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# HW11 - Cheung Anthony Homework 11 Due midnight Inst McCord...

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Cheung, Anthony – Homework 11 – Due: Nov 26 2006, midnight – Inst: McCord 1 This print-out should have 30 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. Remember our old book (and others) refer to internal energy with an E, not U. So assume that Δ E = Δ U for those older questions. 001 (part 1 of 1) 10 points In the manufacture of nitric acid by the oxi- dation of ammonia, the first product is nitric oxide. The nitric oxide is then oxidized to nitrogen dioxide: 2 NO(g) + O 2 (g) -→ 2 NO 2 (g) Calculate the standard reaction enthalpy for the reaction above (as written) using the following data: N 2 (g) + O 2 (g) -→ 2 NO(g) Δ H = 180 . 5 kJ N 2 (g) + 2 O 2 (g) -→ 2 NO 2 (g) Δ H = 66 . 4 kJ 1. - 252 . 4 kJ/mol rxn 2. - 114 . 1 kJ/mol rxn 3. - 128 . 2 kJ/mol rxn 4. - 520 . 2 kJ/mol rxn 5. - 100 . 3 kJ/mol rxn 6. - 690 . 72 kJ/mol rxn 7. - 975 . 0 kJ/mol rxn 002 (part 1 of 1) 10 points Calculate the reaction enthalpy for the forma- tion 2 Al(s) + 3 Cl 2 (g) -→ 2 AlCl 3 (s) , of anhydrous aluminum chloride using the data 2 Al(s) + 6 HCl(aq) -→ 2 AlCl 3 (aq) + 3 H 2 (g) Δ H = - 1049 kJ HCl(g) -→ HCl(aq) Δ H = - 74 . 8 kJ H 2 (g) + Cl 2 (g) -→ 2 HCl(g) Δ H = - 185 kJ AlCl 3 (s) -→ AlCl 3 (aq) Δ H = - 323 kJ 1. - 1406 . 8 kJ 2. - 1450 . 85 kJ 3. - 1502 . 4 kJ 4. - 1100 . 36 kJ 5. - 1225 . 7 kJ 6. - 1883 . 5 kJ 7. - 1826 . 2 kJ 003 (part 1 of 1) 10 points Calculate the standard reaction enthalpy for the reaction. CH 4 (g) + H 2 O(g) CO(g) + 3 H 2 (g) given 2 H 2 (g) + CO(g) CH 3 OH( ) Δ H = - 128 . 3 kJ · mol - 1 2 CH 4 (g) + O 2 (g) 2 CH 3 OH( ) Δ H = - 328 . 1 kJ · mol - 1 2 H 2 (g) + O 2 (g) 2 H 2 O(g) Δ H = - 483 . 6 kJ · mol - 1 1. +412 . 1 kJ · mol - 1 2. +42 . 0 kJ · mol - 1 3. +216 kJ · mol - 1 4. +206 . 1 kJ · mol - 1 5. +155 . 5 kJ · mol - 1 004 (part 1 of 1) 10 points Calculate the standard enthalpy change for the reaction 2 HCl(g) + F 2 (g) 2 HF( ) + Cl 2 (g) given

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Cheung, Anthony – Homework 11 – Due: Nov 26 2006, midnight – Inst: McCord 2 4 HCl(g) + O 2 (g) 2 H 2 O( ) + 2 Cl 2 (g) Δ H 0 = - 202 . 4 kJ/mol rxn 1 2 H 2 (g) + 1 2 F 2 (g) HF( ) Δ H 0 = - 600 . 0 kJ/mol rxn H 2 (g) + 1 2 O 2 (g) H 2 O( ) Δ H 0 = - 285 . 8 kJ/mol rxn 1. Δ H 0 = - 1015 . 4 kJ/mol rxn 2. Δ H 0 = +1116 . 6 kJ/mol rxn 3. Δ H 0 = - 1587 . 2 kJ/mol rxn 4. Δ H 0 = +1587 . 2 kJ/mol rxn 5. Δ H 0 = - 1088 . 2 kJ/mol rxn 6. Δ H 0 = - 1116 . 6 kJ/mol rxn 7. Δ H 0 = - 516 . 6 kJ/mol rxn 8. Δ H 0 = +516 . 6 kJ/mol rxn 9. Δ H 0 = +1015 . 4 kJ/mol rxn 10. Δ H 0 = +1088 . 2 kJ/mol rxn 005 (part 1 of 1) 10 points Consider the reaction 4 FeO(s) + O 2 (g) -→ 2 Fe
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HW11 - Cheung Anthony Homework 11 Due midnight Inst McCord...

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