{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Homework 1-solutions-2

# Homework 1-solutions-2 - hayes(snh457 Homework 1...

This preview shows pages 1–3. Sign up to view the full content.

hayes (snh457) – Homework 1 – Sutcli ff e – (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. spontaneous, greater 2. spontaneous, less 3. None of these; Δ G is not directly related to K . 4. nonspontaneous; greater 5. nonspontaneous; less correct Explanation: A positive Δ G rxn (standard reaction free energy) denotes an endothermic reation, which is nonspontaneous. Also, Δ G rxn = - R T 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 0 at 775 C for a gas phase reac- tion for which K p = 5 . 51 × 10 4 at 775 C. Correct answer: - 95 . 1198 kJ / mol rxn. Explanation: K p = 5 . 51 × 10 20 T = 775 C + 273 = 1048 K Δ G 0 = - R T ln K p = - (8 . 314 J / mol · K) (1048 K) × ln(5 . 51 × 10 4 ) = - 95 . 1198 kJ / mol rxn 003 10.0 points Calculate the equilibrium constant at 310 K for the reaction HgO(s) Hg( ) + 1 2 O 2 (g) , given the data S Δ H f ( J K · mol ) ( kJ mol ) HgO(s) 70 . 29 - 90 . 83 Hg( ) 76 . 02 0 O 2 (g) 205 . 14 0 Correct answer: 2 . 24861 × 10 - 10 . Explanation: HgO(s) Hg( ) + 1 2 O 2 (g) , Δ H r = - [ Δ H f , HgO(s) ] = - ( - 90 . 83) = 90 . 83 kJ / mol Δ S r = S Hg( ) + 1 2 S O 2 (g) - S HgO(s) = 76 . 02 J K · mol + 1 2 205 . 14 J K · mol - 70 . 29 J K · mol = 108 . 3 J K · mol At 310 K, Δ G r(310 K) = 90 . 83 K - (310 K) × 108 . 3 J K · mol 1 kJ 1000 J = 57 . 257 kJ / mol Δ G r(310 K) = - R T ln K ln K = - Δ G r(310 K) R T = - 57 . 257 kJ (8 . 314 J / K)(310 K) × 1000 J 1 kJ = - 22 . 2155 K = e - 22 . 2155 = 2 . 24861 × 10 - 10 .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
hayes (snh457) – Homework 1 – Sutcli ff e – (51060) 2 004 10.0 points Which combination of Δ G and K is possible at standard conditions? 1. Δ G = - 56 . 8 J, K = 3 . 29 × 10 - 12 2. Δ G = - 60 . 9 kJ, K = 0 . 982 3. Δ G = 45 . 3 J, K = 3 . 8 × 10 10 4. Δ G = 47 . 8 kJ, K = 4 . 97 × 10 - 9 cor- rect 5. Δ G = 66 . 1 kJ, K = 1 . 02 Explanation: If Δ G is positive, K must be less than 1; if Δ G is negative, K must be greater than 1.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 6

Homework 1-solutions-2 - hayes(snh457 Homework 1...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online