CHEM 103 Module 3 Exam Questions and Answers | Latest Update | Portage Learning

Show the calculation of the heat of reaction (ΔHrxn) for the reaction: 2 C2H6 (g) + 5 O2 (g) → 4 CO (g) + 6 H2O (l) by using the following thermochemical data: ΔHf 0 C2H6 (g) = -84.0 kJ/mole, ΔHf 0 CO (g) = -110.5 kJ/mole, ΔHf 0 H2O (l) = -285.8 kJ/mole Your Answer: 2 C2H6 (g) + 5 O2 (g) → 4 CO (g) + 6 H2O (l) ΔHrxn = ∑ n ΔHf 0 (products) - ∑ m ΔHf 0 (reactants) ΔHrxn = 4 ΔHf 0 (CO) + 6 ΔHf 0 (H2O) - 2 ΔHf 0 (C2H6 ) - 5 ΔHf 0 (O2) ΔHrxn = 4 (-110.5 kJ/mole) + 6 (-285.8 kJ/mole) - 2 (-84.0 kJ/mole) - 5 (0) ΔHrxn = (-442 kJ/mol) + (-1714.8 kJ/mol) - (-168) ΔHrxn = -2,156.8 - (-168) ΔHrxn = -1.988.8 kJ/mol 2 C2H6 (g) + 5 O2 (g) → 4 CO (g) + 6 H2O (l) ΔHf 0 C2H6 (g) = -84.0 kJ/mole, ΔHf 0 CO (g) = -110.5 kJ/mole, ΔHf 0 H2O (l) = -285.8 kJ/mole ΔHrxn = 2(+84.0) + 5(0) + 4(-110.5) + 6(-285.8) = - 1988.8 kJ/mole Question 6 Not yet graded / 10 pts Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the new pressure of a gas sample which has an original volume of 560 ml when collected at 1.05 atm and 32 oC when the volume becomes 1.35 liters at 50oC. Your Answer: P1V1/T1 = P2V2/T2 V1 = 0.560 L T1 = 32 = 273 = 305 K P1 = 1.05 atm V2 = 1.35 L P2 = ? T2 = 50 + 273 = 323 K P1V1/T1 = P2V2/T2 [(1.05) x (0.560)] / 305 = [(P2) x (1.35)] / 323 0.00193 = 1.35P2 /323 P2 = 0.4613 atm (Pi x Vi ) / Ti = (Pf x Vf ) / Tf 560 ml/1000 = 0.560 liters = Vi 1.05 atm = Pi 1.35 liters = Pf 32oC + 273 = 305oK = Ti