Hess’ Law of Heat Summation

1. Hess’ Law of Heat Summation

2. Scenario #1 – Personal Wealth • Bill works diligently his whole life and retires at age 55, with a personal wealth of 20 million dollars. • Donald makes and loses a fortune of 20 million dollars several times through his life and eventually at age 55, ends up with a personal wealth of 20 million dollars • Conrad lives in poverty until the day he retires at age 55, and then wins a lottery of 20 million dollars Their wealth is independent of how their lives were lived!

3. Scenario #2 – Test Marks • Paris has a midterm average of 52% and finally decides to stop partying and aces the remainder of her tests and assignments. She receives a final course mark of 82%. • Britney has a midterm average of 96%. Feeling overly confident, she decides not to put forth the time and effort to studying and finishes the course with an 82% • Hilary works diligently throughout the entire course maintaining an “A” average and finally finishes the course with an 82%. Their final marks do not indicate how much effort they put in throughout the year!

4. Scenario #3 – Displacement • Angelina travels from Brampton to Ajax to pick-up a friend to go see a concert in Barrie. Their final displacement is 150km from their origin. • Brad travels from Brampton to Mississauga to drop off the kids at the babysitters, then drives to North York to pick up his friend. They stop for a bite in Woodbridge then make it to the concert in Barrie. Their final displacement is 150km from their origin. Displacement is independent of path taken. It depends only on beginning and end point!

5. Hess’ Law of Heat Summation: • States that then enthalpy change of a physical or chemical process depends only on the beginning conditions (reactants) and the end conditions (products). • The enthalpy change is independent of the pathway of the process and the number of intermediate steps in the process.

6. Hess’ Law

7. Using Hess’ Law to Calculate Enthalpy Change Part 1: Combining Chemical Equations Algebraically

8. Hess’s Law Example Problem CH4 C + 2H2 +74.80 kJ Step #1: CH4 must appear on the reactant side, so we reverse reaction #1 and change the sign on H.

9. Hess’s Law Example Problem CH4 C + 2H2 +74.80 kJ C + O2 CO2 -393.50 kJ Step #2: Keep reaction #2 unchanged, because CO2 belongs on the product side

10. Hess’s Law Example Problem CH4 C + 2H2 +74.80 kJ C + O2 CO2 -393.50 kJ 2H2 + O2 2 H2O -571.66 kJ Step #3: Multiply reaction #3 by 2 because 2 H20 required on product side

11. Hess’s Law Example Problem CH4 C + 2H2 +74.80 kJ C + O2 CO2 -393.50 kJ 2H2 + O2 2 H2O -571.66 kJ CH4 + 2O2 CO2 + 2H2O -890.36 kJ Step #4: Sum up reaction and H

12. When Combining Chemical Equations Algebraically… • There are two key ways in which you can manipulate an equation: • Reverse an equation • Multiply each coefficient in an equation

13. Using Hess’ Law to Calculate Enthalpy Change Part 2: Using Enthalpy of Formation Reactions

14. Standard molar enthalpy of formation ∆Hf: • is the quantity of energy that is absorbed or released when one mole of a compound is formed directly from its elements in their standard states

15. Standard Molar Enthalpies of Formation • Enthalpy of formation of an element in its standard state is zero. • Standard state (25°C and 100kPa) of an element is usually most stable form • Elements may exist in more than one form at standard state. • One forms is more stable and has ∆Hf =0 • Graphite versus diamond • Oxygen versus ozone

16. Formation reactions • substance formed from elements in their standard states H2(g) + ½ O2(g) H2O(l) ∆H°f = -285.8 kJ • Example 1 • Liquid sulfuric acid has a ∆Hf= -814.0 kJ/mol. Write an equation to show formation of liquid sulfuric acid. The standard stage of sulfur is rhombic sulfur, S(s). • Example 2 • Write a thermochemical equation for the formation of gaseous cesium. The standard enthalpy of formation of Cs(g) is 76.7kJ/mol

17. Calculating Enthalpy Changes Hrxn =  nHf(products) -  nHf(reactants)

18. Calculation of Heat of Reaction Hrxn =  nHf(products) -  nHf(reactants) Hrxn = [-393.50kJ + 2(-285.83kJ)] – [-74.80kJ] Hrxn = -890.36 kJ