Practice Problem 5

Calculate H° and S° for the following reaction and decide in which direction each of these factors will drive the reaction.

N₂(g) + 3 H₂(g) 2 NH₃(g)

Solution

Using a standard-state enthalpy of formation and absolute entropy data table, we find the following information:

Compound              H_f^o(kJ/mol)         S°(J/mol-K)

N₂(g)                                 0                 191.61

H₂(g)                                 0                 130.68

NH₃(g)                             -46.11         192.45

The reaction is exothermic ( H° < 0), which means that the enthalpy of reaction favors the products of the reaction:

H^o = H_f^o(products) - H_f^o(reactants)

        = [2 mol NH₃ x 46.11 kJ/mol] - [1 mol N₂ x 0 kJ/mol + 3 mol H₂ x 0 kJ/mol]

        = -92.22 kJ

The entropy of reaction is unfavorable, however, because there is a significant increase in the order of the system,  when  N₂ and H₂ combine to form NH₃.

S^o = S^o(products) - S^o(reactants)

        = [2 mol NH₃ x 192.45 J/mol-K] - [1 mol N₂ x 191.61 J/mol-K + 3 mol H₂ x 130.68 J/mol-K]

        = -198.75 J/K