AP Notes, Outlines, Study Guides, Vocabulary, Practice Exams and more!

Enthalpy

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
Subject: 
Chemistry
Subject X2: 
Chemistry

Need Help?

We hope your visit has been a productive one. If you're having any problems, or would like to give some feedback, we'd love to hear from you.

For general help, questions, and suggestions, try our dedicated support forums.

If you need to contact the Course-Notes.Org web experience team, please use our contact form.

Need Notes?

While we strive to provide the most comprehensive notes for as many high school textbooks as possible, there are certainly going to be some that we miss. Drop us a note and let us know which textbooks you need. Be sure to include which edition of the textbook you are using! If we see enough demand, we'll do whatever we can to get those notes up on the site for you!