Thermochemical I Practice Problems with Answers

Thermochemical I Practice Problems with Answers Part I: Heat capacity, Specific heat (SpH) and Molar heat capacity 1. It takes 3.34 x 10-1 kJ to melt 1.00 gram of ice. If you have 100.00 grams of ice in a glass containing 250.0 grams of lemonade, how much ice must melt to lower the temperature of the lemonade from 25.0 oC to 10.0 oC. Useful information: Specific heat of lemonade = 4.00 J/g oC Specific heat of water = 4.184 J/g oC q = SpH x mass x T q = 4.0 J/goC x 250.0 g x 15.0 oC q = 1.50 x 101 kJ (15.0 kJ/x grams) = (0.334 kJ/1.00 g) x = 45.0 grams of ice must melt 2. A gold ring has a mass of 5.00 grams. The ring is heated from a temperature of 25.0 oC to 28.0 oC. Given that the specific heat of gold is 0.129 J/goC, what is the amount of heat absorbed by the ring. q = SpH x mass x T q = 0.129 J/goC x 5.00 g x 3.0 oC q = 1.9 J 3. It takes 18.7 cal of energy to raise the temperature of 45.6 g of lead by 13.3 oC. Find the specific heat capacity of lead in J/g oC. Determine the molar heat capacity of lead. 18.7 cal x (4.184 J/1 cal) = 78.2 J q = SpH x mass x T 78.2 J = SpH x 45.6 g x 13.3 oC SpH = 0.129 J/goC 45.6 g Pb x (1 mole Pb/207 g/mol) = 0.220 mol Pb 78.2 J/0.220 mole Pb = 355 J/mol 4. The heat of combustion for camphor (C10H16O) is 5903. 6 kJ/mol. When 0.1204 grams of camphor is burnt in a bomb calorimeter the temperature raises from 25.20 oC to 27.48 oC. Find the heat capacity of camphor. Heat gained by calorimeter = heat lost by combustion of camphor 0.1204 g x (1 mole camphor/152 g) = 7.920 x 10-4 mol camphor 5903.6 kJ/mol x (1000 J/1 kJ) = 5.9036 x 106 J 5.9036 x 106 J x 7.920 x 10-4 mol = 4.676 x 103 J heat lost by camphor 4.676 x 103 J = SpH x 2.28 oC = 2.05 X103 J/oC 5. A 50.0 mL liquid sample is heated from 15.0 oC to 20.0oC. Determine the amount of energy absorbed by the liquid (in joules) given that its specific heat is 0.809 cal/goC and its density is 0.897 g/cm3. (x/50.0 mL) = 0.897 g/mL x = 44.9 gram of sample q = 0.809 cal/goC x 5.0 oC x 44.9 g q = 180 cal 180 cal x (4.184 J/1 cal) = 760 J 6. The specific heat for water is 4.1796 J/goC. What is water’s molar heat capacity? 18.0 g /mol x 4.1796 J/goC = 75.2 J/mol oC 7. In an experiment, 45.0 mL of 1.00 M HCl and 55.0 mL of 0.85 M NaOH are mixed together. The initial temperature of the two solutions prior to mixing was 24.0 oC. After the two solutions have been mixed the temperature of the solution was recorded at 30.5 oC. If the density of the two solutions is 1.03 g/mL determine the molar heat (J/mol) for the neutralization of this acid and base. HCl (aq) + NaOH (aq → H2O (l) = NaCl (aq) T = 6.5 oC 0.045 L x 1.00 M = 0.045 mol HCl 0.055 L x 0.85 M = 0.047 mol NaOH 45.0 mL + 55.0 mL = 100.0 mL of solution (x /100.0 mL) = 1.03 g/mL x = 103.0 grams of solution q = SpH x mass x T q = 4.184 J x 103.0 g x 6.5 oC q = 2.8 x 103 J Limiting reactant problem Have a 1: 1 stoichiometric relationship between HCl and NaOH 0.045 mol HCl x (1 mole NaOH/1 mol HCl) = 0.045 mol NaOH needed to neutralize 0.045 mol of HCl 0.047 mole NaOH x (1 mole HCl/ 1 mole NaOH) = 0.047 mol HCl needed to neutralize 0.047 mol NaOH HCl is the limiting reactant 2.8 x 103 J / 0.045 mol = 6.2 x 104 J/mol 8. Steam burns: A cup of boiling water (100.00 grams) is spilled on your hand. Determine the amount of energy your hand just absorbed. The temperature of body is 37.0 oC. The Hvap for water = 540 cal/g SpH for water is 1.00 cal/goC q = SpH x mass x T q = 1.00 cal/goC x 100.00 g x 63.0 oC q = 6.30 x 103 cal heat absorbed by hand when hot water lands on hand 540 g/cal x 100.00 g = q q = 5.40 x 104 cal heat absorbed by hand when steam → water Total heat absorbed by hand 6.3 x 103 cal + 5.40 x 104 cal = 60,300 cal 9. The combustion of hydrogen gas provides 3X as much energy