Enthalpy Changes

1. Enthalpy Changes • Measuring and Expressing ∆H • ☾ Thermochemical Equations ☽

2. Thermochemical Equations • In a thermochemical reaction, ∆H for the reaction can be written either as a reactant or product.

3. Thermochemical Equations • In a thermochemical reaction, ∆H for the reaction can be written either as a reactant or product. • CaO(s) + H2O(l) → Ca(OH)2(s) + 65.2 kJ

4. Thermochemical Equations • In a thermochemical reaction, ∆H for the reaction can be written either as a reactant or product. • CaO(s) + H2O(l) → Ca(OH)2(s) + 65.2 kJ • When 1 mol of CaO reacts with 1 mol of H2O, 65.2 kJ of heat are produced at standard temperature, 298 K, and standard pressure, 1 atm.

5. Thermochemical Equations • In a thermochemical reaction, ∆H for the reaction can be written either as a reactant or product. • CaO(s) + H2O(l) → Ca(OH)2(s) + 65.2 kJ • When 1 mol of CaO reacts with 1 mol of H2O, 65.2 kJ of heat are produced at standard temperature, 298 K, and standard pressure, 1 atm. • More often we write the equation like this ...

6. Thermochemical Equations • In a thermochemical reaction, ∆H for the reaction can be written either as a reactant or product. • CaO(s) + H2O(l) → Ca(OH)2(s) + 65.2 kJ • When 1 mol of CaO reacts with 1 mol of H2O, 65.2 kJ of heat are produced at standard temperature, 298 K, and standard pressure, 1 atm. • More often we write the equation like this ... • CaO(s) + H2O(l) → Ca(OH)2(s) ∆H = −65.2 kJ

7. Thermochemical Equations • CaO(s) + H2O(l) → Ca(OH)2(s) ∆H = −65.2 kJ

8. Thermochemical Equations • CaO(s) + H2O(l) → Ca(OH)2(s) ∆H = −65.2 kJ CaO(s) + H2O(l)

9. Thermochemical Equations • CaO(s) + H2O(l) → Ca(OH)2(s) ∆H = −65.2 kJ CaO(s) + H2O(l) The reactants lose 65.2 kJ of heat to the surroundings.

10. Thermochemical Equations • CaO(s) + H2O(l) → Ca(OH)2(s) ∆H = −65.2 kJ CaO(s) + H2O(l) The reactants lose 65.2 kJ of heat to the surroundings. ∆H = −65.2 kJ

11. Thermochemical Equations • CaO(s) + H2O(l) → Ca(OH)2(s) ∆H = −65.2 kJ CaO(s) + H2O(l) ∆H = −65.2 kJ Ca(OH)2(s)

12. Thermochemical Equations • CaO(s) + H2O(l) → Ca(OH)2(s) ∆H = −65.2 kJ CaO(s) + H2O(l) The reaction is exothermic. ∆H = −65.2 kJ Ca(OH)2(s)

13. Thermochemical Equations • 2 NaHCO3(s) → Na2CO3(s) + H2O(g) + CO2(g) ∆H = +129 kJ

14. Thermochemical Equations • 2 NaHCO3(s) → Na2CO3(s) + H2O(g) + CO2(g) ∆H = +129 kJ NaHCO3(s)

15. Thermochemical Equations • 2 NaHCO3(s) → Na2CO3(s) + H2O(g) + CO2(g) ∆H = +129 kJ The reactant gains 129 kJ of heat from the surroundings. NaHCO3(s)

16. Thermochemical Equations • 2 NaHCO3(s) → Na2CO3(s) + H2O(g) + CO2(g) ∆H = +129 kJ The reactant gains 129 kJ of heat from the surroundings. ∆H = +129 kJ NaHCO3(s)

17. Thermochemical Equations • 2 NaHCO3(s) → Na2CO3(s) + H2O(g) + CO2(g) ∆H = +129 kJ Na2CO3(s) + H2O(g) + CO2(g) ∆H = +129 kJ NaHCO3(s)

18. Thermochemical Equations • 2 NaHCO3(s) → Na2CO3(s) + H2O(g) + CO2(g) ∆H = +129 kJ Na2CO3(s) + H2O(g) + CO2(g) The reaction is endothermic. ∆H = +129 kJ NaHCO3(s)

19. Thermochemical Equations • The ∆H of reactions is an extensive variable.

20. Thermochemical Equations • The ∆H of reactions is an extensive variable. • The value of ∆H depends on the amount of material.

