WA 7 Dearing

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Thomas Edison State College *

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Chemistry

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Apr 3, 2024

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Name: Eric Dearing College ID: 0550796 Thomas Edison State College General Chemistry I with Labs (CHE-121) Section no.: Semester and year: May 2014 Written Assignment 7: Energy and Thermochemistry Answer all assigned questions and problems, and show all work. 1. Consider this reaction: 2CH 3 OH( l ) + 3O 2 ( g ) 4H 2 O( l ) + 2CO 2 ( g ) H = –1452.8 kJ/mol a. Is this reaction endothermic or exothermic? (2 points) exothermic b. What is the value of ∆ H if the equation is multiplied throughout by 2? (2 points) x2 throughout would double H output, H = –1452.8 x2 = –2905.6 kJ/mol c. What is the value of ∆ H if the direction of the reaction is reversed so that the products become the reactants and vice versa? (2 points) Reaction becomes endothermic and ∆ H becomes +, H = 1452.8 kJ/mol d. What is the value of ∆ H if water vapor instead of liquid water is formed as the product? (2 points) an additional 44kJ per mole of H2O are required. If water vapor is produced instead of liquid water: H2O (l) -> H2O (g) = 44 kJ/mol H2O x 4 mol = -1452.8 + 44 x 4 mol = -1276.8 kJ/mol 2. The first step in the industrial recovery of zinc from the zinc sulfide ore is roasting, that is, the conversion of ZnS to ZnO by heating: 2ZnS( s ) + 3O 2 ( g ) 3ZnO( s ) + 2SO 2 ( g ) H = –879 kJ/mol Calculate the heat evolved (in kJ) per gram of ZnS roasted. (5 points) Find ∆ H for 1 mol ZnS, ∆ H = –879 kJ/mol / 2 = ∆ H = –437.5 kJ/mol ZnS amu 97, –437.5 / 97amu = -4.51 kJ/g 3. A 6.22 kg piece of copper metal is heated from 20.5 °C to 324.3 °C. Calculate the heat absorbed (in kilojoules) by the metal. Specific heat Cu = 0.385 J/g°C. (5 points) ∆t = 324.8°C – 20.5°C = 303.8°C q = ms∆t, q = m 6220g x s 0.385 J/g°C x ∆t 303.8°C = 7.28x10 5 kJ 1
4. A sheet of gold weighing 10.0 g and at a temperature of 18.0 °C is placed flat on a sheet of iron weighing 20.0 g and at a temperature of 55.6 °C. What is the final temperature of the combined metals? Assume that no heat is lost to the surroundings. (5 points) Specific heat Au = 0.129 J/g o C, Specific heat Fe = 0.444 J/g o C 10g x 0.129J x (t-18C) = 20g x 0.444J x (t-55.6) 10g x 0.129 = 1.29 x -18 = -23.2C, 20 x 0.444 = 8.88 x 55.6 = 494 1.29 t -23.2 = -8.88 t + 494, 1.29 t -23.2 + (-494) = 517 -8.88 + -1.29 = 10.17 10.17 t = 517, t = 517/10.17 = Final Temp 50.8C 5. A 0.1375 g sample of solid magnesium is burned in a constant-volume bomb calorimeter that has a heat capacity of 3024 J/°C. The temperature increases by 1.126°C. Calculate the heat given off by the burning Mg, in kJ/g and in kJ/mol. (5 points) (Reference: Chang 6.37) 3024 J/°C x 1.126°C = 3405kJ / 0.1375g = - 24.76kJ/g 24.76kJ/g / Mg 24.3 amu = - 602kJ/mol 6. Calculate the heat of decomposition for this process at constant pressure and 25°C: (10 points) CaCO 3 ( s ) CaO( s ) + CO 2 ( g ) (1 mol CaO -635.6) + (1 mol CO 2 -393.5) – (1 mol CaCO 3 -1206.9) = 177.8 kJ Substance ∆H° f , kJ/mol CaCO 3 ( s ) –1206.9 CaO( s ) –635.6 CO 2 ( g ) –393.5 (Reference: Chang 6.51) 7. Methanol, ethanol, and n -propanol are three common alcohols. When 1.00 g of each of these alcohols is burned in air, heat is liberated as shown by the following data: (a) methanol (CH 3 OH), –22.6 kJ/g; (b) ethanol (C 2 H 5 OH), –29.7 kJ/g; (c) n - propanol (C 3 H 7 OH), –33.4 kJ/g. Calculate the heats of combustion of these alcohols in kJ/mol. (12 points) (Reference: Chang 6.55) (a) methanol (CH 3 OH), –22.6 kJ/g CH 3 OHamu 32, 1g / 32 = 0.031mol, -22.6 kJ/g / 0.031mol = -729 kJ/mol (b) ethanol (C 2 H 5 OH), –29.7 kJ/g C 2 H 5 OHamu 46, 1g / 46 = 0.022mol, -29.7 kJ/g / 0.022mol = -1350 kJ/mol (c) n -propanol (C 3 H 7 OH), –33.4 kJ/g C 3 H 7 OHamu 60, 1g / 60 = 0.017mol, -33.4 kJ/g /0.017mol = -1965kJ/mol 2
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