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Unformatted text preview: 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 kJmol 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 61 ( 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 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 Jmol 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 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  = = = + = + = + ( 29 ( 29 1 1 3 Treating He as an ideal gas, , ,100 bar , ,1 bar 100 bar 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....
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 Fall '09
 Abra
 Equilibrium

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