CK-12-Chemistry Intermediate

20.1. Entropy http://www.ck12.org

The entropy is decreasing because a gas is becoming a liquid.

CaCO 3 (s)→CaO(s) + CO 2 (g)

The entropy is increasing because a gas is being produced, and the number of molecules is increasing.

N 2 (g) + 3H 2 (g)→2NH 3 (g)

The entropy is decreasing because four total reactant molecules are forming two total product molecules.

All are gases.

AgNO 3 (aq) + NaCl(aq)→NaNO 3 (aq) + AgCl(s)

The entropy is decreasing because a solid is formed from aqueous reactants.

H 2 (g) + Cl 2 (g)→2HCl(g)

The entropy change is unknown (but likely not zero) because there are equal numbers of molecules on

both sides of the equation, and all are gases.

Standard Entropy

All molecular motion ceases at absolute zero (0 K). Therefore, the entropy of a pure crystalline substance at absolute
zero is defined to be equal to zero. As the temperature of the substance increases, its entropy increases because of
an increase in molecular motion. The absolute or standard entropy of substances can be measured. The symbol for
entropy is S, and the standard entropy of a substance is given by the symbol S°, indicating that the standard entropy
is determined under standard conditions. The units for entropy are J/K•mol. Standard entropies for a few substances
are shown below (Table20.1).

TABLE20.1: Standard Entropy Values at 25°C

Substance S° (J/K•mol)

H 2 (g) 131.0

O 2 (g) 205.0

H 2 O(l) 69.9

H 2 O(g) 188.7

C(graphite) 5.69

C(diamond) 2.4

The standard entropy change (∆S°) for a reaction can be calculated from the absolute entropies of the various reaction
components, analogous to the way that the enthalpy change of a reaction can be calculated from standard heat of
formation values.

∆S° =∑nS°(products) -∑nS°(reactants)

The standard entropy change is equal to the sum of all the standard entropies of the products minus the sum of all
the standard entropies of the reactants. The symbol “n” signifies that each entropy must first be multiplied by its
coefficient in the balanced equation.

For example, the entropy change for the vaporization of water can be found as follows:

∆S° = S°(H 2 O(g)) - S°(H 2 O(l))

∆S° = 188.7 J/K•mol - 69.9 J/K•mol = 118.8 J/K•mol

The entropy change for the vaporization of water is positive because the gas state has higher entropy than the liquid
state.

The entropy change for the formation of liquid water from gaseous hydrogen and oxygen can also be calculated
using this equation.