Practice Problem 5

Calculate deltaH and deltaS for the following reaction and decide in which direction each of these factors will drive the reaction.

N2(g) + 3 H2(g) equilibr.gif (63 bytes) 2 NH3(g)


Solution

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

Compound              deltaHfo(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 ( deltaH < 0), which means that the enthalpy of reaction favors the products of the reaction:

deltaHo = sigmaHfo(products) - sigmaHfo(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.

deltaSo = sigmaSo(products) - sigmaSo(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

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