3. Hesss law Hesss Law states that the heat of a whole reaction is equivalent to the sum of its steps. For example: C + O 2 CO 2 (pg. 165) The book tells us that this can occur as 2 steps C + ½O 2 CO H = – 110.5 kJ CO + ½O 2 CO 2 H = – 283.0 kJ C + CO + O 2 CO + CO 2 H = – 393.5 kJ I.e. C + O 2 CO 2 H = – 393.5 kJ Hesss law allows us to add equations. We add all reactants, products, & H values. We can also show how these steps add together via an enthalpy diagram …

4. Steps in drawing enthalpy diagrams 1.Balance the equation(s). 2.Sketch a rough draft based on H values. 3.Draw the overall chemical reaction as an enthalpy diagram (with the reactants on one line, and the products on the other line). 4.Draw a reaction representing the intermediate step by placing the relevant reactants on a line. 5.Check arrows:Start: two leading away Finish: two pointing to finish Intermediate: one to, one away 6.Look at equations to help complete balancing (all levels must have the same # of all atoms). 7.Add axes and H values.

5. C + O 2 CO 2 H = – 393.5 kJ Reactants Intermediate Products C + O 2 CO 2 CO Enthalpy Note: states such as (s) and (g) have been ignored to reduce clutter on these slides. You should include these in your work. H = – 110.5 kJ H = – 283.0 kJ H = – 393.5 kJ + ½ O 2 C + ½ O 2 CO H = – 110.5 kJ CO + ½ O 2 CO 2 H = – 283.0 kJ

6. Practice Exercise 6 (pg. 167) with Diagram Using example 5.6 as a model, try PE 6. Draw the related enthalpy diagram. C 2 H 4 (g) + 3O 2 (g) 2CO 2 (g) + 2H 2 O(l) H = –1411.1 kJ 2CO 2 (g) + 3H 2 O(l) C 2 H 5 OH(l) + 3O 2 (g) H = +1367.1 kJ C 2 H 4 (g) + H 2 O(l) C 2 H 5 OH(l) Reactants Products Intermediate C 2 H 4 (g) + H 2 O(l) C 2 H 5 OH(l) 2CO 2 (g) + 3H 2 O(l) Enthalpy H = – 1411.1 kJ H = +1367.1 kJ H = – 44.0 kJ + 3O 2 (g) H = – 44.0 kJ

7. 5.51 (pg. 175) GeO(s) Ge(s) + ½ O 2 (g) H = + 255 kJ Ge(s) + O 2 (g) GeO 2 (s) H = – 534.7 kJ GeO(s) + ½ O 2 (g) GeO 2 (s) Reactants Products Intermediate GeO(s) + ½ O 2 (g) GeO 2 (s) Ge(s) + O 2 (g) Enthalpy H = + 255 kJ H = – 534.7 kJ H = – 280 kJ H = – 279.7 kJ

8. 5.52 (pg. 175) NO(g) ½ N 2 (g) + ½ O 2 (g) H = – 90.37 kJ ½ N 2 (g) + O 2 (g) NO 2 (g) H = + 33.8 kJ NO(g) + ½ O 2 (g) NO 2 (g) Reactants Products Intermediate NO + ½ O 2 (g) NO 2 (g) ½ N 2 (g) + O 2 (g) Enthalpy H = – 90.37 kJ H = +33.8 kJ H = – 56.6 kJ H = – 56.57 kJ

9. Hesss law: Example 5.7 (pg. 166) We may need to manipulate equations further: 2Fe + 1.5O 2 Fe 2 O 3 H =?, given Fe 2 O 3 + 3CO 2Fe + 3CO 2 H = – 26.74 kJ CO + ½ O 2 CO 2 H = – 282.96 kJ 1: Align equations based on reactants/products. 2: Multiply based on final reaction. 3: Add equations. 2Fe + 1.5O 2 Fe 2 O 3 3CO + 1.5 O 2 3CO 2 H = – 848.88 kJ 2Fe + 3CO 2 Fe 2 O 3 + 3CO H = + 26.74 kJ CO + ½ O 2 CO 2 H = – 282.96 kJ Dont forget to add states. Try 5.55, 5.57, 5.58, 5.61 (pg. 175) H = – 822.14 kJ For more lessons, visit www.chalkbored.com www.chalkbored.com