CHEM 103 MODULE 3 EXAM

ivy [COMPANY NAME] [Company address] CHEM 103 MODULE 3 EXAM Question 1 CHEM 103 MODULE 3 EXAM 2022 Click this link to access the Periodic Table. This may be helpful throughout the exam. A reaction between HCl and NaOH is being studied in a styrofoam coffee cup with NO lid and the heat given off is measured by means of a thermometer immersed in the reaction mixture. Enter the correct thermochemistry term to describe the item listed. 1. The type of thermochemical process 2. The amount of heat released in the reaction of HCl with NaOH 1. Heat given off = Exothermic process 2. The amount of heat released = Heat of reaction Question 2 Click this link to access the Periodic Table. This may be helpful throughout the exam. 1. Show the calculation of the final temperature of the mixture when a 22.8 gram sample of water at 74.6 oC is added to a 14.3 gram sample of water at 24.3 oC in a coffee cup calorimeter. c (water) = 4.184 J/g oC 2. Show the calculation of the energy involved in freezing 54.3 grams of water at 0 oC if the Heat of Fusion for water is 0.334 kJ/g 1. - (mwarn H2O x cwarn H2O x ∆twarn H2O) = (mcool H2O x ccool H2O x ∆tcool H2O) - [22.8 g x 4.184 J/g oC x (Tmix - 74.6 oC)] = [14.3 g x 4.184 J/g oC x (Tmix - 24.3 oC)] - [95.3952 J/ oC x (Tmix - 74.6 oC)] = [59.8312 J/ oC x (Tmix - 24.3 oC)] Tmix = 55.2 oC 2. ql↔s = m x ∆Hfusion = 54.3 g x 0.334 kJ/g = 18.14 kJ (since heat is removed) = - 18.14 kJ Question 3 Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the amount of heat involved if 35.6 g of H2S is reacted with excess O2 to yield sulfur trioxide and water by the following reaction equation. Report your answer to 4 significant figures. 2 H2S (g) + 3 O2 (g) → 2 SO2 (g) + 2 H2O (g) ΔH = - 1124 kJ 1 mol H2s = 34.1 g of H2S = 603.2 kj (35.6g/34.1 g) x -603.2 kJ = (1.0439) x (-603.2 kJ) = -629. 7 kj 2 H2S (g) + 3 O2 (g) → 2 SO2 (g) + 2 H2O (g) ΔH = - 1124 kJ ΔHrx is for 2 mole of H2S reaction uses 35.6 g of H2S = 35.6/34.086 = 1.044 mole of H2S q = ΔHrx x new moles / original moles q = -1124 kJ x 1.044 mole of H2S / 2 mole H2S = 586.7 given off Question 4 Click this link to access the Periodic Table. This may be helpful throughout the exam. f 4 f f 2 Show the calculation of the heat of reaction (ΔHrxn) for the reaction: 3 C (graphite) + 4 H2 (g) → C3H8 (g) by using the following thermochemical data: C (graphite) kJ + O2 (g) → CO2 (g) ΔH = - 393.51 2 H2 (s) kJ + O2 (g) → 2 H2O(l) ΔH = - 571.66 C3H8 (g) + 5 O2 (g) → 3 CO2 (g) + 4 H2O(l) ΔH = - 2220.0 kJ Your Answer: 3 (C (graphite) + O2 (g) → CO2 (g) ΔH = - 393.51 kJ) 2 (2 H2 (s) + O2 (g) → 2 H2O(l) ΔH = - 571.66 kJ) 3 CO2 (g) + 4 H2O(l) → C3H8 (g) + 5 O2 (g) ΔH = + 2220.0 kJ 3 C (graphite) + 4 H2 (g) → C3H8 (g) ΔHrxn = - 103.85 kJ ΔHrxn = 3(- 393.51) + 2(- 571.66) + 2220.0 = - 103.85 Question 5 Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the heat of reaction (ΔHrxn) for the reaction: 2 CH4 (g) + 3 O2 (g) → 2 CO (g) + 4 H2O (l) by using the following ΔHf 0data: ΔH0CH (g) = -74.6 kJ/mole, ΔH 0CO (g) = -110.5 kJ/mole, ΔH 0H O (l) = -285.8 kJ/mole 2 CH4 (g) + 3 O2 (g) → 2 CO (g) + 4 H2O (l) ΔHf 0CH4 (g) = -74.6 kJ/mole, ΔHf 0CO (g) = -110.5 kJ/mole, ΔHf 0H2O (l) = -285.8 kJ/mole ΔHrxn = 2(+74.6) + 3(0) + 2(-110.5) + 4(-285.8) = - 1215.0 kJ/mole Question 6 Click this link to access the Periodic Table. This may be helpful throughoutthe exam. Show the calculation of the new temperature of a gas sample has an original volume of 740 ml when collected at 710 mm and 35 oC when the volume becomes 460 ml at 1.20 atm. P1V1/T1 = P2V2/T2 T2 = 1.2 x 460x308 / 0.934 x 740 = 245.98 kelvin = -27.17oC (Pi x Vi) / Ti = (Pf x Vf ) / Tf 740 ml/1000 = 0.740 liters = Vi 710 mm/760 = 0.934 atm = Pi 460 ml/1000 = 0.460 liters = Vf 1.20 atm = Pf 35 oC + 273 = 308 oK = Ti (0.934) x (0.740) / 308 = (1.20) x (0.460) / Tf Tf= 246 oK Question 7 Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the volume occupied by a gas sample containing 0.632 mole collected at 710 mm and 35 oC. P x V = n x R x T Question 9 0.632 mole = n R = 0.0821 710 mm/760 = 0.934 atm = P 35oC + 273 = 308oK = T (0.934) x V = (0.632) x (0.0821) x (308) V = 17.1 liters Question 8 Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the volume of CO2 gas formed by the combustion of 25.5 grams of C6H6 at 20 oC and 1.25 atm. The combustion of benzene (C6H6) takes place by the following reaction equation. 2 C6H6 (g) + 15 O2 (g) → 12 CO2 (g) + 6 H2O (g) (MW = 78) (MW = 32) (MW = 44) (MW = 18) 2 C6H6 (g) + 15 O2 (g) → 12 CO2 (g) + 6 H2O (g) 25.5 grams 37.75 liters ↓ ↑ by V = nRT / P = (1.9614)(0.0821)(293)/1.25 0.3269 mol → 12/2 x 0.3269 mol Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the mole fraction of each gas in a 1.00 liter container holding a mixture of 7.60 g of N2 and 8.40 g of O2 at 25 oC. nN2 = gN2 / (MWN2) = 7.60 g / 28.02 = 0.2712 mol nO2 = gO2 / (MWO2) = 8.40 g / 32.00 = 0.2625 mol XN2 = 0.2712 / (0.2712 + 0.2625) = 0.5082 XO2 = 0.2625 / (0.2712 + 0.2625) = 0.4918 Question 10 Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the molecular weight of an unknown gas if the rate ofeffusion of carbon dioxide gas (CO2) is 1.83 times faster than that of an unknown gas. (rN2 /runknown) 2 = MWunknown / MWCO2 (1.83/1) 2 = MWunknown / 44.01 MWunknown = (1.83) 2 x 44.01 = 147.4 calculation error; minus 2.5