1Week 12 CHEM 1310 - Sections L and M 1Energy, Enthalpy, & Thermochemistry 9.1 The Nature of Energy 9.2 Enthalpy 9.3 Thermodynamics of Ideal Gases 9.4 Calorimetry 9.5 Hess’s Law 9.6 Standard Enthalpies of Formation 9.7 Present Sources of Energy 9.8 New Energy SourcesWeek 12 CHEM 1310 - Sections L and M 2Example Problems What is ∆E in kJ? ∆E = q + w Recall: q = -10 kJ (because heat is released) Recall: w = 125 L x atm Must convert units in order to add…A gas is compressed from 40L to 15L at a constantpressure of 5 atm. In the course of this compression10 kJ of energy is released.Show on boardWeek 12 CHEM 1310 - Sections L and M 3Yesterday’s Chemical ExplosionDue to Unforeseen Reactivity in Calorimetry Experiment WSBtv.com2Week 12 CHEM 1310 - Sections L and M 4What is Calorimetry? Measurement of amounts of heat flow andthe accompanying temp changes.In “bomb calorimetry”, a combustion reactionoccurs inside of the bomb.Reaction chamber has a fixed volume, so P∆V is 0.Week 12 CHEM 1310 - Sections L and M 5What is Calorimetry? Measurement of amounts of heat flow andthe accompanying temp changes.A simple design uses astyrofoam cup sealed.Heat lost to the styrofoamitself, the thermometer,and to the surrounding airis negligible.Week 12 CHEM 1310 - Sections L and M 6Calorimetry Problem A student placed 50.0 mL of 1.00 M HCl at 25.5°Cinto a styrofoam cup calorimetry. To this, sheadded 50.0 mL of 1.00 M NaOH at 25.5°C. Themixture was stirred and the temp increased to32.2°C. What is the energy evolved in J/mol of HCl? Assume specific heat for reaction is that ofwater (4.18 Jg-1°C-1) Density of HCl = 1.02 g/mL Density of NaOH = 1.04 g/mLShow on board3Week 12 CHEM 1310 - Sections L and M 7Hess’s Law If two or more chemical equations are added togive a new equation, then adding the enthalpies ofthe reactions that they represent gives theenthalpy of the new reaction.Significance: Experimental determination of Δ H for some rxns!Recall: Enthalpy is a state function! Path is irrelevant. Week 12 CHEM 1310 - Sections L and M 8Hess’s LawEnthalpy changeis the same whether 1 step or 2.Week 12 CHEM 1310 - Sections L and M 9Standard Enthalpy of Formationa A + b B → c C + d DΔ H°f = cΔ H°f (C) + dΔ H°f (D) - aΔ H°f (A) - bΔ H°f (B)Products ReactantsNotice:(1) Coefficients in balanced chemical eqn are a part of the molar enthalpy term.(2) Magnitude of contribution from reactants is subtracted from that of the products to denote “unformation”.4Week 12 CHEM 1310 - Sections L and M 10Example ProblemN2H4(l) + 3 O2(g) → 2 NO2(g) + 2 H2O(l)ΔH°f = cΔH°f (C) + dΔH°f (D) - aΔH°f (A) - bΔH°f (B)Calculate ΔH°fFind standard enthalpy of formation for individualmolecules in AppendixWeek 12 CHEM 1310 - Sections L and M 11Example ProblemN2H4(l) + 3 O2(g) → 2 NO2(g) + 2 H2O(l)ΔH°f = cΔH°f (C) + dΔH°f (D) - aΔH°f (A) - bΔH°f (B)ΔH°fN2H4(l) = 50.63 kJ/molO2(g) = 0 kJ/molNO2(g) = 33.18 kJ/molH2O(l) = -285.83 kJ/molFrom Reference sourceWeek 12 CHEM 1310 - Sections L and M 12Example ProblemN2H4(l) + 3 O2(g) → 2 NO2(g) + 2 H2O(l)ΔH°f = 2ΔH°f (NO2) + 2ΔH°f (H2O)- 1ΔH°f (N2H4) - 3ΔH°f (O2)ΔH°f = 2 (33.18) + 2 (-285.83)- 1 (50.63) - 3 (0)ΔH°f = 66.36 - 571.66 - 50.63 = -555.93
View Full Document
Unlocking...