solution3_pdf

# solution3_pdf - spano velasco(ds28282 – hw 03 –...

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Unformatted text preview: spano velasco (ds28282) – hw 03 – Fakhreddine – (52395) 1 This print-out should have 30 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 1.0 points Consider the reaction 2 HgO(s) ⇀ ↽ 2 Hg( ℓ ) + O 2 (g) . What is the form of the equilibrium constant K c for the reaction? 1. K c = [O 2 ] [HgO] 2 2. K c = [Hg] 2 [O 2 ] 3. K c = [O 2 ] correct 4. K c = [Hg] 2 [O 2 ] [HgO] 2 5. None of the other answers is correct. Explanation: Solids and liquids are not included in the K expression. 002 1.0 points Write the equilibrium constant for 2 NaBr(aq) + Pb(ClO 4 ) 2 (aq) → PbBr 2 (s) + 2 NaClO 4 (aq) . 1. K = [PbBr 2 ] [Pb 2+ ][Br − ] 2 2. K = 1 [Pb(ClO 4 ) 2 ][NaBr] 2 3. K = 1 [Pb 2+ ][Br − ] 2 correct 4. K = [Pb 2+ ][Br − ] 2 5. K = [NaClO 4 ] 2 [NaBr] 2 [Pb(ClO 4 ) 2 ] Explanation: 003 1.0 points Consider the famous ammonia preparation 3 H 2 (g) + N 2 (g) ⇀ ↽ 2 NH 3 (g) The equation K = [ x ] 2 [0 . 1- 3 x ] 3 [0 . 7- x ] is not a possible correct description of the equilibrium situation because 1. the denominator and numerator should be inverted. 2. The equation is correct. 3. the 0.1 and 0.7 in the denominator are incompatible. 4. [0 . 7- x ] in the denominator should be [0 . 7- 3 x ]. 5. [ x ] in the numerator should be [2 x ]. cor- rect Explanation: 3 H 2 (g) + N 2 (g) ⇀ ↽ 2 NH 3 (g) ini, M 0.1 0.7 Δ, M- 3 x- x +2 x eq, M . 1- 3 x . 7- x 2 x K = [NH 3 ] 2 [H 2 ] 3 [N 2 ] = (2 x ) 2 (0 . 1- 3 x ) 3 (0 . 7- x ) 004 1.0 points Given the following equilibrium data at 973 K MgCl 2 (s) + 1 2 O 2 (g) ⇀ ↽ MgO(s) + Cl 2 (g) K p = 2 . 95 atm MgCl 2 (s) + H 2 O(g) ⇀ ↽ MgO(s) + 2 HCl(g) K p = 8 . 40 atm Calculate the equilibrium constant K p at 973 K for 2 Cl 2 (g) + 2 H 2 O(g) ⇀ ↽ O 2 (g) + 4 HCl(g) . 1. 24.8 2. 11.4 spano velasco (ds28282) – hw 03 – Fakhreddine – (52395) 2 3. 2.85 4. 5.50 5. 8.11 correct Explanation: T = 973 K K p = 8.40 atm MgCl 2 (s) + 1 2 O 2 (g) ⇀ ↽ MgO(s) + Cl 2 (g) K 1 = P Cl 2 P 1 / 2 O 2 = 2 . 95 MgCl 2 (s) + H 2 O(g) ⇀ ↽ MgO(s) + 2 HCl(g) K 2 = P 2 HCl P H 2 O = 8 . 4 Twice eq(2) minus twice eq(1) yields 2 MgCl 2 (s) + 2 H 2 O(g) + 2 MgO(s) + 2 Cl 2 (g) ⇀ ↽ 2 MgO(s) + 4 HCl(g) + 2 MgCl 2 (s) + O 2 (g) 2 H 2 O(g) + 2 Cl 2 (g) ⇀ ↽ 4 HCl(g) + O 2 (g) Adding, K p = P 4 HCl · P O 2 P 2 H 2 O · P 2 Cl = K 2 2 K 2 1 = parenleftbigg 8 . 4 2 . 95 parenrightbigg 2 = 8 . 10801 005 1.0 points A: An equilibrium reaction is not affected by increasing the concentrations of products. B: If one starts with a higher pressure of reactant, the equilibrium constant will be larger. C: If one starts with higher concentrations of reactants, the equilibrium concentrations of the products will be larger....
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## This note was uploaded on 04/08/2009 for the course CH 302 taught by Professor Holcombe during the Spring '07 term at University of Texas.

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solution3_pdf - spano velasco(ds28282 – hw 03 –...

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