#33a The reaction: 2Mg(s) + O_{2}(g) --> 2MgO(s) ΔH = -1204 kJ The negative sign means that energy is lost by the reacting particles so the reaction is exothermic. #33b The number by the side of the equation refers to the amount of energy per mol. Mol of Mg = [2.40 g] / [24.305 g/mol] = 0.0987 mol There are two mol of Mg per every 1204 kJ of energy (2:1 ratio). Heat transferred = [-1204 kJ/mol] X [(1/2) X 0.0987 mol] = 59.4 kJ #33c Notice that the ratio MgO:energy is 2:1 Mol of MgO = [2 X 96.0 kJ] / [1204 kJ/mol] = 0.159 mol grams of MgO = [0.159 mol] X [40.3 g/mol] = 6.43 g #33d Notice that the ratio MgO:energy is 2:1 Mol of MgO = [7.50 g] / [40.3 g/mol] = 0.186 mol energy absorbed = [(1/2) X 0.186 mol] X [1204 kJ/mol] = 112 kJ #49 mcΔT = Q m = 100.0 g of water + 9.55 g of sodium hydroxide = 109.55 g ΔT = T_{2} – T_{1} = [47.4 – 23.6] ^{0}C = 23.8^{0}C [109.55 g] X [4.184 J/g x ^{0}C] X [23.8^{0}C] = 10900 J Mol of NaOH = [9.55 g] / [40.0] = 0.23875 mol ΔH = [-Q] / [n] = [-10900 J] / [0.23875 mol] = -45654 J/mol --> -45.65 kJ/mol #77a C_{8}H_{18}(liquid) + (25/2)O_{2}(g) --> 8CO_{2}(g) + 9H_{2}O(g) ΔH = -5069 kJ #77b The reaction for the formation of the octane from its elements is: 8C(s) + 9H_{2}(g) --> C_{8}H_{18}(liquid) Carbon is graphite because diamond is too expensive. ΔH^{0} = Σ ΔH_{f}^{0} Products - ΣΔH_{f}^{0}Reactants Σ ΔH_{f}^{0} Products = ΣΔH_{f}^{0}Reactants + ΔH^{0} Σ ΔH_{f}^{0} C_{8}H_{18}(liquid) = [8 X (-393.5 kJ)] + [9 X (-241.82 kJ)] + 5069 kJ = -255 kJ #97a ΔT for the copper block = [30.1 – 100.4] ^{0}C = 70.3 ^{0}C Q = mcΔT = [121.0 g] X [0.385 J/g x ^{0}C] X [70.3 ^{0}C] = -3.27 X 10^{3} J Heat is lost because Q is negative. #97b ΔT for water = [30.1 – 25.1] ^{0}C = 5.00 ^{0}C Q = mcΔT = [150.0 g] X [4.184 J/g x ^{0}C] X [5.00 ^{0}C] = 3.14 X 10^{3} J Heat is gained because Q is positive. #97c Energy difference = [3.27 X 10^{3} J – 3.14 X 10^{3} J] = 130 J Heat Capacity = Q / [ΔT] = [130 J] / [5.00 ^{0}C] = 26.0 J/deg deg applies to both Kelvin(K) and Celcius(^{0}C) since they have the same magnitude and ΔT only measures the difference between two numbers. ^{0}C = K + 273 #97d If the cup is made of a perfectly insulating material, all the heat from copper gets transferred only to the water. Q / [mc] = ΔT = [3.27 X 10^{3} J] / [150 g X 4.184 J/g x ^{0}C] = 5.21 ^{0}C T_{2} = T_{1} + 5.21 ^{0}C = 30.3 ^{0}C #107a You can find this equation in any general chemistry textbook: CH_{4}(g) + 2O_{2}(g) --> CO_{2}(g) + H_{2}(liquid) ΔH = -890.36 kJ/mol CH_{4} But there is a more difficult way of doing it. First get the enthalpies of formation of reactants and products: ΔH_{f}^{0} CO_{2}(g) = -393.5 kJ/mol ΔH_{f}^{0} H_{2}(liquid) (g) = -285.83 kJ/mol ΔH_{f}^{0} O_{2}(g) = 0.0 kJ/mol ΔH_{f}^{0} CH_{4}(g) = -74.8 kJ/mol Then solve the equation: ΔH^{0} = Σ ΔH_{f}^{0} Products - ΣΔH_{f}^{0}Reactants ΔH^{0} = -393.5 kJ + [2 X (-285.83 kJ)] – [-74.8 kJ - 2(0) kJ] = 890.36 kJ That is the energy per 6.022 X 10^{23} molecules of CH_{4} reacted. Energy per molecule = 890.36 kJ / 6.022 X 10^{23} = 1.479 X 10^{-21} kJ = 1.479 X 10^{-18} J (There are 1000 Joules in one kiloJoule) #107b From the back of the book: 1.0 keV = 96.485 X 10^{6} J/mol, and since each mol