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Thermochemistry Part 3 - Enthalpy and Thermochemical Equations.

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Presentation on theme: "Thermochemistry Part 3 - Enthalpy and Thermochemical Equations."— Presentation transcript:

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 = 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 = kJ-890. kJ 1 mol CH 4 2 mol O 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 = kJ When 1 mole of CH 4 is burned, 890. kJ of heat are released. When 2 moles of CH 4 are burned, 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 = kJ Calculate H when 1.00 g of Cl 2 reacts. ΔH = 1.00 g Cl 2 x 1 mol Cl 2 x 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 = 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 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 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 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 = 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 –( kJ) H = 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 = kJ 2) 2CO + O 2  2CO 2 H = kJ  2 flip C + ½O 2  CO H = 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 = kJ ½N 2 + O 2  NO 2 H = flip NO  ½N 2 + ½O 2 H = kJ NO + ½O 2  NO 2 H = kJ


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