13_new - Homework#1 holcombe(51160 This print-out should...

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Homework #1 – holcombe – (51160) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points What factor determines whether a reaction will be exothermic or endothermic? 1. the di f erence in energy between reactants and products correct 2. the speed of the reaction 3. the heat properties of the products 4. the relative concentrations of reactants and products in solution 5. the activation energy of the reaction Explanation: H is the energy di f erence between the products and reactants; H for exothermic reactions is negative, and H for endother- mic reactions is positive. 002 10.0 points would be a chemical equation with A moles of pentane, B moles of oxygen, C moles of carbon dioxide, and D moles of water. What are the values for A, B, C, and D, respectively? 1. 1; 8; 5; 6 correct 2. 2; 4; 2; 4 3. 2; 4; 4; 4 4. 1; 6; 5; 6 5. 1; 6; 4; 6 6. 2; 4; 5; 4 7. 2; 6; 2; 6 8. 1; 4; 4; 6 Explanation: 004 10.0 points Determine the standard reaction enthalpy for the partial combustion of methane to carbon monoxide according to the following reaction: 2 CH 4 (g) + 3 O 2 (g) −→ 2 CO(g) + 4 H 2 O( ) A reaction which is exothermic will be sponta-neous at (all/no) temperatures if S is (posi-tive/negative). 1. all, negative 2. no, negative 3. all, positive correct 4. no, positive Explanation: If H is negative (i.e. the reaction is exothermic) it will be spontaneous at all tem-peratures as long as S is positve. 003 10.0 points Pentane (C 5 H 12 ) is burned in oxygen to produce carbon dioxide and water. The heat of combustion of pentane is -4112 kJ/mole. The standard combustion reaction of pentane The standard heat of combustion of CH 4 (g) is 890 kJ / mol and of CO(g) 283 kJ / mol. 1. − 1000 . 3 kJ/mol of rxn 2. − 1214 kJ / mol correct 3. − 975 . 0 kJ/mol of rxn 4. − 690 . 72 kJ/mol of rxn 5. − 520 . 2 kJ/mol of rxn 6. − 1110 . 2 kJ/mol of rxn 7. − 2521 . 4 kJ/mol of rxn Explanation: The balanced equations for standard heats of combustion are CH 4 (g) + 2 O 2 (g) −→ CO 2 (g) + 2 H 2 O( ) H = 890 kJ / mol
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Homework #1 – holcombe – (51160) 2 CO(g) + 1 2 O 2 (g) −→ CO 2 (g) H = 283 kJ / mol The desired reaction can be obtained by doubling the first reaction and then adding it to twice the reverse of the second reaction: 2 CH 4 (g) + 4 O 2 (g) −→ 2 CO 2 (g) + 4 H 2 O( ) H = 2( 890 kJ / mol) = 1780 kJ / mol 2 CO 2 (g) −→ 2 CO(g) + O 2 (g) H = 2(283 kJ / mol) = 566 kJ / mol 2 CH 4 (g) + 3 O 2 (g) −→ 2 CO(g) + 4 H 2 O( ) H = 1214 kJ / mol 005 10.0 points Calculate the standard reaction enthalpy for the oxidation of nitric oxide to nitrogen dioxide 2 NO(g) + O 2 (g) 2 NO 2 (g) given N 2 (g) + O 2 2 NO(g) H = +180 . 5 kJ · mol −1 2 NO 2 (g) N 2 (g) + 2 O 2 (g) H
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13_new - Homework#1 holcombe(51160 This print-out should...

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