CHhmwk3 - moore (jwm2685) – Homework 3 – vanden bout...

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Unformatted text preview: moore (jwm2685) – Homework 3 – vanden bout – (51640) 1 This print-out should have 41 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points What is the main overall driving force for any spontaneous reaction or change? 1. To maximize electrostatic interactions. 2. To release heat energy. 3. To maximize entropy. 4. To obey the laws of gravity. 5. To lower the available free energy. cor- rect Explanation: 002 10.0 points When the chemical reaction A + B ⇀ ↽ C + D is at equilibrium, 1. the sum of the concentrations of A and B equals the sum of the concentrations of C and D. 2. all four concentrations are equal. 3. the reverse reaction has stopped. 4. the forward reaction has stopped. 5. neither the forward nor the reverse reac- tions have stopped. correct 6. both the forward and reverse reactions have stopped. Explanation: Chemical equilibrium is dynamic equilib- rium, with both foward and reverse processes occuring at the same rate. 003 10.0 points Equilibria involving species in more than one phase are called 1. reversible reaction. 2. homogeneous equilibria. 3. heterogeneous equilibria. correct 4. chemical equilibria. Explanation: If all components of the equilibrium system are in the same phase, equilibrium is homoge- nous; if more than one phase is represented in the system, equilibrium is heterogenous. 004 10.0 points Write the equilibrium expression for the fol- lowing reaction: 2 NO ( aq ) + 4 H 2 O( ℓ ) ←→ 3 H 2 ( aq ) + 2 H + ( aq ) + 2 NO − 3 ( aq ) 1. K = [ NO ] 2 [ H 2 ] 3 · [ H + ] 2 · [ NO − 3 ] 2 2. K = [ NO ] 2 [ H + ] 2 · [ NO − 3 ] 2 3. K = [[ H + ] 2 · [ NO − 3 ] 2 [ NO ] 2 4. K = [ H 2 ] 3 · [ H + ] 2 · [ NO − 3 ] 2 [ NO ] 2 · [ H 2 O ] 4 5. K = [ H 2 ] 3 · [ H + ] 2 · [ NO − 3 ] 2 [ NO ] 2 correct Explanation: Set up K , products in the numerator, reac- tants in the denominator, all raised to respec- tive stoichiometric coefficients. 005 10.0 points The following equilibrium exists in a closed container: N 2 ( g ) + O 2 ( g ) ←→ 2 NO( g ) moore (jwm2685) – Homework 3 – vanden bout – (51640) 2 If the partial pressures of N 2 , O 2 and NO at equilibrium are 1 . 5 atm, 2 . 5 atm and . 75 atm respectively, what is the equilbrium constants? 1. K = 0.2 2. K = 5 3. K = 6.67 4. K = 0.15 correct Explanation: K = P 2 NO P N 2 · P O 2 = . 75 2 1 . 5 · 2 . 5 = 0 . 15 006 10.0 points K c = 4 . 9 × 10 5 for the reaction A(g) ⇀ ↽ 2 B(g) . What is K c for the reaction 2 B(g) ⇀ ↽ A(g) ? 1. 2 . 04 × 10 − 6 correct 2. 9 . 81 × 10 − 6 3. 2 . 45 × 10 − 6 4. 3 . 40 × 10 − 6 5. 1 . 59 × 10 − 6 Explanation: 007 10.0 points Given the following equilibria and equilibrium constants K 1 CO(g) + H 2 O(g) ⇀ ↽ CO 2 (g) + H 2 (g) K 2 CH 4 (g) + H 2 O(g) ⇀ ↽ CO(g) + 3 H 2 (g) K 3 CH 4 (g) + 2 H 2 O(g) ⇀ ↽ CO 2 (g) + 4 H 2 (g) The correct expression for K 3 in terms of K 1 and K 2 is 1. K 3 = K 1 K 2 2.2....
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This note was uploaded on 05/02/2011 for the course CH 302 taught by Professor Holcombe during the Spring '07 term at University of Texas.

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CHhmwk3 - moore (jwm2685) – Homework 3 – vanden bout...

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