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Unformatted text preview: x 10-4 does not change.
Short explanation of re-establishment of equilibrium:
After reduction in volume, Q<K. Products will be
favored as equilibrium is re-established. b.) What are the partial pressures of all the species in the reaction at the new equilibrium?
After reduction in volume:
P*CO2 = P*H2 = 2.0 atm
P*CO = 6.4 x 10-4 atm
K = (6.4 x 10-4 + x) / (2.0 - x)(2.0 - x) When equilibrium is re-established:
P*CO2 = P*H2 = 2.0 atm - x
P*CO = 6.4 x 10-4 atm + x
PCO2: 2.0 atm ≈ (6.4 x 10-4 + x) / 4.0 = 3.2 x 10-4
! x = 6.4 x 10-4 PH2: 2.0 atm
PCO: 1.3 x 10-3 atm Page 6 of 6 c.) Name:_________________________________________ Using the information on page 1, determine ∆H° for the reaction from ∆Hf°’s and determine ∆S°
for the reaction from S°’s. Explain the sign of ∆S°.
∆H° = ∆Hf°(H2O) + ∆Hf°(CO) - ∆Hf°(CO2) - ∆Hf°(H2)
= -285.8 kJ/mol –110.5 kJ/mol + 393.5 kJ/mol + 0 kJ/mol
∆S° = S°(H2O) + S°(CO) - S°(CO2) - S°(H2)
= 70.0 J/mol·K +197.7 J/mol·K - 213.6 J/mol·K - 130.6 J/mol·K ∆H°: -2.8 kJ/mol ∆S°: -76.5 J/mol·K Explanation: ∆S < 0 as more disordered reactants
(all gas) are converted to more ordered products
(both gas and liquid). d.) Determine ∆G° for the reaction at 25 oC.
∆G° = ∆H° - T∆S°
= (-2.8 kJ/mol) – (298K)(-76.5 J/mol·K) e.) ∆G°: 20.0 kJ/mol Below what temperature does the reaction become spontaneous?
∆G° = ∆H° - T∆H°< 0
T < ∆H° / ∆S° T: 36.6 K Page 7 of 7 Name:_________________________________________ (47 pts)
For questions a – d consider the following: Different hydrocarbon structural isomers burn differently in
oxygen. This is the basis for the ‘octane’ rating of gasoline. Consider three structural isomers, X,Y and
Z, of a certain hydrocarbon. The hydrocarbon isomers can react as follows at 298 K. 1) X K1 = 20.0 ∆H° = -40.0 kJ/mol ∆S° = -120 J/K mol 2) Y a.) Y
Z K2 = 0.4 ∆H° = 20.0 kJ/mol ∆S° = 60 J/K mol What is the equilibrium constant for X Z? Keq = K1 • K2 Keq: b.) Arrange the compounds X, Y, and Z on the relative enthalpy scale below? _______ X
∆Hf° _______ Z
_______ Y 8 Page 8 of 8
Upon combustion, which isomer, X or Z, burns hotter? Circle one and explain.
X Z Same Explanation: X is less stable as shown by its higher heat of formation. d.) Sketch a plot of ∆G° vs. T for the equilibrium X Z. ∆H° < 0 and ∆S° < 0, so ∆G° is negative at low T and positive at high T....
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This note was uploaded on 09/11/2009 for the course CHEM 1A taught by Professor Nitsche during the Spring '08 term at University of California, Berkeley.
- Spring '08