HW4 - Homework 4 Feb. 15, 2007 10. 1 CO(g) + O2 (g) CO2 (g)...

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Homework 4 Feb. 15, 2007

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10. CO(g) + 1 2 O 2 (g) CO 2 (g) P O 2 = 0 . 2 atm , P CO 2 = 3 × 10 - 4 atm From Table A.5, Δ G 0 298 = Δ G 0 f (CO 2 ) - G 0 298 (CO) + Δ G 0 298 (O 2 )) = - 394 . 359 - ( - 137 . 168 + 0) = - 257 . 191 kJ mol - 1 ln( K ) = - Δ G 0 298 RT = - 257 . 191 0 . 008314 × 298 = 103 . 808 Thus, K = a CO 2 a CO · a 1 / 2 O 2 = 1 . 2 × 10 45 For ideal gases, a i = P i 1 atm , so K = P O 2 P CO · P 1 / 2 O 2 = 1 . 2 × 10 45 P CO = 5 . 5 × 10 - 49 atm . Thus, no need to worry about the equilibrium concentration of CO. 12. a. From Table 4.2, pK = - log K = 4 . 76 K = 10 - 4 . 76 , so Δ G 0 = - RT ln K = - 8 . 314 × 10 - 3 × 298 × ln(10 - 4 . 76 ) = 27 . 15 kJ mol - 1 . 1
b. At equilibrium, the Gibbs free-energy change is 0. c. Δ G = Δ G 0 + RT ln Q Q = a HOAc a H + · a OAc - . Since γ = 1 for all species, a HOAc a H + · a OAc - = [HOAc] [H + ][OAc - ] = 1 10 - 4 · 10 - 2 = 10 6 . From part a, Δ G 0 = - 27 . 15 kJ mol - 1 Δ G = - 27 . 15 + 0 . 008314 × 298 × ln(10 6 ) = 7 . 07 kJ mol

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This note was uploaded on 08/03/2009 for the course BME 100 taught by Professor Yuan during the Spring '07 term at Duke.

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HW4 - Homework 4 Feb. 15, 2007 10. 1 CO(g) + O2 (g) CO2 (g)...

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