RelativeResourceManager 2

# RelativeResourceManager 2 - CHEM 6B PRACTICE EXAM 3-2(Prof...

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Chem 6B Practice Exam 3-2 1 Prof. Crowell CHEM 6B PRACTICE EXAM 3-2 (Prof. Crowell) Spr 2010 May 17, 2010 *** FILL IN YOUR STUDENT ID NUMBER , NAME , and TEST FORM = B *** Bubble in the correct response on scantron (25 questions, worth 4 pts each) 1. Given the equilibrium constants for reactions 1 and 2 below at 450K, calculate the equilibrium constant for reaction 3 at this same temperature: Rxn 1: N 2 O 5 (g) 2 NO 2 (g) + ½ O 2 (g), K 1 = 3.3 x 10 4 . Rxn 2: 2 NO (g) + O 2 (g) 2 NO 2 (g), K 2 = 6.4 x 10 3 . Rxn 3: 4 NO (g) + 3 O 2 (g) 2 N 2 O 5 (g), K 3 = ?? a) 5.16 b) 26.6 c) 0.5 d) 0.51 *e) 0.038 To get Rxn 3, we need to rearrange Rxns 1 and 2 to have 2 moles of N 2 O 5 as a product, and 4 moles of NO as a reactant. As a consequence, we need to reverse Rxn 1 and multiply it by 2, and add this result to Rxn 2 multiplied by 2. Consequently, the new reactions and associated equilibrium constants are: 22 2 5 1 2 1 2 2 2 2 2 5 3 1 2 2 1 2 2 31 2 2 11 1 Rxn-1 : 4 NO (g) + O (g) 2 N O (g), K = K Rxn-2 : 4 NO (g) + 2 O (g) 4 NO (g), K = K K Rxn-3: 4 NO (g) + 3 O (g) K K K K 6 KK K        2 3 2 4 .4 10 0.194 0.038 3.3 10 x x 2. At 2000 C, K p = 0.01 for the reaction N 2 (g) + O 2 (g) 2 NO (g). Predict how the concentrations will change to reach equilibrium at 2000 C if 0.4 moles of N 2 , 0.1 moles of O 2 , and 0.08 moles of NO are placed in a 1.0 L container. a) The concentrations of NO, N 2 and O 2 will remain unchanged. b) The concentration of NO will increase; the concentrations of N 2 and O 2 will decrease. c) The concentration of NO will decrease; the concentrations of N 2 and O 2 will remain unchanged. *d) The concentration of NO will decrease; the concentrations of N 2 and O 2 will increase. e) The concentrations of NO, N 2 and O 2 will all increase. We are given K P = 0.01, but we need to determine K C since our concentrations are given in moles per liter.: 0.01 since 0 for this reaction. n PC C RTc K n P To determine which way the reaction will shift, we need to determine Q C and compare it to K C :  NO 0.08 Q= 0.16 0.01 NO 0 . 40 . 1 Rxn shifts back to the LEFT NO decreases, N increases, and O increases. CC QK  

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Chem 6B Practice Exam 3-2 2 Prof. Crowell 3. Consider the following reaction (assume an ideal gas mixture): 2 NOBr (g) 2 NO (g) + Br 2 (g) A 1.0 L vessel was initially filled with pure NOBr, at a pressure of 5.0 bar, at 300K. After equilibrium was established, the partial pressure of NOBr was 3.5 bar. What is K p for this reaction? *a) 0.14 b) 0.75 c) 0.48 d) 0.27 e) 0.32 2 2 NOBr (g) 2 NO (g) + Br (g) 5.0 bar 0 0 -2x +2x +x initial change final 2 2 2 2 22 5.0-2x +2x +x We are told ( ) 3.5 bar 5 2 (5 3.5)/ 2 0.75 bar ( ) 0.75 , ( ) 2 2(0.75) 1.50 bar (1.50) (0.75) (3.50) NO Br P NOBr PNOB r x x P Br x bar P NO x PP K P   0.14 4. Consider the following reaction: 3 N 2 H 4 (g) + 4 ClF 3 (g) 12 HF(g) + 3 N 2 (g) + 2 Cl 2 (g) A mixture, initially consisting of 0.880 M N 2 H 4 (g) and 0.880 M ClF 3 (g), reacts at a certain temperature. At equilibrium, the concentration of N 2 (g) is 0.525 M. Calculate the concentration of ClF 3 (g) at equilibrium.
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RelativeResourceManager 2 - CHEM 6B PRACTICE EXAM 3-2(Prof...

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