Hess's Law

Hess’s Law Germain Henri Hess

A. Hess’s Law Hess’s law states that if you can add two or more thermochemical equations to produce a final equation for a reaction, then the sum of the enthalpy changes for the individual reactions is the enthalpy change for the final reaction 2S(s) + 3O2(g)  2SO3(g) H = ?

A. Hess’s Law

Hess’s law Examples Hess’s Law states that the heat of a whole reaction is equivalent to the sum of it’s steps. For example: C + O2  CO2 (pg. 165) The book tells us that this can occur as 2 steps C + ½O2  CO H = – 110.5 kJ CO + ½O2  CO2 H = – 283.0 kJ C + CO + O2  CO + CO2 H = – 393.5 kJ I.e. C + O2  CO2 H = – 393.5 kJ Hess’s 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” …

C + ½ O2  CO H = – 110.5 kJ C + O2  CO2 H = – 393.5 kJ C + O2 CO + ½ O2  CO2 H = – 283.0 kJ C + O2  CO2 H = – 393.5 kJ C + O2 Reactants H = – 110.5 kJ CO + ½ O2 Intermediate Enthalpy H = – 393.5 kJ H = – 283.0 kJ Products CO2 Note: states such as (s) and (g) have been ignored to reduce space on these slides.

Practice Exercise 6 with Diagram Using example as a model, Draw the related enthalpy diagram. C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l) H= –1411.1 kJ 2CO2(g) + 3H2O(l)  C2H5OH(l) + 3O2(g) H= +1367.1 kJ C2H4(g) + H2O(l)  C2H5OH(l) H= – 44.0 kJ C2H4(g) + H2O(l) + 3O2(g) Reactants Products C2H5OH(l) + 3O2(g) H= – 44.0 kJ Enthalpy – 1411.1 kJ H = H = +1367.1 kJ 2CO2(g) + 3H2O(l) Intermediate

Practice Exercise with Diagram NO(g)  ½ N2(g) + ½ O2(g) H= – 90.37 kJ ½ N2(g) + O2(g)  NO2(g) H= + 33.8 kJ H= – 56.57 kJ NO(g) + ½ O2(g)  NO2(g) NO + ½ O2(g) Reactants H= – 56.6 kJ NO2(g) Enthalpy – 90.37 kJ H = Products H = +33.8 kJ ½ N2(g) + O2(g) Intermediate

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