Homework 04

# Homework 04 - ycs73 – Homework 4 – Fakhreddine –...

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Unformatted text preview: ycs73 – Homework 4 – Fakhreddine – (53605) 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 1.0 points The reaction A + B ⇀ ↽ C + 2 D has an equilibrium constant of 3 . 7 × 10 − 3 . Consider a reaction mixture with [A] = 2 . × 10 − 2 M [B] = 1 . 7 × 10 − 4 M [C] = 2 . 4e- 06 M [D] = 3 . 5 × 10 − 3 M Which of the following statements is defi- nitely true? 1. Heat will be evolved. 2. The reverse reaction can occur to a greater extent than the forward reaction until equilibrium is established. 3. The system is at equilibrium. 4. No conclusions about the system can be made without additional information. 5. The forward reaction can occur to a greater extent than the reverse reaction until equilibrium is established. correct Explanation: Q = [C] [D] 2 [A] [B] = (2 . 4e- 06 M) (0 . 0035 M) 2 (0 . 02 M) (0 . 00017 M) = 8 . 64706 e- 06 Since Q < K the foward reaction is favored. 002 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. None of the other answers is correct. 2. K c = [O 2 ] correct 3. K c = [Hg] 2 [O 2 ] [HgO] 2 4. K c = [O 2 ] [HgO] 2 5. K c = [Hg] 2 [O 2 ] Explanation: Solids and liquids are not included in the K expression. 003 1.0 points It is possible to calculate the equilibrium con- stant for a reaction at a chosen temperature if we know 1. the entropy change for the reaction. 2. the enthalpy change for the reaction. 3. the Gibbs free energy change for the reac- tion. 4. the standard molar enthalpy change for the reaction. 5. the standard Gibbs free energy change for the reaction. correct Explanation: The equation is Δ G =- RT ln K 004 1.0 points For the reaction 2 ICl(g) ⇀ ↽ I 2 (s) + Cl 2 (g) , Δ H =- 35 . 56 kJ/mol and Δ G = 11 . 04 kJ/mole at 25 ◦ C. Calculate K p for this reac- tion at 79 . 8 ◦ C. Correct answer: 0 . 00124905. Explanation: Δ H =- 35 . 56 kJ/mol R = 8 . 314 J/mol · K Δ G = 11 . 04 kJ/mol T = 25 ◦ C + 273 = 298 K T f = 79 . 8 ◦ C + 273 = 352 . 8 K ycs73 – Homework 4 – Fakhreddine – (53605) 2 Δ G 298 = Δ H- T Δ S Δ S 298 = Δ H- Δ G T = (- 35 . 56- 11 . 04) kJ / mol 298 K =- . 156376 kJ / mol · K Δ G 352 . 8 K =- 35 . 56 kJ / mol- (352 . 8 K)(- . 156376 kJ / mol · K) = 19 . 6094 kJ / mol Δ G =- RT ln K K p = e − Δ G / ( R T ) = exp bracketleftbigg- 19609 . 4 J / mol (8 . 314 J / mol · K)(352 . 8 K) bracketrightbigg = 0 . 00124905 005 1.0 points A 0 . 847 mole sample of HI is placed in a 5.000 liter container and held at a constant temperature until the gas-phase reaction 2 HI ⇀ ↽ H 2 + I 2 comes to equilibrium. At equilibrium, it is found that the concentration of I 2 is 3 . 2 × 10 − 2 mol/L. What is the numerical value of the equilibrium constant K c at this tempera- ture?...
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## This note was uploaded on 04/15/2009 for the course CH 53625 taught by Professor Fakhreddine during the Spring '09 term at University of Texas at Austin.

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Homework 04 - ycs73 – Homework 4 – Fakhreddine –...

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