Practice Problem 6

Practice Problem 6

Calculate H° and S° for the following reaction:

NH₄NO₃(s) + H₂O(l) NH₄⁺ (aq) + NO₃^- (aq)

Use the results of this calculation to determine the value of G^ofor this reaction at 25^oC, and explain why NH₄NO₃ spontaneously dissolves is water at room temperature.

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)

NH₄NO₃(s) -365.56 151.08

NH₄⁺ (aq) -132.51 113.4

NO₃^- (aq) -205.0 146.4

This reaction is endothermic, and the enthalpy of reaction is therefore unfavorable:

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

= [1 mol NH₄ x 132.51 kJ/mol + 1 mol NO₃^- x -205.0 kJ/mol] - [1 mol NH₄NO₃ x -365.56 kJ/mol]

= 28.05 kJ

The reaction leads to a significant increase in the disorder of the system, however, and is therefore favored by the entropy of reaction:

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

= [1 mol NH₄ x 113.4 J/mol-K + 1 mol NO₃^- x 146.4 J/mol-K] - [1 mol NH₄NO₃ x 151.08 J/mol-K]

= 108.7 J/K

To decide whether NH₄NO₃ should dissolve in water at 25^oC we have to compare the H^o and TS^o to see which is larger. Before we can do this, we have to convert the temperature from ^oC to kelvin:

T_K = 25^oC + 273.15 = 298.15 K

We also have to recognize that the units of H^o for this reaction are kilojoules and the units of S^o are joules per kelvin. At some point in this calculation, we therefore have to convert these quantites to a consistent set of untis. Perhaps the easiest way of doing this is to convert H^o to joules. We then multiply the entropy term by the absolute temperature and subtract this quantity from the enthalpy term:

G^o = H^o - T S^o

= 28,050 J - (298.15 K x 108.7 J/K)

= 28,050 J - 32,410 J

= -4360 J

At 25^oC, the standard-state free energy for this reaction is negative because the entropy term at this temperature is larger that the enthalpy term:

G^o= -4.4 kJ

The reaction is therefore spontaneous at room temperature.