1. Thermochemistry Part 3 - Enthalpy and Thermochemical Equations

2. ENTHALPY  Enthalpy = a type of chemical energy (thermodynamic potential), sometimes referred to as “heat content”

3. Enthalpies of Reaction  The enthalpy change (in kJ/mol) that accompanies a chemical reaction is called the enthalpy of reaction  H rxn Also called heat of reaction

4. Enthalpies of Reaction  If  H rxn = negative exothermic  heat is given off  Under conditions of constant pressure, q = ΔH < 0 (negative sign) R PH time

5. Enthalpies of Reaction  If  H rxn = positive endothermic  heat must be added  Under conditions of constant pressure, q = ΔH > 0 (positive sign) R P H time

6. Thermochemical Equations  Thermochemical equations are balanced chemical equations that show the associated enthalpy change (H) balanced equation enthalpy change (H rxn )

7. Thermochemical Equations  An example of a thermochemical equation: CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(l) H = -890. kJ  The coefficients in the balanced equation show the # moles of reactants and products that produced the associated H. If the number of moles of reactant used or product formed changes, then the H will change as well.

8. Thermochemical Equations  For the following reaction: CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (l) H = -890. kJ-890. kJ 1 mol CH 4 2 mol O 2-890. kJ 1 mol CO 2 2 mol H 2 O

9. Thermochemical Equations  The thermochemical equation for burning 1 mole of CH 4 (g): CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (l) H = -890. kJ When 1 mole of CH 4 is burned, 890. kJ of heat are released. When 2 moles of CH 4 are burned, 1780. kJ of heat are released.

10. Rules of Thermochemistry Rule #1 - The magnitude of H is directly proportional to the amount of reactant consumed and product produced.

11. Rules of Thermochemistry Example 1: H 2 + Cl 2  2HClH = - 185 kJ Calculate H when 1.00 g of Cl 2 reacts. ΔH = 1.00 g Cl 2 x 1 mol Cl 2 x 70.90 g Cl 2 This thermochemical equation says that there is 185 kJ of energy released per how many moles of Cl 2 ? (look at the balanced equation) For every 1 mole (coefficient) Cl 2 = -185 kJ (released) -185 kJ = -2.61 kJ 1 mol

12. Rules of Thermochemistry Example 2: When an ice cube weighing 24.6 g of ice melts, it absorbs 8.19 kJ of heat. Calculate H when 1.00 mol of solid water melts. ΔH = 1.00 mol x 18.02 g x 1 mol 8.19 kJ = 6.00 kJ 24.6 g

13. Rules of Thermochemistry Example 3: Methanol burns to produce carbon dioxide and water: 2CH 3 OH + 3O 2  2CO 2 + 4H 2 O + 1454 kJ What mass of methanol is needed to produce 1820 kJ? Does “produce” mean exo or endo? + or – sign?

14. Rules of Thermochemistry Example 4: How much heat is produced when 58.0 liters of hydrogen (at STP) are also produced? Zn + 2HCl  ZnCl 2 + H 2 + 1250 kJ

15. Rules of Thermochemistry Rule #2 - H for a reaction is equal in the magnitude but opposite in sign to H for the reverse reaction. (If 6.00 kJ of heat absorbed when a mole of ice melts, then 6.00 kJ of heat is given off when 1.00 mol of liquid water freezes)

16. Rules of Thermochemistry 2 H 2 O 2 (l) 2 H 2 O (l) + O 2 (g) H = -196 kJ 2 H 2 O (l) + O 2 (g) 2 H 2 O 2 (l) H = +196 kJ Remember “elephant’s toothpaste?”

17. Rules of Thermochemistry Example: Given: H 2 + ½O 2  H 2 OH = -285.8 kJ Calculate H for the equation: 2H 2 O  2H 2 + O 2 Flip eq’n and x 2 …so do the same to ∆H H = 2 x –(-285.8 kJ) H = 571.6 kJ

18. Rules of Thermochemistry Rule #3) The value of H for a reaction is the same whether it occurs in one step or in a series of steps. H for the overall equation is the sum of the H’s for the individual equations: Hess’s Law:H = H 1 + H 2 + …

19. Example 1: Calculate H for the reaction: C + ½O 2  CO Given: 1) C + O 2  CO 2 H = -393.5 kJ 2) 2CO + O 2  2CO 2 H = -566.0 kJ  2 flip C + ½O 2  CO H = -110.5 kJ

20. Hess’s Law:  H =  H 1 +  H 2 + … Example 2: Find the heat of reaction (enthalpy) for the following reaction NO + ½O 2  NO 2 H = ? Given the following equation…. ½N 2 + ½O 2  NOH = +90.4 kJ ½N 2 + O 2  NO 2 H = +33.6 flip NO  ½N 2 + ½O 2 H = -90.4 kJ NO + ½O 2  NO 2 H = -56.8 kJ