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116 Chapter 6 Thermochemistry Rearrange to solve for CA. CA (161.13 = 176.904J = = 1.097896 J = 1.10 °C Check: The units °C) are correct. The magnitude of the answer °C) makes physical sense because the mass of substance B is greater than the mass of substance A by a factor of ~4.1 and the temperature change for sub- stance A is greater than the temperature change of substance B by a factor of ~4.4; so the heat capacity of substance A will be slightly smaller. Calorimetry 6.71 = and = qv = AH PAV. Because combustions always involve expansions, expansions do work and therefore have a negative value. Combustions are always exothermic and therefore have a negative value. This means that is more negative than so A (-25.9 kJ) is the constant-volume process, and B kJ) is the constant-pressure process. Given: 0.514 g bomb calorimeter; Tᵢ = 25.8 °C; = 29.4 °C; = 5.86 kJ/°C Find: Conceptual Plan: AT then then g C₁₂H₁₀ mol C₁₂H₁₀ 1 mol then mol Solution: = Tᵢ = 29.4 °C 25.8 = 3.6 °C then = = = 21.096 kJ then = = -21.096 kJ and 0.514 X 1 mol C₁₂H₁₀ = 0.00333333 mol C₁₂H₁₀ then = 0.00333333 -21.096kJ = -6.3 X kJ/mol Check: The units (kJ/mol) are correct. The magnitude of the answer (-6000) makes physical sense because such a large amount of heat is generated from a very small amount of biphenyl. 6.75 Given: 0.103 g zinc; coffee-cup calorimeter; Tᵢ = 22.5 °C; Tf = 23.7 °C; 50.0 mL solution; (solution) = 1.0 g/mL; = kJ/g °C Find: Conceptual Plan: Tᵢ, and mL soln g soln then T, then = g Zn mol Zn then mol Zn 1 mol Solution: = - Tᵢ = 23.7 22.5 °C = and 50.0 mL X 1.0 mL then = = 50.0 J = 250.8 J then = = -250.8 J and 0.103 X 65.38 1 mol = 0.00157540 mol Zn then = mol 0.00157540 -250.8J mol Zn = 1.6 X = 1.6 X kJ/mol Check: The units (kJ/mol) are correct. The magnitude of the answer (-160) makes physical sense because such a large amount of heat is generated from a very small amount of zinc. Copyright © 2017 Pearson Education, Inc.