CHEM103 /CHEM 103 Module 3 Exam Questions and Answers | Latest 2026–2027 Update | General Chemistry

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CHEM103 / CHEM 103 Module 3 Exam Questions & Answers | Latest 2026–2027 Update | General Chemistry I with Lab | Portage Learning | Verified Solutions ... | Grade A 2026 / 2027 Academic Year Q: Heat temp change = qtemp change = m x c x ∆t Answer -(40.5 x 4.184 x (∆t - 85.7)) = 36.8 x 4.184 x (∆t - 26.3) -(169.452 x (∆t - 85.7)) = 153.9712 x (∆t - 26.3) -(169.452∆t - 14522.0364) = 153.9712∆t - 4049.44256 -169.452∆t + 14522.0364 = 153.9712∆t - 4049.44256 18571.479 = 323.423∆t 57.4 C = ∆t 1. Show the calculation of the final temperature of the mixture when a 40.5 gram sample of water at 85.7C is added to a 36.8 gram sample of water at 26.3C in a coffee cup calorimeter. c (water) = 4.184 J/g C Q: qs↔l = mass x Heat of Fusion = m x ∆Hfusion Answer 120 x 0.334 = 40.08 kJ 2. Show the calculation of the energy involved in melting 120 grams of ice at 0oC if the Heat of Fusion for water is 0.334 kJ/g. Q: moles = grams/molecular weight Answer moles (S) = 42.8/32.07 = 1.335 mols S = ΔHrx x new moles / original moles q = -792 x ( 1.335/2 )= -526.7 kJ 3. Sulfur undergoes combustion to yield sulfur trioxide by the following reaction equation: 2 S + 3 O2 → 2 SO3 ΔH = - 792 kJ If 42.8 g of S is reacted with excess O2, what will be the amount of heat given off? Q: moles = grams/molecular weight Answer moles (H2S ) = 26.2/34.086 = 0.7686 mols H2S ΔHrx = q / (new moles / original moles) q reaction = - 58,791 J + (-295,365 J) = - 354,156 J = - 354,156 J x 1 kJ / 1000 J = - 354.16 kJ -431.8 / ( 0.7686/2) = -1123.6 kJ 4. Hydrosulfuric acid (H2S) undergoes combustion to yield sulfur dioxide and water by the following reaction equation: 2 H2S + 3 O2 → 2 SO2 + 2 H2O What is the ΔH of the reaction if 26.2 g of H2S reacts with excess O2 to yield 431.8 kJ? 1) q water = s (specific heat of water) x mass x Δt = 4.18 J / g / K x 500 g x (53.13 C - 25.00 C) = - 58,791 J q calorimeter = heat capacity x Δt = (10.5 kJ/C) x (53.13 C - 25.00 C) = - 295.365 kJ x 1000 J/1 kJ= - 295,365 J (new) moles C6H6 = 7.05 g / 78 = 0.0904 mole C6H6 ΔH = q / (new moles / original moles) ΔH = - 354.16 / (0.09048 / 2) = - 7828 kJ / mole (2) Isolated system (bomb calorimeter) (3) Exothermic (temperature of water rises due to heat given off by combustion reaction) 5. A sample of benzene (C6H6), weighing 7.05 g underwent combustion in a bomb calorimeter by the following reaction: 2 C6H6 (l) + 15 O2 (g) → 12 CO2 (g) + 6 H2O (l) Q: The heat given off was absorbed by 500 g of water and caused the temperature of the water and the calorimeter to rise from 25.00 to 53.13 C. The heat capacity of water = 4.18 J/g/C and the heat capacity of the calorimeter = 10.5 kJ/C. (1) what is the ΔH of the reaction? Using the definitions at the beginning of the module describe (2) the calorimeter + contents, (3) the type of process. Answer q water = s (specific heat of water) x mass x Δt = 4.18 J / g / K x 100 g x 5.14 K = 2148.52 J q calorimeter = heat capacity x Δt = 150 x 5.14 K = 771 J q reaction = 771 J + 2148.52 J = + 2919.52 J = + 2919.52 J x 1 kJ / 1000 J = 2.92 kJ (1) ΔH reaction = 2.92 kJ / 14.5 g NaHCO3 / 84 g NaHCO3 / mol NaHCO3 = + 16.9 kJ / mol (2) The calorimeter + contents (no lid) = open system, (3) Endothermic process 6. A sample of 14.5 g of sodium bicarbonate (NaHCO3) was dissolved in 100 ml of water in a coffee-cup calorimeter with no lid by the following reaction Answer NaHCO3 (s) → Na+ (aq) + HCO3- (aq) Q: If the temperature of the water and the calorimeter (heat capacity of calorimeter = 150 J/C) decreases from 25.00oC to 19.86C, (1) what is the ΔH of the reaction? Using the definitions at the beginning of the module describe (2) the calorimeter + contents, (3) the type of process. Answer 2 (2 NH3 (g) → 3 H2 (g) + 2 N2 (g) ΔH = + 91.8 kJ) 3 (2 H2 (g) + O2 (g) → 2 H2O (g) ΔH = - 483.7 kJ) 2 (N2 (g) + O2 (g) → 2 NO (g) ΔH = 180.6 kJ) 4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (g) ΔHrxn = - 906.3 kJ ΔHrxn = 2 (+ 91.8) + 3 (-483.7) + 2 (180.6) = - 906.3 kJ 7. The combustion of ammonia by the following reaction yields nitric oxide and water Answer 4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (g) Determine the heat of reaction (ΔHrxn) for this reaction by using the following thermochemical data: N2 (g) + O2 (g) → 2 NO (g) ΔH = 180.6 kJ N2 (g) + 3 H2 (g) → 2 NH3 (g) ΔH = - 91.8 kJ 2 H2 (g) + O2 (g) → 2 H2O (g) ΔH = - 483.7 kJ Q: C2H5OH (l) → 2 C (graphite) + 3 H2 (g) + 1/2 O2 (g) ΔHf0 (C2H5OH) = -

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