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thermo_ism_ch06 - Chapter 6 Chemical Equilibrium P6.1...

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Chapter 6: Chemical Equilibrium P6.1 ) C 6 H 6 ( l ) + 15/2 O 2 ( g ) → 6CO 2 ( g ) + 3H 2 O( l ) ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 2 2 2 6 6 1 1 1 1 15 3 H O, 6 CO , O , C H , 2 15 3 237.1 kJ mol 6 394.4 kJ mol 0 124.5 124.5 kJ mol 2 3202 kJ mol 3202 kJ mol max nonexpansion R f f f f max nonexpansion w G G l G g G g G l w - - - - = ∆ = + - - ∆ = × - + × - - × - - = - = - o o o o o 1 1 1 mol 40.96 kJg 78.18 g - - × = - H 2 ( g ) + 1/2 O 2 ( g ) → H 2 O( l ) On a per gram basis, nearly three times as much work can be extracted from the oxidation of hydrogen than benzene. P6.2) At constant , we consider the reversible process. Because is a state function, any path whether reversible or irreversible, between the same initial and final states will give the same r dA SdT PdV T A = - - 1 1 3 esult. 12.0 L ln 2.00 mol 8.314 J mol K 298 K ln 5.30 10 J 35.0 L f i V f i V V A PdV nRT V - - = - = - = - × × × = × P6.3 ) Calculate G for the isothermal expansion of 2.50 mol of an ideal gas at 350 K from an initial pressure of 10.5 bar to a final pressure of 0.500 bar. At constant , we consider the reversible process. Because is a state function, any path between the same initial and final states will give the same result. ln 2.50 mol f i P f i P dG SdT VdP T G P G VdP nRT P = - + = = = 1 1 3 0.500 bar 8.314 J mol K 350 K ln 22.1 10 J 10.5 bar - - × × × = - × P6.4) a) for the isothermal reversible path 6-1 ( 29 ( 29 ( 29 2 2 2 1 1 1 1 1 H O, O , H , 2 237.1 kJ mol 0 0 237.1 kJ mol 1 mol 237.1 kJ mol 117.6 kJg 2.016 g max nonexpansion R f f f max nonexpansion w G G l G g G g w - - - - = ∆ = ∆ - - ∆ = - - - = - = - × = - o o o o
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Chapter 6/Chemical Equilibrium 1 1 3 ln ln 10.0 L 2.50 mol 8.314 J mol K 298 K ln 9.97 10 J 50.0 L f i P f i i f P P V G VdP nRT nRT P V - - = = = = × × × = - × 1 1 3 ln 50.0 L 2.50 mol 8.314 J mol K 298 K ln 9.97 10 J 10.0 L f i V f i V V A PdV nRT V - - = - = - = - × × × = - × b) Because A and G are state functions, the answers are the same as to part (a) because the systems go between the same initial and final states, T , V i T , V f . G A = H U = ∆( PV ) = ∆( nRT ). Therefore, G = A for an ideal gas if T is constant. P6.5 ) ( 29 ( 29 ( 29 3 5 3 For a solid or liquid, ( , ,100 bar) ( , ,1 bar) ( , ,1 bar) 12.011 10 kg 0 99.0 10 Pa = 52.8 J 2250 kg m f i P f i P m m m f i m f i G VdP V P P M G C s G C s V P P G C s P P ρ - - = = - = + - = + - × = + × × ( 29 ( 29 1 1 3 Treating He as an ideal gas, , ,100 bar , ,1 bar 100 bar 0 ln 1 mole 8.314 J mol K 298.15 K ln =11.4 10 J 1 bar f i P m m P f i G He g G He g VdP P RT P - - = + = + = × × × × This result is a factor of 216 greater than that for graphite. P6.6) Assuming that f H o is constant in the interval 275 K – 600 K, calculate f G o ( H 2 O, g , 525 K). ( 29 ( 29 ( 29 ( 29 1 2 2 1 1 2 1 3 1 3 1 3 1 1 1 228.6 10 Jmol 1 1 525 K 241.8 10 Jmol 298.15 K 525 K 298.15 K 525 K 218.5 10 Jmol f f f f G T G T T H T T T T G - - - = + ∆ - - × = × - × × - = - × o o o o 6-2
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Chapter 6/Chemical Equilibrium P6.7) ( 29 ( 29 ( 29 ( 29 2 2 3 1 3 1 3 1 1 298.15 K CO , CO, O , 2 394.4 10 J mol + 137.2 10 J mol 0 257.2 10 J mol reaction f f f G G g G g G g - - - = ∆ - ∆ - = - × × - = - × o o o o ( 29 ( 29 ( 29 ( 29 2 2 3 1 3 1 3 1 1 298.15 CO , CO, O , 2 393.5 10 J mol + 110.5 10 J mol 0 283.0 10 J mol reaction f f f H K H g H g H g - - - = ∆ - ∆ - = - × × - = - × o o o o ( 29 ( 29 ( 29 1 2 2 1 1 2 1 1 1 reaction reaction reaction G T G T T H T T T T = + ∆ -
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