CHEM 103 MODULE 3 EXAM WITH ANSWERS

CHEM 103 MODULE 3 EXAM WITH ANSWERS 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 a 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 calorimeter and mixture of HCl + NaOH Your Answer: 1. The type of thermochemical process is an acid base reaction or neutralization reaction where heat is given off called a exothermic reaction. 2. The calorimeter and mixture of HCl + NaOH would be described as enthalpy or the total energy of the system. 1. Heat given off = Exothermic process 2. With lid = closed system 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 31.6 gram sample of water at 92.3oC is added to a 40.5 gram sample of water at 25.1oC in a coffee cup calorimeter. c (water) = 4.184 J/g oC 2. Show the calculation of the energy involved in melting 85 grams of ice at 0oC if the Heat of Fusion for water is 0.334 kJ/g. Your Answer: 1. - (mwarm H2o x Cwarm H2O x ∆twarm H2O) = (mcool H2O x ccool H2O x ∆tcool H2O) - [31.6 g x 4.184 J/g°C x (Tmix - 92.3°C)] = [40.5g x 4.184 J/g°C x (Tmix - 25.1°C)] - [132.2144 J/°C x (Tmix - 92.3°C)] = [169.452 J/°C x (Tmix - 25.1°C)] -132.2144Tmix + 12,203.3891 = 169.452Tmix - 4,253.2452 16,456.6343 = 301.6664Tmix Tmix = 54.55°C 2. qI↔s = m x ∆Hfusion qI↔s = 85 g x 0.334 kJ/g qI↔s = 28.39 kJ and because heat is added + 28.39 kJ 1. - (mwarn H2O x cwarn H2O x ∆twarn H2O) = (mcool H2O x ccool H2O x ∆tcool H2O) - [31.6 g x 4.184 J/g oC x (Tmix - 92.3oC)] = [(40.5 g x 4.184 J/g oC x (Tmix - 25.1oC)] - [132.2144 J/oC x (Tmix - 92.3oC)] = [(169.452 J/oC x (Tmix - 25.1oC)] Tmix = 54.6oC 2. ql↔s = m x ∆Hfusion = 85 g x 0.334 kJ/g = 28.39 kJ (since heat is added) = + 28.39 kJ 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 18.3 g of S is reacted with excess O2 to yield sulfur trioxide by the following reaction equation. Report your answer to 4 significant figures. 2 S (s) + 3 O2 (g) → 2 SO3 (g) ΔH = - 792 kJ Your Answer: 2 S (s) + 3 O2 (g) → 2 SO3 (g) ΔH = - 792 kJ ∆Hrx = 2 moles of S reaction uses 18.3 g S = 18.3/32.07 = 0.5706 mole S q = ∆Hrx x new moles/original moles q = -792kJ x 0.5706 mole S / 2 mole S q = -225.9576 kJ q = -226.0 2 S (s) + 3 O2 (g) → 2 SO3 (g) ΔH = - 792 kJ ΔHrx is for 2 mole of S reaction uses 18.3 g S = 18.3/32.07 = 0.5706 mole S q = ΔHrx x new moles / original moles q = -792 kJ x 0.5706 mole S / 2 mole S = 226.0 kJ given off 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 Ca (s) + 2 C (graphite) + 3 O2 (g) → 2 CaCO3 (s) by using the following thermochemical data: 2 Ca (s) + O2 (g) → 2 CaO (s) ΔH = - 1270.18 kJ C (graphite) + O2 (g) → CO2 (g) ΔH = - 393.51 kJ CaO (s) + CO2 (g) → CaCO3(s) ΔH = - 178.32 kJ Your Answer: 2 Ca (s) + O2 (g) → 2 CaO (s) ΔH = - 1270.18 kJ C (graphite) + O2 (g) → CO2 (g) ΔH = - 393.51 kJ CaO (s) + CO2 (g) → CaCO3(s) ΔH = - 178.32 kJ 2 Ca (s) + 2 C (graphite) + 3 O2 (g) → 2 CaCO3 (s) ΔHrxn ΔHrxn = (1270.18) + 2 x (-393.51) + 2 x (-178.32) ΔHrxn = 2413.84 kJ 2 Ca (s) + O2 (g) → 2 CaO (s) ΔH = - 1270.18 kJ 2 (C (graphite) + O2 (g) → CO2 (g) ΔH = - 393.51 kJ) 2 (CaO (s) + CO2 (g) → CaCO3(s) ΔH = - 178.32 kJ) 2 Ca (s) + 2 C (graphite) + 3 O2 (g) → 2 CaCO3 (s) ΔHrxn= - 2413.84 kJ ΔHrxn = -1270.18 + 2(-393.51) + 2(-178.32) = - 2413.84 Click this link to access the Periodic Table. This may be helpful throughout the exam.