Lecture27 - Chem 162 Lect 27 Spring 2011 Free Energy and...

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Free Energy and Non-standard Conditions; Equilibrium Jumping to the conclusion: G = G o + RTlnQ G o = change in free energy under standard conditions Q = mass action expression (gases, solutions etc. .) G = change in free energy under non-standard conditions (pressures not equal to 1 atm; concentrations not equal to 1 M) Example: H 2 (g) 2H(g) G o f (H 2 ) = 0.0 G o f (H) = 203 kJ/mol Example: T = 1,000 K P H 2 = 10 atm P H = 10 -5 atm G = G o + RTlnQ = [ G o + 2.303RTlogQ] = 2 x203 kJ – 1x0.0 + (8.31J/K)x(1000K)xln(10 -5 ) 2 /(10) = 406 kJ + (-210 kJ) = 196 kJ G o is very unfavorable . ...... the correction term (low pressure on the right, high on the left) favors the reaction (Le Chatelier) and makes overall G less unfavorable. Free Energy and Equilibrium At equilibrium, G = 0 and Q = K, so G = 0 = G prod – G react = G o + RTlnQ = G o + RTlnK => G o = - RTlnK !!!!!! 1
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Lecture27 - Chem 162 Lect 27 Spring 2011 Free Energy and...

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