1. the heat evolved or absorbed in a chemical process is the same whether the process takes place in one or several steps.

2. if two or more chemical equations can be added together to produce an overall equation, the sum of the enthalpy equals the enthalpy change of the overall equation. This is called the Heat of Summation, ∆H

3. Analogy for Hess's Law There is an old Chinese proverb which says: There are many ways to the top of a mountain, but the view from the top is always the same.

4. Develop an analogy for soccer and scoring a goal.

6. Hess’s Law Read through the whole question Plan a Strategy Evaluate the given equations. Rearrange and manipulate the equations so that they will produce the overall equation. Add the enthalpy terms.

7. H 2 O(g) + C(s) → CO(g) + H 2 (g) Use these equations to calculate the molar enthalpy change which produces hydrogen gas. C(s) + ½ O 2 (g) → CO(g) ∆H = -110.5kJ H 2 (g) + ½ O 2 (g) → H 2 O(g) ∆H = -241.8kJ Example 1

8. H 2 O(g) + C(s) → CO(g) + H 2 (g) Use these equations to calculate the molar enthalpy change which produces hydrogen gas. C(s) + ½ O 2 (g) → CO(g) ∆H = -110.5kJ H 2 O(g) → H 2 (g) + ½ O 2 (g) ∆H = +241.8kJ _____________________________________ C(s) + H 2 O(g) → H 2 (g) + CO(g) ∆H=+131.3kJ

9. 4C(s) + 5H 2 (g) → C 4 H 10 (g) Use these equations to calculate the molar enthalpy change which produces butane gas. C 4 H 10 (g) + 6 ½ O 2 (g) → 4CO 2 (g) + 5H 2 O(g) ∆H= -2657.4kJ/mol C(s) + O 2 (g) → CO 2 (g) ∆H= -393.5kJ/mol H 2 (g) + ½O 2 (g) → H 2 O(g) ∆H= -241.8kJ/mol Read through the whole question Plan a Strategy Evaluate the given equations. Rearrange and manipulate the equations so that they will produce the overall equation. Add the enthalpy terms. REWRITE THE CHANGES. Example 2

10. 4C(s) + 5H 2 (g) → C 4 H 10 (g) Use these equations to calculate the molar enthalpy change which produces butane gas. C 4 H 10 (g) + 6 ½ O 2 (g) → 4CO 2 (g) + 5H 2 O(g) ∆H= -2657.4kJ/mol C(s) + O 2 (g) → CO 2 (g) ∆H= -393.5kJ/mol H 2 (g) + ½O 2 (g) → H 2 O(g) ∆H= -241.8kJ/mol Example 2

11. 4C(s) + 5H 2 (g) → C 4 H 10 (g) Use these equations to calculate the molar enthalpy change which produces butane gas. 5H 2 O(g) + 4CO 2 (g) → 6 ½ O 2 (g) + C 4 H 10 (g) ∆H= +2657.4kJ/mol C(s) + O 2 (g) → CO 2 (g) ∆H= -393.5kJ/mol H 2 (g) + ½O 2 (g) → H 2 O(g) ∆H= -241.8kJ/mol Example 2

12. 4C(s) + 5H 2 (g) → C 4 H 10 (g) Use these equations to calculate the molar enthalpy change which produces butane gas. 5H 2 O(g) + 4CO 2 (g) → 6 ½ O 2 (g) + C 4 H 10 (g) ∆H= +2657.4kJ/mol 4(C(s) + O 2 (g) → CO 2 (g)) ∆H= 4(-393.5kJ/mol) H 2 (g) + ½O 2 (g) → H 2 O(g) ∆H= -241.8kJ/mol Example 2

13. 4C(s) + 5H 2 (g) → C 4 H 10 (g) Use these equations to calculate the molar enthalpy change which produces butane gas. 5H 2 O(g) + 4CO 2 (g) → 6 ½ O 2 (g) + C 4 H 10 (g) ∆H= +2657.4kJ/mol 4C(s) + 4O 2 (g) → 4CO 2 (g) distribute the 4 ∆H= 4(-393.5kJ/mol) H 2 (g) + ½O 2 (g) → H 2 O(g) ∆H= -241.8kJ/mol Example 2

14. 4C(s) + 5H 2 (g) → C 4 H 10 (g) Use these equations to calculate the molar enthalpy change which produces butane gas. 5H 2 O(g) + 4CO 2 (g) → 6 ½ O 2 (g) + C 4 H 10 (g) ∆H= +2657.4kJ/mol 4C(s) + 4O 2 (g) → 4CO 2 (g) distribute the 4 ∆H= 4(-393.5kJ/mol) 5(H 2 (g) + ½O 2 (g) → H 2 O(g)) distribute the 5 ∆H= 5(-241.8kJ/mol) Example 2

15. 4C(s) + 5H 2 (g) → C 4 H 10 (g) Use these equations to calculate the molar enthalpy change which produces butane gas. 5H 2 O(g) + 4CO 2 (g) → 6 ½ O 2 (g) + C 4 H 10 (g) ∆H= +2657.4kJ/mol 4C(s) + 4O 2 (g) → 4CO 2 (g) ∆H= 4(-393.5kJ/mol) 5H 2 (g) + 2½O 2 (g) → 5H 2 O(g) ∆H= 5(-241.8kJ/mol) Example 2

16. 4C(s) + 5H 2 (g) → C 4 H 10 (g) Use these equations to calculate the molar enthalpy change which produces butane gas. 5H 2 O(g) + 4CO 2 (g) → 6 ½ O 2 (g) + C 4 H 10 (g) ∆H= +2657.4kJ/mol 4C(s) + 4O 2 (g) → 4CO 2 (g) ∆H= 4(-393.5kJ/mol) 5H 2 (g) + 2½O 2 (g) → 5H 2 O(g) ∆H= 5(-241.8kJ/mol) _____________________________________________________ ∆H = -125.6kJ/mol Example 2