A at equilibrium). Calculate the equilibrium constant Kp and the standard free energy change ∆G◦^ for the dissociation of A at 80◦C. ( 6 ) We can set up the following table to keep track of the equilibrium amounts of each substance A (g) ⇀↽ B (g) + C (g)
Initial number of moles n 0 0
Moles at equilibrium n − αn αn αn
Total number of moles n − αn + αn + αn = n + αn at equilibrium
Mole fraction at equilibrium nn−+αnαn nαn+αn n+αnαn
Partial pressure at equilibrium
( n−αn n+αn
) P
( αn n+αn
) P
( αn n+αn
) P (Pi = XiP )
We are told the total pressure (P = 1 bar) and that α = 0. 4. We can use this information along with the equilibrium expression to find Kp.
Kp =
PB PC
PA
( αn n+αn
) P
( αn n+αn
) P ( n−αn n+αn
) P
= P
α^2 n^2 (n + αn)(n − αn)
Kp = P α^2 n^2 n^2 (1 + α)(1 − α)
= P
α^2 1 − α^2 Inserting P = 1 bar and α = 0. 4 we get Kp = 0. 190
We could have also used the relationship: Kp = P ∆ν^ Kx where ∆ν is the difference in the stoichiometric coefficients (products - reactants) and is equal to 1 in this case, Kx is the equilibrium constant written in terms of mole fraction and Kp is the equilibrium constant in terms of partial pressures.
We can now solve for ∆G◦ rxn using the relationship between ∆G and Kp along with the equilibrium constant solve for above and the temperature:
ln Kp = −∆G◦ rxn RT ∆G◦ rxn = +4874 J/mole
(b) Calculate the pressure at which the fraction α that is dissociated is 0.9 at 80◦C. ( 6 )
We can use the expression for Kp we solved for above to find P if α = 0. 9. We should expect that the pressure required should decrease since if α increases, more products are formed. The product side of the reaction contains more moles of gas which would be favored only if the pressure was reduced.
Kp = P α^2 1 − α^2
0 .190 = P
P = 0. 045 bar
(c) At 320◦C, Kp = 1.076. Calculate the standard heat of reaction ∆H◦^ for the dissociation of “A” if ∆H◦^ is assumed to be independent of temperature. [If you were unable to do part (a), then use (incorrectly) Kp(80◦C) = 0.30]. ( 6 ) To find the enthalpy of the reaction we can employ the van’t Hoff Equation since we know Kp at two temperatures. Since the equilibrium constant increases with temperature, we should expect the reaction to be endothermic.
ln Kp(T 2 ) = ln Kp(T 1 ) +
∆H rxn◦ R
( 1 T 1
T 2
)
Inserting values for Kp for each temperature T 1 and T 2 we find: ∆H rxn◦ = 12. 6 kJ/mol
(c) If P is decreased from P = 4.5 kbar to P = 4.0 kbar, how will the temperatures of the horizontal regions of the T vs. t plot change? What do these changes tell you about the sign of ∆V for the II → V and V → liquid phase transitions? ( 4 ) According to the phase diagram, if we reduce the pressure to 4 kbar, T 1 will increase and T 2 will decrease.
We know that changes of P and T along the coexistence curve are governed by the Clapeyron equation: (^) ( dP dT
)
coexist
∆H
T ∆V
For both transitions, dP < 0 and ∆H > 0. Therefore, for the II → V transition since dT > 0 , ∆V must be negative to make dPdT negative. For the V → liquid, ∆V must be positive since dT < 0 and dPdT > 0.
This should also make sense by looking at the slopes of the coexistence curves. For II → V the coexistence curve has a positive slope, while for V → liquid it has a negative slope.
(d) Sketch a similar heating curve for a pressure of 2.5 kbar. Indicate any relevant trans- formation temperatures on the sketch. ( 4 ) The heating curve would have the same form as the one given at the beginning of the problem. However, at 2.5 kbar we cross different two phase equilibria: II → III at approximately -30◦C and III → L at around -20◦C.
(e) Why do the non-horizontal portions of the heating curve have different slopes? ( 3 ) The non-horizontal regions indicate a single phase is present. The slopes are difference since each phase has a different heat capacity. The heating rate is constant, so each phase will require a different amount of heat transfered to raise the temperature. Hence the slopes are different.
- ( 16 points) Please identify each of the following statements as true or false. Give a brief justification for each answer.
(a) The mixing of ideal gases is spontaneous at high temperatures but non-spontaenous at low temperatures. ( 4 ) FALSE. The mixing of ideal gases is spontaneous at any temperature since entropy is the only driving force for mixing.
(b) The entropy of a system is always maximized at equilibrium. ( 4 ) FALSE. This statement is only true in the case of an isolated system. In general, the entropy of the universe is maximized at equilibrium.
(c) The chemical potential of 12 grams of ice at 0◦C and 1 atm is equal to the chemical potential of 12 grams of liquid water at 0◦C and 1 atm. ( 4 ) TRUE. At 0◦C and 1 atm, liquid water and ice are in equilibrium (at the melting point). Therefore the chemical potential of the two phases are equal.
(d) Increasing the pressure on a gas-phase reaction will always shift equilibrium towards the products. ( 4 ) FALSE. This is not true in general. The impact of changes in pressure on equilibrium depends on the particular gas stoichiometry of the reaction of interest. We demonstrated in class that as you increase the pressure on a gas-phase reaction the equilibrium will shift towards this side with less moles of gas.
