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The Richard Stockton College of New Jersey
Chemistry Program, Division of Natural Sciences and Mathematics PO Box 195, Pomoma, NJ
CHEM 3410: Physical Chemistry I — Fall 2007
Exam 2
Model Solutions
November 16, 2006
8:30–9:45 AM
Name: Key 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.
- You have 75 minutes to complete the exam.
- 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 15
2 15
3 20
4 20
Total 70
- ( 15 points) For the isomeric conversion reaction n-butane ⇀↽ iso-butane, the equilibrium constant Kp is 0.686 at 450 K and 0.858 at 500 K.
(a) Without doing any computations, do you expect this reaction to be exothermic or endothermic? Briefly explain your answer. ( 8 ) From the data, it appears the Kp increases as the temperature increases. This indicates that the reaction is endothermic. An increase in temperature shifts the reaction towards the products.
(b) Use this data to calculate the enthalpy of this reaction. State any assumptions you make to solve this problem. ( 7 ) We have the equilibrium constant at two temperatures. Therefore we can apply the van’t Hoff equation to determine the enthalpy of the reaction. This relationship can only be applied if we assume that the enthalpy is independent of temperature in this temperature range.
ln Kp(T 2 ) = ln Kp(T 1 ) −
∆H
R
( 1 T 2
T 1
)
Rearranging things to solve for ∆H:
∆H =
ln Kp(T 1 ) − ln Kp(T 2 ) ( 1 T 2 −^
1 T 1
) R
∆H =
ln(0.686) − ln(0.858) ( 1 500 −^
1 450
) (^) (8. 314 J/molK)
∆H = +8370 J/mol
The sign for ∆H is consistent with the prediction from part (a).
- ( 20 points) Consider the reaction:
CO(g) + 2 H 2 (g) ⇀↽ CH 3 OH(g)
You are given the following data:
∆H rxn◦ = − 90 .2 kJ/mol ∆S rxn◦ = − 219 .1 J/mol·K
(a) Find ∆G◦ rxn and Kp at 500 K for the reaction. You may assume that both ∆H rxn◦ and ∆S◦ rxn are independent of T. ( 5 ) Since we are told that ∆H rxn◦ and ∆S rxn◦ are independent of temperature, we can solve for ∆G◦ rxn : ∆G◦ rxn = ∆H rxn◦ − T ∆S rxn◦
∆G◦ rxn = − 90. 2 × 103 J/mol − (500 K)(− 219. 1 J/molK)
∆G◦ rxn = +19350 J/mol
ln Kp =
−∆G◦ rxn RT
→ Kp = e
−∆G◦ rxn RT (^) = e
− 19350 (500)(8.314)
Kp = 0. 00952
(b) Write an expression of Kp in terms of the partial pressures of the species involved in the reaction. Feel free to omit the reference pressure for convenience. ( 3 )
Kp =
PCH 3 OH
PCOP H^22
(c) Starting from your expression in part (b), write Kp in terms of the mole fraction of each species and the total pressure, Ptot. You may use a relationship we derived in class or the following relationship between the partial pressure (Pi) and mole fraction (Xi) of a species may be useful:
Pi = XiPtot
Using the relationship derived in class, with ∆ν = − 2 ( 1 − 3 = − 2 ): ( 3 )
Kp = P (^) tot∆ν Kx = P (^) tot−^2
XCH 3 OH
XCOX H^22
or using the result for part (b):
Kp =
PCH 3 OH
PCOP H^22
(XCH 3 OH Ptot) (XCOPtot)(XH 2 Ptot)^2
= P (^) tot−^2
XCH 3 OH
XCOX H^22
(d) In a particular experiment, you begin with 1 mole of CO(g) and 1 mole of H 2 (g). The reactants are allowed to react to form CH 3 OH(g) at 500 K and at constant total pressure Ptot. What is Ptot if at equilibrium 0.25 moles of CH 3 OH(g) are formed? ( 5 ) Setting up and ICE table in terms of the number of moles of each species. But we are told the number of moles of CH 3 OH formed at equilibrium is 0.25 so we can fill the actual values for number of moles of each species at equilibrium:
CO(g) + 2 H 2 (g) ⇀↽ CH 3 OH(g) I 1 1 0 C −x − 2 x +x E 1 − x = 0. 75 1 − 2 x = 0. 5 x = 0. 25
mole fractions: X 01.^75. 5 = 0. 5 01 ..^55 = 0. 333 01.^25. 5 = 0. 1667
Applying the relationship from part (c) and the equilibrium constant from (a):
Kx =
0 .5(0.333)^2
Kp =
P (^) tot^2
Kx
Ptot =
√ Kx Kp
√
- 001
. 00952
Ptot = 17.75 bar
(e) Suppose that the yield of CH 3 OH(g) needs to be increased to 0.3 mole. Should the pressure be increased, decreased, or not changed compared to the pressure in part c. Explain briefly. (No calculation is required). ( 4 ) We want to increase the production of products. To shift the equilibrium further towards the products we should increase the pressure. By increase the pressure we will shift the equilibrium towards the side of the reaction with more moles of gas, which in this case is the products. We can also think of this quantitatively using the expression from part(b). If we want to increase the amount of product, we want to increase Kx. To keep the product (^) P^1 tot
Kx constant, we would need to increase the total pressure.
Potentially useful information
R = 0.0821 L·atm/mol·K = 8.314 J/mol·K
P V = nRT for an ideal gas
∆G = ∆H − T ∆S
For the equation ax^2 + bx + c = 0 solutions have the form:
x =
−b ±
b^2 − 4 ac 2 a
