lee (ml37757) – Ch 5 Thermochemistry – tetzlaff – (201201) This print-out should have 27 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 difference in energy between reactants and products 2. the speed of the reaction 3. the relative concentrations of reactants and products in solution 4. the heat properties of the products 1. +350 kJ 5. the activation energy of the reaction 2. −50 kJ 002 10.0 points Refer to the potential energy diagram shown below. 10 350 9
8 7 6 300 5 4 3 2 250 1 0
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003 10.0 points Refer to the potential ...view middle of the document...
2. Work is done by the system as it occurs. 3. Work may be done on or by the system as the reaction occurs, depending upon the temperature. 4. Work is done on the system as it occurs. 007 10.0 points A reaction under certain conditions has an enthalpy change of ∆H = −118 kJ. This means that 118 kJ of heat would be liberated 1. no matter how the reaction were carried out. 2. if the reaction were carried out at constant pressure. 3. if the reaction were carried out at constant concentrations. 4. if the reaction were carried out in any type of calorimeter. 5. if the reaction were carried out at constant volume. 008 10.0 points The heat of combustion of butane (C4 H10 ), is −2623 kJ/mol. How much heat would be absorbed or evolved when 90 g butane is burned? 1. 2623 kJ evolved 2. 4070 kJ absorbed 3. 2.36 × 105 kJ evolved 4. 2.36 × 105 kJ absorbed 5. 2623 kJ absorbed 6. 4070 kJ evolved
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009 10.0 points The heat of combustion of propene (C3 H6 ) is −1911 kJ/mol. What is the change in energy when 84 grams of propene is combusted? 1. 160524 kJ released 2. 1911 kJ absorbed 3. 955.5 kJ released 4. 955.5 kJ absorbed 5. 22.75 kJ released 6. 160524 kJ absorbed 7. 1911 kJ released 8. 22.75 kJ absorbed 9. 3822 kJ released 010 10.0 points How much heat is given off when 11.0 grams of manganese is converted to Mn2 O3 (s) at standard state conditions? ∆Hf of Mn2 O3 (s) is −962.3 kJ/mol. 1. 96.4 kJ 2. 289 kJ 3. 964 kJ 4. 192 kJ
lee (ml37757) – Ch 5 Thermochemistry – tetzlaff – (201201) 5. 460 kJ 011 10.0 points How does the heat produced when 1 mole of C is burned directly to CO2 by the reaction C(s) + O2 (g) → CO2 (g) compare with that produced when 1 mole of C undergoes the two-step process: C(s) + 1 O2 (g) → CO(g) 2 3. −1502.4 kJ 4. −1450.85 kJ 5. −1100.36 kJ 6. −1826.2 kJ 7. −1225.7 kJ
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013 10.0 points Calculate the standard enthalpy change for the reaction 2 HCl(g) + F2 (g) → 2 HF( ) + Cl2 (g) given 4 HCl(g) + O2 (g) → 2 H2 O( ) + 2 Cl2 (g) ∆H 0 = −202.4 kJ/mol rxn 1 1 H2 (g) + F2 (g) → HF( ) 2 2 ∆H 0 = −600.0 kJ/mol rxn H2 (g) + 1 O2 (g) → H2 O( ) 2 ∆H 0 = −285.8 kJ/mol rxn 1. ∆H 0 = +1587.2 kJ/mol rxn 2. ∆H 0 = −1015.4 kJ/mol rxn 3. ∆H 0 = +1088.2 kJ/mol rxn 4. ∆H 0 = +1116.6 kJ/mol rxn 5. ∆H 0 = −1088.2 kJ/mol rxn 6. ∆H 0 = −516.6 kJ/mol rxn 7. ∆H 0 = +516.6 kJ/mol rxn 8. ∆H 0 = −1587.2 kJ/mol rxn 9. ∆H 0 = −1116.6 kJ/mol rxn
1 O2 (g) → CO2 (g) 2 if all reactants and products are at the same temperature and pressure? CO(g) + 1. More heat is produced in the two-step process. 2. The heat produced is the same either way. 3. More heat is produced in the one-step process. 4. Cannot tell without calculating the heat produced. 012 10.0 points Calculate the reaction enthalpy for the formation 2 Al(s) + 3 Cl2(g) −→ 2 AlCl3(s) , of anhydrous aluminum chloride using the data 2 Al(s) + 6 HCl(aq) −→ 2 AlCl3(aq) + 3 H2(g) ∆H ◦ = −1049 kJ HCl(g) −→ HCl(aq) ∆H ◦ = −74.8 kJ H2 (g) + Cl2 (g) −→ 2 HCl(g) ∆H ◦ = −185 kJ AlCl3(s) −→ AlCl3(aq) ∆H ◦ = −323 kJ...