Hess’s Law

1. Hess’s Law Determining Enthalpy Changes for Chemical Reactions

2. Review • Enthalpy is a measure of the total heat content of a system. • Exothermic reactions cause a system to lose heat to its surroundings. • H is negative. • Surroundings warm up. • Endothermic reactions cause a system to pull heat from its surroundings. • H is positive. • Surroundings cool down.

3. Potential Energy Diagrams • Shows the change in potential energy in a chemical reaction. • x-axis = Reaction Coordinate • Not the same thing as time. • y-axis = Potential Energy

4. Activation Energy Reactants Potential Energy  H Products Rxn Coordinate  Potential Energy Diagrams H is negative, so this reaction is exothermic.

5. Activation Energy Products Potential Energy  H Reactants Rxn Coordinate  Potential Energy Diagrams H is positive, so this reaction is endothermic.

6. State Functions • Enthalpy is a state function. • State functions – depend on the starting and ending points of a reaction, not on the pathway. • The H is the same regardless of how the reactants change into products.

7. State Functions • An analogy: • Little Red Riding Hood wants to visit her grandmother, who lives at the top of the mountain. • Her grandmother’s house is 2000 feet above Red’s house. • Regardless of how Red climbs the mountain, her change in altitude will be the same: 2000 feet. • Change in altitude is therefore a state function.

8. State Functions The pathway doesn’t matter; the altitude change is the same. Grandma’s House Red’s House

9. State Functions • Two teams are playing a basketball game. They finish the regular game with a tie score: • Tigers • 1st quarter: 24 pts • 2nd quarter: 17 pts • 3rd quarter: 21 pts • 4th quarter: 18 pts • Total: 80 pts • Lions • 1st quarter: 15 pts • 2nd quarter: 25 pts • 3rd quarter: 30 pts • 4th quarter: 10 pts • Total: 80 pts The teams both arrived at 80 points by different pathways. The final score of a basketball game is a state function because it does not matter when the team scored the points.

10. State Functions • John works all his life and retires at 55 with a personal wealth of $2 million. • Eric makes and loses fortunes until he’s 55, when his personal wealth is $2 million. • Bill is poor his entire life, but on his 55th birthday he wins a $2 million lottery. • All three men end up with the same amount of wealth at age 55, independent of the course they took to get it.

11. Hess’s Law • If you reverse a reaction, the sign of H changes. • 2H2O2(l) 2H2O(l) + O2(g) • H = -196.4 kJ • 2H2O(l) + O2(g) 2H2O2(l) • H = +196.4 kJ

12. Fractional coefficients are okay in thermochemical equations! Hess’s Law • If you multiply the coefficients of a thermochemical equation by some number, you must multiply H by the same number. • 2H2O2(l) 2H2O(l) + O2(g) • H = -196.4 kJ • 4H2O2(l)  4H2O(l) + 2O2(g) • H = -392.8 kJ • H2O2(l) H2O(l) + ½O2(g) • H = -98.2 kJ

13. Hess’s Law • Thermochemical equations can be combined to give new thermochemical equations. • N2 + 2O22NO2 • H = 67.8 kJ • N2O4N2 + 2O2 • H = -9.6 kJ • N2 + 2O2 + N2O4 2NO2 + N2 + 2O2 • N2O4  2NO2 • H = (67.8 kJ) + (-9.6 kJ) = 58.2 kJ

14. Hess’s Law • We can use known H values to find H for unknown reactions.

15. Hess’s Law • Ex: Methane combines with oxygen to form ketene and water. • 2CH4 + 2O2 CH2CO + 3H2O • H = ??? • We know the H of the following reactions: • CH2CO + 2O2 2CO2 + H2O • H = -981.1 kJ • CH4 + 2O2 CO2 + 2H2O • H = -802.3 kJ • We will manipulate the thermochemical equations to calculate the H of the first reaction.

16. 2CH4 + 2O2 CH2CO + 3H2O CH2CO + 2O2 2CO2 + H2O H = -981.1 kJ CH4 + 2O2 CO2 + 2H2O H = -802.3 kJ Hess’s Law 2CH4 CH2CO H = -623.5 kJ 2CO2 + H2O  CH2CO + 2O2H = +981.1 kJ CH2CO 2CH4 + 4O2 2CO2 + 4H2O H = -1604.6 kJ CH4 2 3

17. Hess’s Law • Calculate H for the following reaction: • CaO(s) + SO3(g)  CaSO4(s) • Using the following thermochemical equations: • H2O + SO3 H2SO4H = -132.5 kJ • H2SO4 + Ca  CaSO4 + H2H = -602.5 kJ • 2Ca + O2 2CaO H = -634.9 kJ • 2H2 + O2 2H2O H = -285.8 kJ H = -560.4 kJ CaO Ca + ½O2 H = +317.5 kJ H = -142.9 kJ H2 + ½O2 H2O

18. Standard Heat of Formation • Std Heat of Formation (Hfº) – energy change when 1 mole of a substance is formed from its elements in their standard states. • Zero for free elements in standard states. • EX: Hfº for O2(g) is 0 kJ. • EX: Hfº for Ag(s) is 0 kJ. • Hfº for CaCO3(s) = -1209.6 kJ • Ca(s) + C(s) + 3/2O2(g)  CaCO3(s) H = -1209.6 kJ

19. Standard Heat of Formation • We can use Hfº to calculate H for a chemical reaction. • 1. Look up the Hfº for each reactant and product. • 2. Multiply the Hfº for each reactant and product by their coefficients. • 3. Add the total Hfº for the reactants. • 4. Add the total Hfº for the products. • 5. H = total Hfº for products – total Hfº for reactants

20. Standard Heat of Formation • What is the H for the following reaction? • 2 CO(g) + O2(g)  2 CO2(g) • Hfº for CO(g) = -110.52 kJ/mol • -110.52 kJ/mol x 2 mol = -221.04 kJ • Hfº for O2(g) = 0 kJ/mol • Hfº for CO2(g) = -393.51 kJ/mol • -393.51 kJ/mol x 2 mol = -787.02 kJ • Total Hfº for reactants: • -221.04 kJ + 0 kJ = -221.04 kJ • Total Hfº for products: • -393.51 kJ • H = (-787.02 kJ) – (-221.04 kJ) = -565.98 kJ