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5 Thermochemistry Solutions to Exercises (b) -1849.5 X 1 mol 1000 = 4.616 kJ/kg 1 mol 40.065 1kg Propylene: -> (a) = 3(-393.5 + 3(-241.82 - 9/2(0) = -1926.4 = (b) -1926.4kJ 1 mol C₃H₆ 1000 C₃H₆ = -4.578 kJ/kg C₃H₆ 1 mol C₃H₆ 42.080 1kg C₃H₆ Propane: C₃H₈(g) -> (a) = + (-103.8 - 5(0) = -2044.0 = (b) 1 mol C₃H₈ 1000 = -4.635 kJ/kg C₃H₈ 1 mol C₃H₈ 44.096 C₃H₈ (c) These three substances yield nearly identical quantities of heat per unit mass, but propane is marginally higher than the other two. 5.88 = -393.5 kJ + 2(-241.82 (-74.8 - 2(0) kJ = -802.3 kJ = = -679.9 kJ + 4(-268.61 kJ) (-74.8 4(0) kJ = -1679.5 kJ The second reaction is twice as exothermic as the first. The "fuel values" of hydrocarbons in a fluorine atmosphere are approximately twice those in an oxygen atmosphere. Note that the difference in values for the two reactions is in the for the products, since the for the reactants is identical. 5.89 Analyze/Plan. Given population, Cal/person/day and kJ/mol glucose, calculate kg glucose/yr. Calculate kJ/yr, then kg/yr. 1 billion = 1 X 365 day = 1 yr. 1 Cal = 1 kcal, 4.184 kJ = 1 kcal = 1 Cal. Solve. 6.8 persons person 1500 Cal day 365 1 yr day 4.184 1Cal = 1.5577 10¹⁶ = 1.6 10¹⁶ kJ/yr 1.5577 X 10¹⁶ kJ 1 mol 2803 C₆H₁₂O₆ X 180.2 1 mol C₆H₁₂O₆ 1kg = 10¹² kg C₆H₁₂O₆/y yr C₆H₁₂O₆ Check. 1 10¹² kg is 1 trillion kg of glucose. 5.90 (a) Use density to change L to molar mass to change g to mol, heat of combustion to change mol to kJ. Ethanol is gasoline is C₈H₁₈. From Exercise 5.79 (c), heat of combustion of ethanol is -1234.8 kJ/mol. 1000 1L mL X 0.79g 1mL X 1 mol 1 mol 1234.81 127