per gram as gasoline. Determine the volume of liquid hydrogen gas required to furnish the equivalent amount of energy that would be gained if 80.0 liters of gasoline is combusted. Useful information: density of hydrogen gas is 0.710 g/ml density of gasoline is 0.740 g/ml [80.0 L x (1000 mL/1 L)] x 0.740 g/mL = 59,200 grams of gasoline H2 gas generates 3 times energy /gram gasoline 59,200 gr gasoline x (0.33/gram gasoline = 19,700 g of H2 (l) needed V = 19,700 g/0.710 g/mL = 2.77 x 105 mL or 277 L of H2 gas needed Part II: Hess's Law 1. Determine the Horxn for the reaction boron (s) + oxygen (g) → diboron trioxide (s) 4 B (s) + 3 O2 (g) → 2 B2O3 (s) Hrxno = [(2 mol x -1263 kJ/mol B2O3)] – [(4 mol x 0 kJ/mol O2) + (3 mol x 0 kJ/mol B)] = -2526 kJ 2. Determine the Horxn for the reaction ____ CH4 (g) + __ O2 (g) → __ CO2 (g) + __ H2O (l) CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (l) Hrxno = [(1 mol x -394 kJ/mol CO2) + (2 mol x -286 kJ/mol H2O)] – [(1 x -75 kJ/mol CH4 + 0 kJ/mol)] = -891 kJ 3. Determine the H for the synthesis of diborane given the following information: Find the Hrxn for the synthesis of diborane 2 B(s) + 3 H2 (g) → B2H6 (g) Useful information: 2 B(s) + 3/2 O2 (g) → B2O3 (g) H = -1273.0 kJ B2H6 (g) 3 O2 (g) → B2O3 (g) + 3 H2O (g) H = -2035.0 kJ H2 (g) + ½ O2 (g) → H2O (l) H = -286.0 kJ H2O (l) → H2O (g) H = 44.0 kJ 2 B(s) + 3/2 O2 (g) → B2O3 (g) B2O3 (g) 3 H2O (g) →B2H6 (g) + 3 O2 (g) 3 (H2O (l) → H2O (g)) 3 (H2 (g) + ½ O2 (g) → H2O (l)) ____________________________ 2 B(s) + 3 H2 (g) → B2H6 (g) H = -1273.0 kJ H = +2035.0 kJ H = 3 (44.0 kJ) H = 3 (-286.0 kJ) _______________ H = 36.0 kJ 4. Using the Ho tables in your text, determine the Hrxno for the following reaction: sulfur (s) + oxygen (g) ----> sulfur dioxide (g) + heat • • • • Is this reaction endothermic or exothermic? Would the system's temperature increase or decrease in temperature? Determine the amount of heat evolved when 275.0 grams of sulfur is burned in excess oxygen gas. Determine the amount of heat released when 150.0 grams of sulfur dioxide gas is formed S (s) + O2 (g) → SO2 (g) Hrxno = [(1 mol x -296.9 kJ/mo SO2)] [ 0 kJ/mol] Hrxno = -296.9 kJ Reaction is exothermic (negative H) & the systems temperature would decrease for energy left the system to go into the surroundings 275.0 g S x (1 mol/32.0 g S) = 8.59 mol S 8.59 mol S x (1 mol SO2/1 mol S) = 8.59 mole SO2 (8.59 mol SO2 / x) = (1 mol SO2/-296.9 kJ) x = - 2.55 x 103 kJ 150.0 g SO2 x (1 mol/64 g SO2) = 2.34 mole SO2 (2.34 mol SO2/x) = (1 mol SO2/-296.9 kJ) x = -6.96 x 102 kJ 5. Hrxn Given the following reactions 2 O3 (g) -----------> 3 O2 (g) O2 (g) -------------> 2 O (g) NO (g) + O3 (g) ---> NO2 (g) + O2 (g) - 427.0 kJ 495.0 kJ - 199.0 kJ Find the Hrxno for the reaction: NO (g) + O (g) ----> NO2 (g) 2 (NO (g) + O3 (g) → NO2 (g) + O2 (g)) 2 O (g) → O2 (g) 3 O2 (g) → 2 O3 (g) _____________________________ 2 NO (g) + 2 O (g) → 2 NO2 (g) 2(- 199.0 kJ) - 495.0 kJ + 427.0 kJ ___________ - 466 kJ Answer: NO (g) + O (g) → NO2 (g) Hrxn = -233 kJ 6. Determine the Hfo for the formation of gasoline (C8H18) given that the Horxn = –9.91 x 103 kJ and the following unbalanced reaction: C8H18 (l) + O2 (g) → CO2 (g) + H2O (l) 2 C8H18 (l) + 25 O2 (g) → 16 CO2 (g) + 18 H2O (l) - 9.91 x 103 kJ = [18 mol x -286 kJ/mol H2O) + (16 mol x -394 kJ/mol CO2)] – [(2x + O)] - 9.91 x 103 kJ = [(-1.15 x 104 kJ) – 2x] 1.54 x 103 kJ = -2x x = - 771 kJ is the Ho for gasoline’s (C8H18 (l)) formation 7. Determine the Hrxn for the following reaction C(s) + ½ O2 (g) → CO (g) given the following information: C(s) + O2 (g) → CO2 (g) CO (g) + ½ O2 (g) → CO2 (g) ½ O2 (g) C(s) + O2 (g) → CO2 (g) CO2 (g) → CO (g) + ½ O2 (g) _____________________ H = -394.0 kJ H = +283.0 kJ _____________ C(s) + ½ O2 (g) → CO (g) - 111 kJ H = -394.0 kJ H = -283.0 kJ