Chapter 20 Lecture 5

Chapter 20 Lecture 5 - Review: Free Energy, Equilibrium and...

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Review: Free Energy, Equilibrium and Reaction Direction G = RT ln Q / K = ln Q - ln K Under standard conditions (1 M concentrations, 1 atm for gases), Q = 1 and ln Q = 0 so The standard free energy change: G o = - ln K The equation is used to calculate G o from K and vice versa 20-55
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Sample Problem Calculating G at Nonstandard Conditions PROBLEM: The oxidation of SO 2 ( g ) 2SO 2 ( g ) + O 2 ( g ) 2SO 3 ( g ) is too slow at 298 K to be useful in the manufacture of sulfuric acid. To overcome this low rate, the process is conducted at an elevated temperature. (a) Calculate K at 298 K and at 973 K, ( G o 298 = -141.6 kJ/mol for reaction as written; using H o and S o values at 973 K. G o 973 = -12.12 kJ/mol for reaction as written.) SOLUTION: (a) Calculating K at the two temperatures: G o = - RT ln K so ) / ( RT G o e K 41 6 kJ/mol)(10 /kJ) At 298, the exponent is - G o / RT = - (-141.6 kJ/mol)(10 3 J/kJ) (8.314 J/mol K)(298 K) = 57.2 7 2 7x10 4 ) / ( RT G o = e 57.2 = 7x10 24 At 973, the exponent is - G o / RT = (-12.12 kJ/mol)(10 3 J/kJ) 314 J/mol )(973 K) = 1.50 e K 20-56 (8.314 J/mol K)(973 K) = e 1.50 = 4.5
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Chapter 20 Lecture 5 - Review: Free Energy, Equilibrium and...

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