ENTHALPY DH = Hfinal - Hinitial Process is ENDOTHERMIC For systems at constant pressure, the heat content is the same as the Enthalpy, or H, of the system. Heat changes for reactions carried out at constant pressure are ths same as changes in enthalpy (∆H) DH = Hfinal - Hinitial If Hfinal > Hinitial then DH is positive Process is ENDOTHERMIC If Hfinal < Hinitial then DH is negative Process is EXOTHERMIC

Standard Enthalpy Values Most DH values are labeled DHo o means measured under standard conditions P = 1 atmosphere ( = 760 torr = 101.3 kPa) Concentration = 1 mol/L T = usually 25 oC with all species in standard states e.g., C = graphite and O2 = gas

Endo- and Exothermic ENDOTHERMIC EXOTHERMIC Heat Heat qsys < 0 Surroundings Heat qsys > 0 System Surroundings System Heat qsys < 0 DEMONSTRATION example of an ENDOTHERMIC reaction - hydration of ammonium nitrate (NH4)+(NO3)- (s) + H2O -> NH4 (aq) + NO3 (aq) (cools surroundings enough to freeze water) WHY SPONTANEOUS ? ENDOTHERMIC EXOTHERMIC

Given the following equation: USING ENTHALPY Given the following equation: H2O2 (aq)  H2O(l) + O2(g) Calculate the heat produced (kJ) when 15.0 grams of H2O2 is used. ΔH-98.2kJ

Using more enthalpy SnO2 + H2  Sn + H2O ΔH = 124Kj What would be the heat released if there were 5 moles of SnO2?

Heat and Changes of State Heat of combustion (∆H)= the heat of reaction for the complete burning of one mole of a substance Molar heat of fusion (∆Hfus)= the heat absorbed by one mole of a substance in melting from a solid to a liquid at a constant temperature Molar heat of solidification (∆Hsolid)= heat lost when one mole of a liquid freezes to a solid at a constant temperature (equal to the negative heat of fusion) Molar heat of vaporization (∆Hvap)= the heat absorbed by one mole of a substance in vaporizing from liquid to a gas Molar heat of condensation (∆Hcond)= heat released by one mole of a vapor as it condenses

Example (Heat of combustion) The standard heat of combustion (∆H°rxn) for glucose (C6H12O6) is 2808 kJ/mol. If you eat and burn 70.g of glucose in one day, how much energy are you getting from the glucose? Step one: convert g of glucose to moles 70. g glucose x 1 mol = 0.28 mol glucose 1 246 g Step two: Use (∆H°rxn) to find amount of kJ gained 0.28 mol glucose x 2808 kJ = 790 kJ gained (+ b/c gained not lost) 1 1 mol