of photon carries 8 keV, Energy per photon = [8 X 96.485 X 10^{6} J] / 6.022 X 10^{23} = 1.479 X 10^{-21} kJ = 1.282 X 10^{-15} J [1.282 X 10^{-15} J from X-ray] / [1.479 X 10^{-18} J from CH_{4}(g)] = 866.6 X-rays have more energy per unit than CH_{4}(g) #111a Mol of CuSO_{4} = Molarity X Volume Mol of CuSO_{4} = [1.00 mol/L] X [50.0 X 10^{-3} L] = 50.0 X 10^{-3} mol Since there is one Cu per every CuSO_{4} Grams of Cu = [50.0 X 10^{-3} mol] X [63.55 g/mol] = 3.177 g of Cu #111b From the equation: Cu^{2+} + SO_{4}^{2-} + 2K^{+}(aq) + 2OH^{-}(aq) --> 2K^{+}(aq) + SO_{4}^{2-}(aq) + Cu(OH)_{2}(s) The precipitate is Cu(OH)_{2} #111c Complete equation from above, net ionic: Cu^{2+} + 2OH^{-}(aq) --> Cu(OH)_{2}(s) #111d ΔT = T_{2} - T_{1} = [27.7 – 21.5] ^{0} C = 6.2 ^{0} C m = 100.0 mL X [1.000 g/mL] = 100.0 g of solution Q = mcΔT = 100.0 g X [4.184 J/g x ^{0} C] X 6.2 ^{0} C = 2584.08 J Mol of KOH = [2.00 mol/L] X [50.0 X 10^{-3} L] = 100.0 X 10^{-3} mol From the stoichiometry get 1:2 ratio for Cu:OH^{-} so there is just enough of both reactants. There is no limiting reactant to worry about in this problem; but this is not always the case. Number of moles of Product = 50.0 X 10^{-3} mol ΔH = -Q / n = [2584.08 J] / [50.0 X 10^{-3} mol] = -51881.6 J/mol #112a Complete equation: Ag^{+} + NO_{3}^{-} + Na^{+}(aq) + Cl^{-}(aq) --> Na^{+}(aq) + NO_{3}^{-}(aq) + AgCl(s) Net ionic: Ag^{+} + Cl^{-}(aq) --> AgCl(s) ΔH_{f}^{0} AgCl(s) = -127.0 kJ/mol ΔH_{f}^{0} Ag^{+}(aq) = 105.90 kJ/mol ΔH_{f}^{0} Cl^{-}(aq) = -167.2 kJ/mol ΔH^{0} = Σ ΔH_{f}^{0} Products - ΣΔH_{f}^{0}Reactants ΔH^{0} = -127.0 kJ -105.90 kJ – [-167.2 kJ] = -65.7 kJ #112b The net ionic equation defines the reaction outcome. The other ions appear on both sides of the equation (spectator ions) so the value obtained above applies to both the complete and the net ionic equation. #112c Now AgNO_{3}(aq) is the product so the equation: ΔH^{0} = Σ ΔH_{f}^{0} Products - ΣΔH_{f}^{0}Reactants becomes: Σ ΔH_{f}^{0} + Σ ΔH_{f}^{0} Products = ΣΔH_{f}^{0}Reactants + ΣΔH^{0} ΔH_{f}^{0} AgNO_{3}(aq) = -446.2 kJ + (-127.0 kJ) – (-407.1 kJ) –(-65.7 kJ) ΔH_{f}^{0} AgNO_{3}(aq) = -100.4 kJ/mol #113a Mol of CO_{2} = [21.83 g] / [44.01 g/mol] = 0.4960 mol From mole ratio CO_{2}:C is 1:1 grams of C = [0.4960 mol] X [12.0 g/mol] = 5.95 g Mol of H_{2}O = [4.47 g] / [18.02 g/mol] = 0.2481 mol From mole ratio H_{2}O:H is 2:1 grams of H = [0.2481 mol] X [1.000 g/mol] = 0.4961 g Mass of compound = [5.95 + 0.4961] g = 6.45 g #113b Carbon --> 5.95 / 12.01 = 0.4960 Hydrogen --> 0.4961 / 1.000 = 0.4960 Carbon --> 0.4960 / 0.4960 = 1 Hydrogen --> 0.4960 / 0.4960 = 1 From 1:1 ratio the empirical formula is CH #113c 6.45 g CH + O_{2}(g) --> 21.83 g CO_{2}(g) + 4.47 g H_{2}O(g) + 311 kJ ΔH^{0} = Σ ΔH_{f}^{0} Products - ΣΔH_{f}^{0}Reactants ΔH_{f}^{0} CH Sample = ΔH_{f}^{0}CO_{2}(g) + ΔH_{f}^{0}H_{2}O(g) + ΣΔH^{0} Use number of mol from part a ΔH_{f}^{0}CO_{2}(g) = [-393.5 kJ/mol] X [0.4960 mol] = 195.2 kJ ΔH_{f}^{0}H_{2}O(g) = [-241.82 kJ/mol] X [0.2481 mol] = 60.00 kJ ΔH_{f}^{0} CH Sample = -195.2 kJ – 60.00 kJ - [-311 kJ] = 55.8 kJ Mol of CH units = [6.45 g] / [13.0 g/mol] = 0.496 mol CH ΔH_{f}^{0} CH Sample = [55.8 kJ] / [0.496 mol] = 112.5 kJ/mol |