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Quiz4_calvin_key

# Quiz4_calvin_key - Chem 125 Spring 2008 Quiz#4 Calvin Dr...

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Chem 125 Quiz #4 Calvin Dr. Retsek Spring 2008 NAME: __ KEY ________________ Computer: ______ 1. (10 points) At 25°C, K c for the following reaction is 4.0 x 10 -7 . N 2 O 4 (g) 2 NO 2 (g) In an experiment, 1.5 moles of N 2 O 4 are placed in a 10.0-L flask. Calculate the concentrations of N 2 O 4 and NO 2 when this reaction reaches equilibrium. [N 2 O 4 ] = 1.5 moles/10.0 L = 0.15 M Need “ICE” table to find out what is happening at equilibrium N 2 O 4 2 NO 2 Initial 0.15 0 Change - x + 2 x (since 2 moles in balanced equation) Equilibrium 0.15 – x 2x K c = 4.0 x 10 -7 = [NO 2 ] 2 /[N 2 O 4 ] = [2x] 2 /[0.15 – x] Since K is so small, there will be very little product (x will be small). Assume that 0.15 – x 0.15 K c = 4.0 x 10 -7 = [2x] 2 /[0.15] x = 1.2 x 10 -4 Check assumption: K’ = (2 x 1.2 x 10 -4 ) 2 /(0.15 – 1.2 x 10 -4 ) = 3.84 x 10 -7 Within 5%? K’/K = (3.84 x 10 -7 /4.0 x 10 -7 ) x 100 = 96% (assumption ok!) [N 2 O 4 ] equilibrium = 0.15 – x = 0.15 – 1.2 x 10 -4 = 0.15 M [NO 2 ] equilbrium = 2x = 2(1.2 x 10 -4 ) = 2.4 x 10 -4 M 2. (5 points) In which direction (if any) will the position of the equilibrium for 2 HI (g) H 2 (g) + I 2 (s) H = -100 kJ be shifted for each of the following changes?

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