hess-law

# 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.
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
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) + ½ Reactants Products Intermediate GeO(s) + ½ 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 Hess’s law: Example 5.7 (pg. 166) We may need to manipulate equations further: Don’t 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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