21. Thermochemical Equations • The ∆H of reactions is an extensive variable. • The value of ∆H depends on the amount of material. • Twice the amount of reactant produces twice the ∆H.

22. Thermochemical Equations • The ∆H of reactions is an extensive variable. • The value of ∆H depends on the amount of material. • Twice the amount of reactant produces twice the ∆H. • CaO(s) + H2O(l) → Ca(OH)2(s) ∆H = −65.2 kJ

23. Thermochemical Equations • The ∆H of reactions is an extensive variable. • The value of ∆H depends on the amount of material. • Twice the amount of reactant produces twice the ∆H. • CaO(s) + H2O(l) → Ca(OH)2(s) ∆H = −65.2 kJ • 2 CaO(s) + 2 H2O(l) → 2 Ca(OH)2(s) ∆H = −130.4 kJ

24. Thermochemical Equations • The ∆H of reactions is an extensive variable. • The value of ∆H depends on the amount of material. • Twice the amount of reactant produces twice the ∆H.

25. Thermochemical Equations • The ∆H of reactions is an extensive variable.

26. Example • What is the ∆H of the decomposition of 105.0g of sodium bicarbonate to sodium carbonate, carbon dioxide, and water. • 2 NaHCO3(s) → Na2CO3(s) + H2O(g) + CO2(g) ∆H = +129 kJ

27. Example • What is the ∆H of the decomposition of 105.0g of sodium bicarbonate to sodium carbonate, carbon dioxide, and water. • 2 NaHCO3(s) → Na2CO3(s) + H2O(g) + CO2(g) ∆H = +129 kJ • Knowns: mNaHCO3 = 105.0 g MNaHCO3 = 84.01 g/mol • ∆H = +129 kJ/2 mol NaHCO3

28. Example • What is the ∆H of the decomposition of 105.0g of sodium bicarbonate to sodium carbonate, carbon dioxide, and water. • 2 NaHCO3(s) → Na2CO3(s) + H2O(g) + CO2(g) ∆H = +129 kJ • Knowns: mNaHCO3 = 105.0 g MNaHCO3 = 84.01 g/mol • ∆H = +129 kJ/2 mol NaHCO3 • Calculation:

29. Example • What is the ∆H of the decomposition of 105.0g of sodium bicarbonate to sodium carbonate, carbon dioxide, and water. • 2 NaHCO3(s) → Na2CO3(s) + H2O(g) + CO2(g) ∆H = +129 kJ • Knowns: mNaHCO3 = 105.0 g MNaHCO3 = 84.01 g/mol • ∆H = +129 kJ/2 mol NaHCO3 • Calculation: 105.0 g NaHCO3 1

30. Example • What is the ∆H of the decomposition of 105.0g of sodium bicarbonate to sodium carbonate, carbon dioxide, and water. • 2 NaHCO3(s) → Na2CO3(s) + H2O(g) + CO2(g) ∆H = +129 kJ • Knowns: mNaHCO3 = 105.0 g MNaHCO3 = 84.01 g/mol • ∆H = +129 kJ/2 mol NaHCO3 • Calculation: 105.0 g NaHCO3 1 mol NaHCO3 × 1 84.01 g NaHCO3

31. Example • What is the ∆H of the decomposition of 105.0g of sodium bicarbonate to sodium carbonate, carbon dioxide, and water. • 2 NaHCO3(s) → Na2CO3(s) + H2O(g) + CO2(g) ∆H = +129 kJ • Knowns: mNaHCO3 = 105.0 g MNaHCO3 = 84.01 g/mol • ∆H = +129 kJ/2 mol NaHCO3 • Calculation: 105.0 g NaHCO3 1 mol NaHCO3 × 1 84.01 g NaHCO3

32. Example • What is the ∆H of the decomposition of 105.0g of sodium bicarbonate to sodium carbonate, carbon dioxide, and water. • 2 NaHCO3(s) → Na2CO3(s) + H2O(g) + CO2(g) ∆H = +129 kJ • Knowns: mNaHCO3 = 105.0 g MNaHCO3 = 84.01 g/mol • ∆H = +129 kJ/2 mol NaHCO3 • Calculation: (105.0)(1) mol NaHCO3 (1)(84.01)

