HW1 solutions

# HW1 solutions - Byers Raymond – Homework 1 – Due noon...

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Unformatted text preview: Byers, Raymond – Homework 1 – Due: Jan 23 2006, noon – Inst: J A Holcombe 1 This print-out should have 27 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Which of the following statements does NOT concern the First Law of Thermodynamics? 1. Energy can be transformed but never made from scratch. 2. Energy cannot be created nor destroyed. 3. The energy of the universe is constant. 4. Spontaneous changes lead to more uni- versal disorder. correct Explanation: The First Law of Thermodynamics states that the total amount of energy in the uni- verse is constant. Energy cannot be created nor destroyed, but it can be transferred and it can change form. The Second Law of Ther- modynamics addresses spontaneity. 002 (part 1 of 1) 10 points When a hydrocarbon is burned in the pres- ence of oxygen, the amount of energy in the universe 1. decreases. 2. stays the same. correct 3. increases. Explanation: 003 (part 1 of 1) 10 points Elemental sulfur exists in several forms, with rhombic sulfur the most stable under normal conditions and monoclinic sulfur slightly less stable. The standard enthalpies of combus- tion of the two forms (to sulfur dioxide) are- 296 . 83 kJ / mol and- 297 . 16 kJ / mol, respec- tively. Calculate the enthalpy of the rhombic → monoclinic transition. 1. . 33 kJ correct 2. 0.8 kJ 3. 0.7 kJ 4. 0.10 kJ 5. 0.15 kJ 6. 0.95 kJ 7. 0.23 kJ 8. 0.25 kJ Explanation: Using Hess’ Law, Δ H ◦ r = X n Δ H ◦ j , prod- X n Δ H ◦ j , reactants The combustion of the monoclinic sulfur is reversed and added to the combustion reac- tion of rhombohedral sulfur: S(rhombic) + O 2 (g)-→ SO 2 (g) Δ H ◦ =- 296 . 83 kJ SO 2 (g)-→ S(mono) + O 2 (g) Δ H ◦ = 297 . 16 kJ S(rhombic)-→ S(mono) Δ H ◦ = . 33 kJ 004 (part 1 of 1) 10 points 3 . 1 g of a hydrocarbon fuel is burned in a calorimeter that contains 273 grams of water initially at 25.00 ◦ C. After the combustion, the temperature is 26.55 ◦ C. How much heat is evolved per gram of fuel burned? The heat capacity of the calorimeter is 92.3 J/ ◦ C. 1. 9828 J/g Byers, Raymond – Homework 1 – Due: Jan 23 2006, noon – Inst: J A Holcombe 2 2. 1913 J/g 3. 617 J/g correct 4. 570 J/g 5. 9257 J/g 6. 1770 J/g 7. 28698 J/g 8. 143 J/g Explanation: m fuel = 3 . 1 g m water = 273 g Δ T = 26 . 55 ◦ C- 26 . 00 ◦ C = 1 . 55 ◦ C The amount of heat evolved by the reaction is equal to the amount of heat gained by the water plus the amount of heat gained by the calorimeter. The specific heat of water is 4.184 J/g · ◦ C, so we have to multiply by grams and temper- ature change in order to obtain Joules: Δ H of water = (SH)(m water )(Δ T ) = (4 . 184 J / g · ◦ C)(273 g) × (1 . 55 ◦ C) = 1770 J The heat capacity of the calorimeter is 92.3 J/ ◦ C. This is not per gram, so we just have to multiply by the temperature change to get Joules: Δ H of calorimeter = (SH)(Δ...
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## This note was uploaded on 02/23/2010 for the course CH 302 taught by Professor Holcombe during the Fall '07 term at University of Texas.

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HW1 solutions - Byers Raymond – Homework 1 – Due noon...

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