1. ☾ Thermochemical Equations ☽
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) 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) kJ
When 1 mol of CaO reacts with 1 mol of H2O, 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) kJ
When 1 mol of CaO reacts with 1 mol of H2O, 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) kJ
When 1 mol of CaO reacts with 1 mol of H2O, 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 = kJ

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

15. Thermochemical Equations
2 NaHCO3(s) → Na2CO3(s) + H2O(g) + CO2(g) ∆H = 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 = 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 = 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 = 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 = − 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 = 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 = kJ
Knowns: mNaHCO3 = g MNaHCO3 = 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 = kJ
Knowns: mNaHCO3 = g MNaHCO3 = 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 = kJ
Knowns: mNaHCO3 = g MNaHCO3 = 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 = kJ
Knowns: mNaHCO3 = g MNaHCO3 = 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 = kJ
Knowns: mNaHCO3 = g MNaHCO3 = 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 = kJ
Knowns: mNaHCO3 = g MNaHCO3 = 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 = kJ
Knowns: mNaHCO3 = g MNaHCO3 = 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 = kJ
Knowns: mNaHCO3 = g MNaHCO3 = 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 = kJ
Knowns: mNaHCO3 = g MNaHCO3 = 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 = kJ
Knowns: mNaHCO3 = g MNaHCO3 = 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 = kJ
Knowns: mNaHCO3 = g MNaHCO3 = 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 = 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 = kJ
H2O(g) → H2(g) + ½ O2(g) ∆H = 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 = kJ
H2O(g) → H2(g) + ½ O2(g) ∆H = 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 = kJ
H2O(g) → H2(g) + ½ O2(g) ∆H = 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.