**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) 2NO_{2}(*g*) N_{2}O_{4}(*g*)

(c) N_{2}(*g*) + O_{2}(*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)**

NO_{2}(*g*)
240.06

N_{2}O_{4}(*g*)
304.29

In this equation, 1 mole of N_{2}O_{4} is formed for every
mole of NO_{2 }consumed, and the value of *S*^{o} is calculated as follows:

*S*^{o}
= *S*^{o}(products)
- *S*^{o}(reactants)

= [1 mol N_{2}O_{4}(g) x
304.29 J/mol-K] - [2 mol NO_{2}(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

N_{2}(*g*)_{
} 191.61_{
}

O_{2}(*g*)_{ }
205.14

The balanced equation for this reaction indicates that 2 mole of NO are
produced when 1 mole of N_{2} reacts with 1 mole of O_{2} .
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 N_{2}
x 191/61 J/mol-K + 1 mol O_{2} 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 ( N_{2} and O_{2}).