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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.

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Presentation on theme: "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."— Presentation transcript:

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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 = – kJ CO + ½O 2 CO 2 H = – kJ C + CO + O 2 CO + CO 2 H = – kJ I.e. C + O 2 CO 2 H = – 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 = – 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 = – kJ H = – kJ H = – kJ + ½ O 2 C + ½ O 2 CO H = – kJ CO + ½ O 2 CO 2 H = – 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 = – kJ 2CO 2 (g) + 3H 2 O(l) C 2 H 5 OH(l) + 3O 2 (g) H = 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 = – kJ H = 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 = kJ Ge(s) + O 2 (g) GeO 2 (s) H = – 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 = kJ H = – kJ H = – 280 kJ H = – kJ

8 5.52 (pg. 175) NO(g) ½ N 2 (g) + ½ O 2 (g) H = – kJ ½ N 2 (g) + O 2 (g) NO 2 (g) H = 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 = – kJ H = kJ H = – 56.6 kJ H = – 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 = – kJ CO + ½ O 2 CO 2 H = – 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 O 2 3CO 2 H = – kJ 2Fe + 3CO 2 Fe 2 O 3 + 3CO H = kJ CO + ½ O 2 CO 2 H = – kJ Dont forget to add states. Try 5.55, 5.57, 5.58, 5.61 (pg. 175) H = – kJ For more lessons, visit


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