1. Communicating Enthalpy Change

2. Method 1: Molar Enthalpies of Reaction, Δ r H m To communicate a molar enthalpy, both the substance and the reaction must be specified. When reactants and products are in their standard state, they are at a pressure of 100 kPa, an aqueous concentration of 1.0 mol/L. and liquids and solids are in their pure state.

3. Δ f H m ° = –239.2 kJ/mol When 1 mol of methanol is formed from its elements when they are in their standard states at SATP, 239.2 kJ of energy is released. Formation Reaction Combustion Reaction Δ c H m ° = –725.9 kJ/mol CH 3 OH The complete combustion of 1 mol of methanol releases 725.9 kJ of energy. Note that the above reactions are balanced for one mole of the compound.

4. Method 2: Enthalpy Changes, Δ r H Write an enthalpy change (Δ r H) beside the chemical equation. CO(g) + 2 H 2 (g) → CH 3 OH(l)Δ r H = –725.9 kJ The enthalpy change is not a molar value, so does not require the “m” subscript and is not in kJ/mol. Δ c H° = –98.9 kJ Δ c H° = –197.8 kJ When 2 moles of sulfur dioxide are burned, twice as much heat energy is released as when 1 mole of sulfur dioxide is burned.

5. 2 SO 2 (g) + O 2 (g)  2 SO 3 (g) Then get the chemical amount of sulfur dioxide from its coefficients in the balanced equation and use Δ c H° = n Δ c H° m Δ c H° = n Δ c H°m = 2 mol x (-98.9 kJ)/1 mol = -197.8 kJ Finish it off by communicating the enthalpy next to a balanced equation 2 SO 2 (g) + O 2 (g)  2 SO 3 (g) Δ c H° = -197.8 kJ Sulfur dioxide and oxygen react to form sulfur trioxide. The standard molar enthalpy of combustion of sulfur dioxide, in this reaction, is -98.9 kJ/mol. What is the enthalpy change for this reaction?

6. Another example... Wild natural gas wells are sometimes lit on fire to eliminate the very toxic hydrogen sulfide gas. The standard molar enthalpy of combustion of hydrogen sulfide is -518.0 kJ/mol. Express this value as a standard enthalpy change for the following: 2 H 2 S (g) + 3 O 2 (g)  2 H 2 O (g) + 2 SO 2 (g) Δ c H °= ? Δ c H= n Δ c H° = 2 mol x -518.0 kJ/mol = -1 036.0 kJ 2 H 2 S (g) + 3 O 2 (g)  2 H 2 O (g) + 2 SO 2 (g) Δ c H°= -1 036.0 kJ

7. Method 3: Energy Terms in Balanced Equations reactants → products + energy reactants + energy → products For endothermic reactions, the energy is listed along with the reactants. For exothermic reactions, the energy is listed along with the products.

8. Method 4: Chemical Potential Energy Diagrams During an exothermic reaction, the enthalpy of the system decreases. Heat flows out of the system and into the surroundings and we observe a temperature increase.

9. Method 4: Chemical Potential Energy Diagrams During an endothermic reaction, the enthalpy of the system increases. Heat flows into the system from the surroundings and we observe a temperature decrease.

12. Read pgs. 495 – 500 Read over the Communication Example Problem 4 on page 500 pg. 501 Section 11.3 Questions # ’ s 1 – 7 Handout