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CHM1 12Example9 Collection

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Oct 18, 2019, 1:17:49 AM
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Oct 18, 2019, 1:17:49 AM
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Example 9

See the calculation on this YouTube video

Solution Using Eq. (4), we have

?Hm = ? ?Hf (products) – ? ?Hf (reactants)

       = [6 ?Hf (H2O) + 4 ?Hf (NO)] – [4 ?Hf (NH3) + 5 ?Hf (O2)]

       = 6(–241.8) kJ mol–1 + 4(90.3) kJ mol–1 – 4(–46.1 kJ mol–1) – 5 × 0

       = –1450.8 kJ mol–1 + 361.2 kJ mol–1 + 184.4 kJ mol–1

       = –905.2 kJ mol–1

Note that we were careful to use ?Hf [H2O(g)] not ?Hf [H2O(l)]. Even though water vapor is not the most stable form of water at 25°C, we can still use its ?Hf value. Also the standard enthalpy of formation of the element O2(g) is zero by definition. Obviously it would be a waste of space to include it in the table above. 

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