33. Example • What is the ∆H of the decomposition of 105.0g of sodium bicarbonate to sodium carbonate, carbon dioxide, and water. • 2 NaHCO3(s) → Na2CO3(s) + H2O(g) + CO2(g) ∆H = +129 kJ • Knowns: mNaHCO3 = 105.0 g MNaHCO3 = 84.01 g/mol • ∆H = +129 kJ/2 mol NaHCO3 • Calculation: (105.0)(1) mol NaHCO3 +129 kJ × (1)(84.01) 2 mol NaHCO3

34. Example • What is the ∆H of the decomposition of 105.0g of sodium bicarbonate to sodium carbonate, carbon dioxide, and water. • 2 NaHCO3(s) → Na2CO3(s) + H2O(g) + CO2(g) ∆H = +129 kJ • Knowns: mNaHCO3 = 105.0 g MNaHCO3 = 84.01 g/mol • ∆H = +129 kJ/2 mol NaHCO3 • Calculation: (105.0)(1) mol NaHCO3 +129 kJ × (1)(84.01) 2 mol NaHCO3

35. Example • What is the ∆H of the decomposition of 105.0g of sodium bicarbonate to sodium carbonate, carbon dioxide, and water. • 2 NaHCO3(s) → Na2CO3(s) + H2O(g) + CO2(g) ∆H = +129 kJ • Knowns: mNaHCO3 = 105.0 g MNaHCO3 = 84.01 g/mol • ∆H = +129 kJ/2 mol NaHCO3 • Calculation: (105.0)(1)(+129) kJ (1)(84.01)(2)

36. Example • What is the ∆H of the decomposition of 105.0g of sodium bicarbonate to sodium carbonate, carbon dioxide, and water. • 2 NaHCO3(s) → Na2CO3(s) + H2O(g) + CO2(g) ∆H = +129 kJ • Knowns: mNaHCO3 = 105.0 g MNaHCO3 = 84.01 g/mol • ∆H = +129 kJ/2 mol NaHCO3 • Calculation: (105.0)(1)(+129) kJ = 80.6 kJ (1)(84.01)(2)

37. Example • What is the ∆H of the decomposition of 105.0g of sodium bicarbonate to sodium carbonate, carbon dioxide, and water. • 2 NaHCO3(s) → Na2CO3(s) + H2O(g) + CO2(g) ∆H = +129 kJ • Knowns: mNaHCO3 = 105.0 g MNaHCO3 = 84.01 g/mol • ∆H = +129 kJ/2 mol NaHCO3 • Calculation: (105.0)(1)(+129) kJ = 80.6 kJ (1)(84.01)(2)

38. Thermochemical Equations • Thermochemical equations also require that we give the physical states of the reactants.

39. Thermochemical Equations • Thermochemical equations also require that we give the physical states of the reactants. • H2O(l) → H2(g) + ½ O2(g) ∆H = 285.8 kJ

40. Thermochemical Equations • Thermochemical equations also require that we give the physical states of the reactants. • H2O(l) → H2(g) + ½ O2(g) ∆H = 285.8 kJ • H2O(g) → H2(g) + ½ O2(g) ∆H = 241.8 kJ

41. Thermochemical Equations • Thermochemical equations also require that we give the physical states of the reactants. • H2O(l) → H2(g) + ½ O2(g) ∆H = 285.8 kJ • H2O(g) → H2(g) + ½ O2(g) ∆H = 241.8 kJ • Even though the stoichiometry is the same, the state of the reactants are different.

42. Thermochemical Equations • Thermochemical equations also require that we give the physical states of the reactants. • H2O(l) → H2(g) + ½ O2(g) ∆H = 285.8 kJ • H2O(g) → H2(g) + ½ O2(g) ∆H = 241.8 kJ • Even though the stoichiometry is the same, the state of the reactants are different. • The difference in ∆H is 44.0 kJ.

43. Summary • In a thermochemical reaction, ∆H for the reaction can be written either as a reactant or product. • More often we write the equation like this ... • CaO(s) + H2O(l) → Ca(OH)2(s) ∆H = −65.2 kJ • Where ∆H is negative, the reaction is exothermic. • Where ∆H is positive, the reaction is endothermic.

44. Summary • The ∆H of reactions is an extensive variable and depends on the amount of material. • Thermochemical equations also require that we give the physical states of the reactants.