Exam 4-solutions - Version 180 Exam 4 McCord (53130) This...

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

View Full Document Right Arrow Icon
Version 180 – Exam 4 – McCord – (53130) 1 This print-out should have 27 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. McCord CH301tt This exam is only For McCord’s Tues/Thur CH301 class. PLEASE carefully bubble in your UTEID and Version Number! You’ll need some (not all) oF the values From this table on the exam. Δ H f Δ G f S Substance ( J mol · K ) ( kJ mol ) ( kJ mol ) CaCO 3 (s) - 1206 . 9 - 1128 . 8 92.9 CaO(s) - 635 . 09 - 604 . 03 39.75 CO 2 (g) - 393 . 51 - 394 . 36 213.74 H 2 O(l) - 285 . 83 - 237 . 13 69.91 H 2 O(g) - 241 . 82 - 228 . 57 188.83 CH 4 (g) - 74 . 81 - 50 . 72 186.26 C 2 H 4 (g) +52 . 26 +68 . 12 219.56 C 3 H 8 (g) - 103 . 85 - 23 . 49 270.2 C 4 H 10 (g) - 126 . 15 - 17 . 03 310.1 C 6 H 6 ( ) +49 . 0 +124 . 5 173.3 C 8 H 18 ( ) - 249 . 9 +6 . 4 358 C 6 H 12 O 6 (s) - 1268 - 910 212 001 10.0 points Carbon monoxide reacts with oxygen to Form carbon dioxide by the Following reaction: 2 CO(g) + O 2 (g) 2 CO 2 (g) Δ H For this reaction is - 135 . 28 kcal. How much heat would be released iF 12.0 moles oF carbon monoxide reacted with su±- cient oxygen to produce carbon dioxide? 1. 811.68 kcal correct 2. 135.28 kcal 3. 541.12 kcal 4. 405.84 kcal 5. 1623.36 kcal 6. 270.56 kcal Explanation: Δ H = - 135 . 28 kcal n CO = 12.0 mol p - 135 . 28 kcal 1 mol rxn Pp 1 mol rxn 2 mol CO P × (12 mol CO) = - 811 . 68 kcal so 811 . 68 kcal were released. 002 10.0 points Heat absorbed by a system at constant vol- ume is equal to 1. Δ H 2. ΔV 3. Δ G 4. Δ E correct 5. Δ S Explanation: Δ E = q + w q = Δ H w = - P Δ V When Δ V = 0, w = 0. When w = 0, Δ E = Δ H . 003 10.0 points Calculate the fnal temperature when 2.50 kJ oF energy is transFerred as heat to 1.50 mol N 2 at 298 K and 1 atm at constant volume. 1. 159 C 2. 145 C 3. 105 C correct 4. 80 C 5. 120 C
Background image of page 1

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

View Full DocumentRight Arrow Icon
Version 180 – Exam 4 – McCord – (53130) 2 Explanation: The contribution of each mode of motion to the total molar internal energy is 1 2 RT . N 2 is a linear molecule so it has three modes of translational motion and two modes of rota- tional motion (assuming no contribution from vibration). Therefore, U m = 5 p 1 2 RT P = 2 . 5 RT , so C V , m = Δ U m Δ T = 2 . 5 R Δ T Δ T = 2 . 5 R q = n C V , m Δ T Δ T = q n C V , m = q ( n )(2 . 5 R ) = +2500 J (1 . 5 mol)(2 . 5)(8 . 314 J · mol 1 · K 1 ) = 80 . 186 K , so T fnal = (298 K + 80 . 186 K) - 273 . 15 = 105 . 036 C 004 10.0 points A reversible reaction is at equilibrium at con- stant pressure and temperature if the free energy of the system 1. decreases. 2.
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 01/12/2010 for the course BIO 49830 taught by Professor Brand during the Spring '09 term at University of Texas.

Page1 / 9

Exam 4-solutions - Version 180 Exam 4 McCord (53130) This...

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

View Full Document Right Arrow Icon
Ask a homework question - tutors are online