hess-law - C 2 H 4 (g) + 3O 2 (g) 2CO 2 (g) + 2H 2 O(l) H =...

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Hess’s law Hess’s Law states that the heat of a whole reaction is equivalent to the sum of it’s steps.
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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 1.Look at equations to help complete balancing (all levels must have the same # of all atoms).
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Reactants Intermediate Products 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 C + ½ O CO H ° = – 110.5 kJ
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Practice Exercise 6 (pg. 167) with Diagram Using example 5.6 as a model, try PE 6. Draw the related enthalpy diagram.
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Unformatted text preview: 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 C 2 H 4 (g) + H 2 O(l) C 2 H 5 OH(l) 2CO 2 (g) + Enthalpy H = 1411.1 kJ H = +1367.1 kJ H = 44.0 kJ + 3O 2 (g) + 3O 2 (g) H = 44.0 kJ 5.51 (pg. 175) GeO(s) Ge(s) + Reactants Products Intermediate GeO(s) + Enthalpy H = + 255 kJ H = 534.7 kJ H = 280 kJ H = 279 .7 kJ 5.52 (pg. 175) Reactants Products Intermediate Enthalpy H = 90.37 kJ H = +33.8 kJ H = 56.6 kJ H = 56.5 7 kJ Hesss law: Example 5.7 (pg. 166) We may need to manipulate equations further: Dont forget to add states. Try 5.55, 5.57, 5.58, H = 822.14 kJ For more lessons, visit www.chalkbored.com...
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hess-law - C 2 H 4 (g) + 3O 2 (g) 2CO 2 (g) + 2H 2 O(l) H =...

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