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)
Using a standard-state enthalpy of formation and absolute entropy data table, we find the following information:
Compound Hfo(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:
Ho = Hfo(products) - Hfo(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.
So = So(products) - So(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