CHEM31PracticeProblemsCh10AK

CHEM31PracticeProblemsCh10AK - CHEM 31 Additional Practice...

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Additional Practice Problems Free Energy, Equilibrium, and Spontaneity I. (Combining free energy, equilibrium, and kinetics) Consider the reaction: 2NO (g) + O 2 (g) 2NO 2 (g) Thermochemical Information (at 298K): G o f (kJ/mol) H o f (kJ/mol) S o (J/K mol) NO 86.7 90.4 210.6 O 2 0 0 205.0 NO 2 51.8 33.85 240.46 a. The forward rate constant for the above reaction is 7.1x10 9 M -2 s -1 at 25 o C. Assuming a simple one-step mechanism, what is the rate constant for the reverse reaction? (Hint: first find K eq at 25 o C and then find k -1 .) Step 1: Find K eq at 25 o C. G o rxn = G o prod - G o reactants =(2)(51.8 kJ/mol) – (2)(86.7 kJ/mol) – 0 = -69.8 kJ/mol G o rxn = -RTlnK -69.8 x 10 3 J/mol = - (8.314 J/mol K)(298 K) ln K K = 1.7 x 10 12 M -1 (Note that we usually take K to be unitless, but when comparing to kinetics it is helpful to keep the units, as seen below.) Step 2: Find k -1 . K = k
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CHEM31PracticeProblemsCh10AK - CHEM 31 Additional Practice...

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