HW8_Answer_ver2

HW8_Answer_ver2 - HW8 Answer Chem340 Spring 2011 (Version 2...

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HW8 Answer Chem340 Spring 2011 (Version 2 Updated 4/11/2011) P6.15) Consider the equilibrium CO g   H 2 O g   ±± ± ± CO 2 g   H 2 g  . At 1000 K, the composition of the reaction mixture is CO 2 H 2 ( g ) CO( g ) H 2 O( g ) Substance ( g ) 27.1 27.1 22.9 22.9 Mole % a. Calculate K P and G reaction at 1000 K. b. Given the answer to part (a), use the H f of the reaction species to calculate G reaction at 298.15 K. Assume that H reaction is independent of temperature. (a) 1 1 80 . 2 ) 4 . 1 ln( 1000 314 . 8 ln 4 . 1 9 . 22 9 . 22 1 . 27 1 . 27 2 2 2 KJmol K JKmol K RT G x x x x K p o reaction O H CO H CO p (b) 1 - 1 - 1 - o reaction f, 1000 o reaction 15 . 298 1 o C f, o H f, o C f, o H f, o reaction f, mol -29.7kJ ) 1000K 1 - 298 1 ( ) mol kJ 2 . 41 ( 15 . 298 ) mol kJ 80 . 2 ( 1000K 298.15K ) 1000K 1 - 298 1 ( Δ H , G 1000K 298.15K , 2 . 41 ) 5 . 110 ( ) 8 . 241 ( 5 . 393 0 Δ H Δ H Δ H Δ H Δ H 2 2 2 K K K G KJmol K K o reaction O O O
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P6.20) Calculate the Gibbs energy change for the protein denaturation described in Problem 5.45 at T = 310.K and T = 340.K. From P5.34 we have:  -1 -1 den mol K J 4 . 1109 K 310 Δ S -1 -1 den mol K kJ 43.9 3 K 310 Δ H K 310 Δ G den is then:   1 - 1 - 1 - 1 - den den den mol kJ 3.3 1 mol K J 4 . 1109 K 98 2 mol kJ 43.9 3 K 310 Δ S T K 310 Δ H K 310 Δ G At 340 K we obtain: 1 - 1 - -1 reaction mol kJ 87 . 47 K 310 1 K 340 1 mol kJ 343.9 K 310 mol J 13300 K 340 K 340 Δ G P6.21) For a protein denaturation at T = 310. K and P = 1.00 atm, the enthalpy change is 911 kJ mol –1 and the entropy change is 3.12 J K –1 mol –1 . Calculate the Gibbs energy change at T = 310. K and P = 1.00 atm. Calculate the Gibbs energy change at T = 310. K and P = 1.00 10 3 bar. Assume for the denaturation V =3.00 mL mol –1 . State any assumptions you make in the calculation. The Gibbs energy change at T = 310. K and P = 1.00 atm is:   den den den -1 -1 -1 -1 Δ G 310 K Δ H 310 K T Δ S 310 K 911.0 kJ mol 310 K 3.12 J K mol 910.0 kJ mol  The Gibbs energy change at T = 310. K and P = 1.00 10 3 bar is:
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     3 den den 0 -1 -6 3 -1 8 1 Δ G 310 K,1.00 10 bar Δ G 310 K,1.00 atm V p p 91.0 kJ mol 3.0 10 m mol 10 Pa 101325 Pa 910.3 kJ mol   We assumed that the volume change is independent over the pressure range of 1000 bar. P6.26) In this problem, you calculate the error in assuming that H reaction is independent of T for a specific reaction. The following data are given at 25°C: CuO( s ) Cu( s ) O 2 ( g ) H f kJ mol 1 –157 G f kJ mol 1 –130 C P , m J K 1 mol 1 42.3 24.4 29.4 a. From Equation (6.71), d ln K P K P T 0 K P T f 1 R H reaction T 2 T 0 T f dT To a good approximation, we can assume that the heat capacities are independent of temperature over a limited range in temperature, giving H reaction T  H reaction T 0  C P T T 0 where C P v i C P , m i i . By integrating Equation (6.71), show that
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ln K P T  ln K P T 0 H
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This note was uploaded on 08/23/2011 for the course CHEM 342 taught by Professor Prestonsnee during the Spring '08 term at Ill. Chicago.

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HW8_Answer_ver2 - HW8 Answer Chem340 Spring 2011 (Version 2...

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