1. Chapter 6B Notes Thermochemistry West Valley High School AP Chemistry Mr. Mata

2. Hess’s Law Germain Henri Hess

3. Hess’ Law Enthalpy is a state function.Enthalpy is a state function. It is independent of the path.It is independent of the path. We can add equations to come up with the desired final product, and add the  HWe can add equations to come up with the desired final product, and add the  H Two rules.Two rules. If the reaction is reversed the sign of  H is changed.If the reaction is reversed the sign of  H is changed. If the reaction is multiplied, so is  H.If the reaction is multiplied, so is  H.

4. Standard Enthalpy The enthalpy change for a reaction at standard conditions (25ºC, 1 atm, 1 M solutions).The enthalpy change for a reaction at standard conditions (25ºC, 1 atm, 1 M solutions). Symbol  Hº.Symbol  Hº. When using Hess’s Law, work by adding the equations up to make it look like the answer.When using Hess’s Law, work by adding the equations up to make it look like the answer. The other parts will cancel out.The other parts will cancel out.

5. Hess’s Law “In going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in one step or a series of steps.”

6. Hess’s Law Example Problem Calculate  H for the combustion of methane, CH 4 : CH 4 + 2O 2  CO 2 + 2H 2 O Reaction  H o C + 2H 2  CH 4 -74.80 kJ C + O 2  CO 2 -393.50 kJ H 2 + ½ O 2  H 2 O-285.83 kJ Step #1: CH 4 must appear on the reactant side, so we reverse reaction #1 and change the sign on  H. CH 4  C + 2H 2 +74.80 kJ

7. Hess’s Law Example Problem Calculate  H for the combustion of methane, CH 4 : CH 4 + 2O 2  CO 2 + 2H 2 O Reaction  H o C + 2H 2  CH 4 -74.80 kJ C + O 2  CO 2 -393.50 kJ H 2 + ½ O 2  H 2 O-285.83 kJ CH 4  C + 2H 2 +74.80 kJ Step #2: Keep reaction #2 unchanged, because CO 2 belongs on the product side C + O 2  CO 2 -393.50 kJ

8. Hess’s Law Example Problem Calculate  H for the combustion of methane, CH 4 : CH 4 + 2O 2  CO 2 + 2H 2 O Reaction  H o C + 2H 2  CH 4 -74.80 kJ C + O 2  CO 2 -393.50 kJ H 2 + ½ O 2  H 2 O-285.83 kJ CH 4  C + 2H 2 +74.80 kJ C + O 2  CO 2 -393.50 kJ Step #3: Multiply reaction #3 by 2 2H 2 + O 2  2 H 2 O -571.66 kJ

9. Hess’s Law Example Problem Calculate  H for the combustion of methane, CH 4 : CH 4 + 2O 2  CO 2 + 2H 2 O Reaction  H o C + 2H 2  CH 4 -74.80 kJ C + O 2  CO 2 -393.50 kJ H 2 + ½ O 2  H 2 O-285.83 kJ CH 4  C + 2H 2 +74.80 kJ C + O 2  CO 2 -393.50 kJ 2H 2 + O 2  2 H 2 O -571.66 kJ Step #4: Sum up reaction and  H CH 4 + 2O 2  CO 2 + 2H 2 O-890.36 kJ

10. Calculation of Heat of Reaction Calculate  H for the combustion of methane, CH 4 : CH 4 + 2O 2  CO 2 + 2H 2 O  H rxn =   H f (products) -   H f (reactants) Substance  H f CH 4 -74.80 kJ O2O2 0 kJ CO 2 -393.50 kJ H2OH2O-285.83 kJ  H rxn = [-393.50kJ + 2(-285.83kJ)] – [-74.80kJ]  H rxn = -890.36 kJ