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Standard Enthalpy Changes = D H o

Standard Enthalpy Changes = D H o. P = 1 bar (0.997 atm) T = 298K, unless otherwise specified n = 1 mole for key compound. Std. Enthalpy. Hess’s Law. Hess’s Law with Non-Standard Temps. Enthalpy of Reaction. Enthalpy is an extensive property It depends upon how much mass you have.

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Standard Enthalpy Changes = D H o

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  1. Standard Enthalpy Changes = DHo • P = 1 bar (0.997 atm) • T = 298K, unless otherwise specified • n = 1 mole for key compound

  2. Std. Enthalpy

  3. Hess’s Law

  4. Hess’s Law with Non-Standard Temps

  5. Enthalpy of Reaction • Enthalpy is an extensive property • It depends upon how much mass you have. • The enthalpy change for a reaction is equal in magnitude, but opposite in sign, to H for the reverse reaction. • Enthalpy change for a reaction depends on the state of the reactants and products. • If a reactant is liquid and the product gas, there is less heat available to transfer to the surroundings • Enthalpy of H2O(g) is greater than enthalpy of H2O(l)

  6. Enthalpy of Reaction • 2H2O(l) 2H2O(g)ΔH = +88kJ • CH4(g) + 2O2(g) CO2(g) + 2H2O(l)ΔH = -890kJ • What if H2O was gas instead of liquid? • Break up the equation into 2 Heat equations: • CH4(g) + 2O2(g) CO2(g) + 2H2O(l)ΔH = -890kJ • 2H2O(l) 2H2O(g)ΔH = +88kJ • Add the 2 equations together keeping track of states: • CH4(g) + 2O2(g) CO2(g) + 2H2O(g) ΔHrxn = -802kJ

  7. Enthalpy of Reactions • Example: H2(g) + 1/2 O2(g) è H2O(l) DH = -285 kJ/mol • This process can be divided into smaller steps such as • H2(g) + 1/2 O2(g) è H2O(g)       DH = -241 kJ/mol • H2O(g) è H2O(l)                        DH = -44 kJ/mol • Add rxns and enthalpys: • H2(g) + 1/2 O2(g) è H2O(l)        DH = -285 kJ/mol

  8. Adjusting for Temperatures: • If you have 2 objects at different temperatures, you have to use specific heat to calculate heat gain or lost: • q = Cs x m x ΔT • q = heat • Cs = specific heat • m = mass • ΔT = temperature change. • Rearrange the equation to solve for temperature change. • Heat lost has to equal heat gained somewhere.

  9. Heat of Reaction and Hess’s Law • Calculate ΔH for the reaction: • 2C(s) + H2(g) C2H2(g) • Givens: • C2H2(g) + 5/2 O2(g) 2CO2(g) + H2O(l) • ΔH = -1299.6kJ • C(s) + O2(g) CO2(g) • ΔH = -393.5 kJ • H2(g) + ½ O2(g) H2O(l) • ΔH = -285.8 kJ • First reaction has to be reversed!

  10. Enthalpy of Reaction and Hess’s Law • 2CO2(g) + H2O(l)  C2H2(g) + 5/2 O2(g) • ΔH = +1299.6kJ • 2(C(s) + O2(g) CO2(g)) • ΔH = -393.5 kJ • H2(g) + ½ O2(g) H2O(l) • ΔH = -285.8 kJ • Net Reaction: • 2C(s) + H2(g) C2H2(g)

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