Practice Problem 4

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

(a) Hg(lequilibr.gif (63 bytes) Hg(g)

(b) 2NO2(g) equilibr.gif (63 bytes) N2O4(g)

(c) N2(g) + O2(g) equilibr.gif (63 bytes) 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:

 deltaSo = sigmaSo(products) - sigmaSo(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  deltaSo 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 deltaSo is calculated as follows:

deltaSo = sigmaSo(products) - sigmaSo(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 deltaSo 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:

deltaSo = sigmaSo(products) - sigmaSo(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

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

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