enthalpy - describes heat flow in chemical changes  

• H = E + PV
• DH = DE + P DV = q + w - w = q = heat change at constant pressure
• DH positive w/ endothermic reactions, negative w/ exothermic reactions
• DH = Hfinal - Hinitial = H(products) - H(reactants)
• enthalpy of reaction - aka heat of reaction, DHrxn; enthalpy change during reaction
• extensive property (depends on moles used in reaction)
• reverse sign of enthalpy if reaction goes in reverse
• depends on state reactants/products (liquid/solid/gas)
  • enthalpy of vaporization (DH for liquid to gas conversion)
  • enthalpy of fusion (DH for solid to liquid conversion)
• enthalpy of combustion (DH for combusting substance in oxygen)

Hess's Law - DH for reaction equals sum of DH for its parts  

• same value of DH regardless of how many steps used for the final reaction

Calculate DH for 2C(s) + H2(g) >> C2H2(g)  

• Given:
  • C2H2 + 5/2 O2 >> 2CO2 + H2O DH = -1299.6 kJ
  • C + O2 >> CO2 DH = -393.5 kJ
  • H2 + 1/2 O2 >> H2O DH = -285.8 kJ
• 2CO2 + H2O >> C2H2 + 5/2 O2 DH = +1299.6 kJ
• 2C + 2O2 >> 2CO2 DH = 2(-393.5) = -787.0 kJ
• 2C + 2O2 + H2 + 1/2 O2 + 2CO2 + H2O >> C2H2 + 5/2 O2 + 2CO2 + H2O DH = 1299.6 - 787.0 - 285.8 = 226.8 kJ
• 2C + 2O2 + H2 + 1/2 O2 + 2CO2 + H2O >> C2H2 + 5/2 O2 + 2CO2 + H2O DH = 1299.6 - 787.0 - 285.8 = 226.8 kJ
• 2C + 2O2 >> C2H2 DH = 226.8 kJ

enthalpy of formation - aka heat of formation, DHf  

• standard state - pure form at atmospheric pressure, 25°C
• standard enthalpy - enthalpy change when reactants/products in standard state
• stardard enthalpy of formation - change in enthalpy for reaction that makes 1 mol of a substance from its elements (everything in standard state)
  • DH°rxn = SDH°f(products) - SDH°f(reactants)
  • equal to 0 for substances in elemental form

Find standard enthalpy change for combustion of C6H6(l) into CO2(g) and H2O(l)  

• Given:
  • DH°f(CO2(g)) = -393.5 kJ/mol
  • DH°f(H2O(l)) = -285.8 kJ/mol
  • DH°f(C6H6(l)) = 49.0 kJ/mol
• C6H6 + 15/2 O2 >> 6CO2 + 3H2O
• DH°rxn = SDH°f(products) - SDH°f(reactants)
• = [6mol x DH°f(CO2) + 3mol x DH°f(H2O)] - [1mol x DH°f(C6H6) + 15/2 mol x DH°f(CO2)
• = [6mol (-393.5 kJ/mol) + 3mol (-285.8 kJ/mol)] - [1mol (49.0 kJ/mol) + 15/2 mol (0 kJ/mol)]
• = -2361 - 857.4 - 49.0 = -3267 kJ