<p> Chapter 5 Thermochemistry</p><p>ENTHALPY H, Enthalpy, is the change in the HEAT CONTENT of a reaction.</p><p>H = Hproducts – Hreactants</p><p>H > 0 (positive) ENDOTHERMIC, Heat is absorbed. H < 0 (negative) EXOTHERMIC, Heat is evolved.</p><p>Example: CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) H = -890.4 kJ</p><p>1 mol CH4 produces 890.4 kJ of energy 1 mol O2 produces 445.2 kJ of energy</p><p>Stoichiometry and Enthalpy Question: For the combustion of 100. g of methane (CH4), how much heat would be generated? Answer: 5570 kJ Solution: CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) H = -890.4 kJ</p><p>(100. g CH4)/(16.0 g/mol) = 6.25 mol CH4</p><p>(6.25 mol CH4)(890.4 kJ) = 5570 kJ of heat produced</p><p>CALORIMETRY</p><p>Heat Capacity … heat required to raise the temperature of an object 1-degree Celsius.</p><p>C = ms or C = Energy/oC</p><p>Specific Heat … heat required to raise the temperature of 1-gm of an object, 1-degree Celsius.</p><p> q = msT T = Tfinal-Tinitial</p><p>Question: 120. gm of metal X is at 98oC. It is lowered in to a Styrofoam cup calorimeter (heat capacity = 0 J/ oC) that has 1200. gm of water at 20. oC. The final temperature is 34oC. What is the specific heat of metal X? Answer: 9.15 J/g oC Solution: Heat lost by metal X = Heat gain by the water </p><p>(mass)(specific heat)( T)metal= (mass)(specific heat)( T)water o o -1 o -1 o o (120. g)(specific heat)metal (98 C -34 C) = (1200. g)(4.184 Jg C )(34 C -20 C) -1 o -1 specific heat metal =9.15 Jg C Question: 1.9862 g of benzoic acid (C6H5COOH) is heated in a constant-volume calorimeter (C = 853 J/oC). What is the expected change in temperature of the calorimeter? o (H benzoic acid = -3226.7 kJ/mol) Answer: 61.6 oC temperature increase Solution: (1.9862 g)/(122.12467 g/mol) = .016264 mol benzoic acid (.016264 mol)x(3226.6 kJ/mol) = 52.476 kJ of heat produced C = Energy/oC oC = Energy/C (52.476 kJ)/(.853 kJ/ oC) = 61.519 oC </p><p>Question: 1.9862 g of benzoic acid (C6H5COOH) is heated in a constant-volume calorimeter o (C = 853 J/ C) that contains .8 liters of H2O. What is the expected change in temperature of the o calorimeter? (H benzoic acid = -3226.7 kJ/mol) Answer: 12.49oC temperature increase Solution: (1.9862 g)/(122.12467 g/mol) = .016264 mol benzoic acid (.016264 mol)(3226.6 kJ/mol) = 52.476 kJ of heat produced 52,476 J of heat = heat absorbed by calorimeter + heat absorbed by water</p><p>= [(C)(T)] + [(mass)(spec. heat)( T)]water = (853 J/ oC)( T) + (800 gm)(4.184 Jg-1 oC -1)( T) T = 12.5 oC</p><p>STANDARD STATE AND ENTHALPY</p><p> o H f is the heat change that results when a compound is formed from its elements at a pressure of 1 atm and at a temperature of 25oC.</p><p> o Convention: H f for an element in its most stable form is zero.(Hint: Look at the periodic table)</p><p>HESS’S LAW</p><p> o o The H rxn may be determined from a list of the H f for the reactants and products. (Hint: Watch the stoichiometry)</p><p> o Question: What is the H for the combustion of C3H8 ? Answer: -2226.9 kJ/mol Solution: C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l)</p><p> o o o H = H products - H reactants</p><p> o o o o o H = [3H CO2(g) + 4H H2O(l)] – [H C3H8(g) + 5H O2(g)]</p><p>Ho = [3(-393.5 kJ/mol) + 4(-285.8 kJ/mol)] – [ (-96.8 kJ/mol ) + 5(0 kJ/mol)] HEAT OF SOLUTION</p><p>The heat associated with the dissolving of a solute in a given solution.</p><p>Examples: (Solvent is water) Endothermic Exothermic NaCl +4 kJ/mol LiCl -37.1 kJ/mol KCl +17.2 kJ/mol CaCl2 -82.8 kJ/mol NH4Cl +15.2 kJ/mol NH4NO3 +26.2 kJ/mol</p>