AP Chem Get Heat HW stamped off Today: Enthalpy Cont., Hess’ Law Unit 4 Quest: next Thurs 11/8
Standard enthalpy of formation, ΔHfo enthalpy change for the formation of 1 mol of compound with all substances in their standard states (1 atm and 25°C) Units of kJ/mol; Values will be given The ΔHfo for the most stable form of any element is 0 (ex. ΔHfo = 0 for H2, O2 etc) Hrxn = Hf°products – Hf° reactants
Calculation of H C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l) H = [3(-393.5 kJ) + 4(-285.8 kJ)] – [1(-103.85 kJ) + 5(0 kJ)] = [(-1180.5 kJ) + (-1143.2 kJ)] – [(-103.85 kJ) + (0 kJ)] = (-2323.7 kJ) – (-103.85 kJ) = -2219.9 kJ 3
Hrxn = Hf°products – Hf° reactants Be careful with negative signs! Make sure to multiply the value in the table by the coefficients in the balanced equation 1) -128.6 kJ 2) -6534.8 kJ 3a) -196.04 kJ 3b) -149.44 kJ 4) Hf for CaC2 = -60.63 kJ/mol
Hess’s Law: If a reaction is carried out in a series of steps, the ΔH for the reaction is the sum of the individual ΔH for each step. We can estimate H using published H values and the properties of enthalpy.
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) Example Calculate ΔH for the reaction: CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) Given: CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) ΔH = -802 kJ H2O(l) → H2O(g) ΔH = +44 kJ CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) ΔH = -802 kJ 2 H2O(g) → 2 H2O(l) ΔH = 2 x - 44 = -88 kJ CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) ΔH = -802 kJ + - 88 kJ = -890 kJ
From the following heats of reaction: 2 H2 (g) + O2 (g) 2 H2O (g) ΔH = -483.6 kJ 3 O2 (g) 2 O3 (g) ΔH = +284.6 kJ Calculate the heat of the reaction for: 3 H2 (g) + O3 (g) 3 H2O (g) To get just one O3 on the reactants side, I need to flip the second equation and divide it by 2 O3 (g) 3/2 O2 (g) ΔH = -284.6/2 = -142.3 kJ 3 H2 (g) + 3/2 O2 (g) 3 H2O (g) ΔH = -483.6 x 3/2 = -725.4 kJ To get 3 H2 and 3 H2O, I need to multiply the first equation by 3/2. ΔH = -142.3 kJ + - 725.4 kJ ΔH = -867.7 kJ
Answers 2) -304.1 kJ 3) -2486.3 kJ 5) 156.1 kJ 6) 171.5 kJ 7b) -401.21 kJ