Practice Problem 4

Calculate the standard-state entropy of reaction for the following reactions and explain the sign of  S° for each reaction.

(a) Hg(l)  Hg(g)

(b) 2NO₂(g) N₂O₄(g)

(c) N₂(g) + O₂(g) 2NO(g)

Solution

(a) Using a standard-state entropy data table, we find the following information:

Compound            S°(J/mol-K)

Hg(l)                         76.02

Hg(s)                      174.96

The balanced equation states that 1 mole of mercury vapor is produced for each mole of liquid mercury that boils.  The standard-state entropy of reaction is therefore calculated as follows:

 S^o = S^o(products) - S^o(reactants)

        = [1 mol Hg(g) x 174.96 J/mol-K] - [1 mol Hg(l) x 76.02 J/mol-K]

        = 98.94 J/K

The sign of  S^o is positive because this process transforms a liquid into a gas, which is inherently more disordered.

(b) Using a standard-state entropy data table, we find the following information:

Compound            S°(J/mol-K)

NO₂(g)                     240.06

N₂O₄(g)                     304.29

In this equation, 1 mole of N₂O₄ is formed for every mole of NO₂ consumed, and the value of S^o is calculated as follows:

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

        = [1 mol N₂O₄(g) x 304.29 J/mol-K] - [2 mol NO₂(l) x 240.06 J/mol-K]

        = -175.83 J/K

The sign of S^o is negative because two molecules combine in this reaction to form a larger, more ordered product.

(c) Using a standard-state entropy data table, we find the following information:

Compound            S°(J/mol-K)

NO(g)                        210.76

                    N₂(g)                              191.61                        

O₂(g)                       205.14

The balanced equation for this reaction indicates that 2 mole of NO are produced when 1 mole of N₂ reacts with 1 mole of  O₂ .   Thus, the standard-state entropy of reaction is calculated as follows:

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

        = [2 mol NO x 210.76 J/mol-K] - [1 mol N₂ x 191/61 J/mol-K + 1 mol O₂ x 205.14 J/mol-K]

        = 24.77 J/K

S^o for this reaction is small but positive because the product of the reaction (NO) is slightly more disordered that the reactants ( N₂ and  O₂).