Homework 1 Sutcliffe answers

# Homework 1 Sutcliffe answers - huynh (hph268) – Homework...

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Unformatted text preview: huynh (hph268) – Homework 1 – Sutcliffe – (51060) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. Note there are a number of numerical (free response) questions on this assignment. I rec- ommend you use the accurate Kelvin - Cel- sius conversion to help avoid rounding errors. Watch your signs and always see if you can predict a qualitative version of the problem to estimate the sign and possible size of your answer. 001 10.0 points If Δ G ◦ rxn is positive, then the forward reaction is (spontaneous / nonspontaneous) and K is (less / greater) than one. 1. nonspontaneous; greater 2. spontaneous, greater 3. nonspontaneous; less correct 4. spontaneous, less 5. None of these; Δ G is not directly related to K . Explanation: A positive Δ G ◦ rxn (standard reaction free energy) denotes an endothermic reation, which is nonspontaneous. Also, Δ G ◦ rxn =- RT ln K , so a positive Δ G ◦ rxn would result in a K that is between the values of zero and one. 002 10.0 points Evaluate Δ G at 245 ◦ C for a gas phase reac- tion for which K p = 8 . 4 × 10 4 at 245 ◦ C. Correct answer:- 48 . 8313 kJ / mol rxn. Explanation: K p = 8 . 4 × 10 20 T = 245 ◦ C + 273 = 518 K Δ G =- RT ln K p =- (8 . 314 J / mol · K) (518 K) × ln(8 . 4 × 10 4 ) =- 48 . 8313 kJ / mol rxn 003 10.0 points Calculate the equilibrium constant at 297 K for the reaction HgO(s) ⇀ ↽ Hg( ℓ ) + 1 2 O 2 (g) , given the data S ◦ Δ H ◦ f ( J K · mol ) ( kJ mol ) HgO(s) 70 . 29- 90 . 83 Hg( ℓ ) 76 . 02 O 2 (g) 205 . 14 Correct answer: 4 . 80822 × 10 − 11 . Explanation: HgO(s) ⇀ ↽ Hg( ℓ ) + 1 2 O 2 (g) , Δ H ◦ r =- [Δ H ◦ f , HgO(s) ] =- (- 90 . 83) = 90 . 83 kJ / mol Δ S ◦ r = S ◦ Hg( ℓ ) + 1 2 S ◦ O 2 (g)- S ◦ HgO(s) = 76 . 02 J K · mol + 1 2 parenleftbigg 205 . 14 J K · mol parenrightbigg- 70 . 29 J K · mol = 108 . 3 J K · mol At 297 K, Δ G ◦ r(297 K) = 90 . 83 K- (297 K) × parenleftbigg 108 . 3 J K · mol parenrightbiggparenleftbigg 1 kJ 1000 J parenrightbigg = 58 . 6649 kJ / mol Δ G ◦ r(297 K) =- RT ln K ln K =- Δ G ◦ r(297 K) RT =- 58 . 6649 kJ (8 . 314 J / K)(297 K) × 1000 J 1 kJ =- 23 . 7581 K = e − 23 . 7581 = 4 . 80822 × 10 − 11 . huynh (hph268) – Homework 1 – Sutcliffe – (51060) 2 004 10.0 points Which combination of ΔG ◦ and K is possible at standard conditions? 1. Δ G ◦ = 53 . 5 J, K = 0 . 979 correct 2. Δ G ◦ =- 45 . 7 kJ, K = 0 . 982 3. Δ G ◦ = 95 kJ, K = 1 . 05 4. Δ G ◦ = 46 . 3 J, K = 8 . 69 × 10 7 5. Δ G ◦ =- 117 J, K = 3 . 14 × 10 − 17 Explanation: If ΔG ◦ is positive, K must be less than 1; if ΔG ◦ is negative, K must be greater than 1.is negative, K must be greater than 1....
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## This note was uploaded on 06/06/2011 for the course CHEM 301 taught by Professor Wandelt during the Spring '08 term at University of Texas.

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Homework 1 Sutcliffe answers - huynh (hph268) – Homework...

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