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Entropy (S)

>> Parts of this equation/concept include:

A -	Predicting Entropy Qualitatively
B -	Calculating Entropy Quantitatively

Entropy is a measure of randomness. Consequently, systems that are more chaotic have a higher entropy value. If a change makes a system more random, S is positive. If a change makes a system more ordered, S is negative.

To qualitatively predict the sign on S for a chemical reaction, consider the following (listed in order of relative importance).

• physical state	gas > aqueous > liquid > solid
• mixtures	mixtures > pure substances
• number of moles of gas	more moles > fewer moles
• bonds in a molecule	more bonds > fewer bonds single bonds > double bonds > triple bonds less rigid > more rigid

>> Example 1

What is the sign on S for the following reactions?

1. 2 C₈H₁₈(l) + 25 O₂(g) 16 CO₂(g) + 18 H₂O(g)
2. NaCl(s) Na⁺(aq) + Cl^–(aq)
3. N₂(g) + 3 H₂(g) 2 NH₃(g)

Solution:

1. In this example a liquid and a gas become gases. There is also an increase in number of moles (27 to 34). Both of these indicate an increase in randomness or +S.
2. One mole of solid makes 2 moles of a mixture. This is also an increase in randomness and +S.
3. In this case the physical state of all materials is a gas, so physical state is not a criterion that can be used. However, the reaction takes 4 moles of gas and creates 2. It also goes from a mixture (two substances) to a pure substance. Therefore randomness decreases and S is negative.

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Use the values of entropy in the appendix and Equation 13.5 to quantitatively calculate S. Normally, values for S are positive, since the lowest (most ordered value) is 0 K. Also, the values are for S, not S, since there is an absolute value given in the third law of thermodynamics. Such a parallel does not exist for enthalpy.

The mathematical procedure is the same as that for calculating H from heats of formation.

S = nS_products – mS_reactants         (Equation 11.13)

Recall that is the summation sign, n is the stoichiometric coefficient of the product, and m is the stoichiometric coefficient of the reactant. This equation assumes that all reactants and products are under the same thermodynamic conditions. If there is a superscript 0, standard conditions of 25°C (298 K) and 100 Pa (about 1 atm) are indicated. Since tables only list values at standard conditions, they are often assumed, even if not explicitly stated.

Fortunately, the values do not change substantially with temperature and small pressure changes. Consequently, standard state values are commonly used even when the system is not at standard state. After all, those are usually the only values available.

>> Example 2

Calculate the value of S° for the following reactions. Compare to the predictions in Example 1.

1. 2 C₈H₁₈(l) + 25 O₂(g) 16 CO₂(g) + 18 H₂O(g)
2. NaCl(s) Na⁺(aq) + Cl^–(aq)
3. N₂(g) + 3 H₂(g) 2 NH₃(g)

Solution:

1. Use the appendix to look up the S° values for each reactant and product. All units are in J/mol•K.

   As long as no other units are used, we can skip them in the calculations and just put the units in at the end.

   S° of C₈H₁₈ = 463.7

   S° of O₂ = 205.0

   S° of CO₂ = 213.6

   S° of H₂O = 188.7

   Note: The appendix lists two physical states for water. Make sure you pick the one consistent with the reaction.

   Use Equation 13.5.

   S = nS_products – mS_reactants

   S = [16(213.6) + 18(188.7)] – [2(483.7) + 25(205.0)]

   S = [3417.6 + 3396.6] – [967.4 + 5125.0]

   S = [6814.2] – [6092.4]

   S = +721.8 J/mol•K

   Following significant figures in these problems can be tricky because both multiplication and addition are involved. Remember that it is best to keep as many units as possible and to round at the final answer.

   In this example, the stoichiometric coefficients are exact and therefore have infinite significant figures. Significant figures are counted in multiplication; decimal places, in addition. The digits that are significant are underlined in each step of the preceding calculation. Therefore the final answer should be

   S = 722 J/mol•K

2. The v alues for each component of the reaction are

   S of NaCl = 72.1

   S° of Na⁺ = 59.0

   S° of Cl^– = 56.5 S = nS_products – mS_reactants

   S = [59.0 + 56.5] – [72.1] = [115.5] – [72.1] = +43.4 J/mol•K

   Since no multiplication is involved, use the least amount of decimal places to determine the appropriate significant figures.

3. The values for each component of the reaction are

   S for N₂ = 191.5

   S° for H₂ = 130.6

   S° for NH₃ = 192.3

   S = nS_products – mS_reactants

   S = [2(192.3)] – [191.5 + 3(130.6)] S = [384.6] – [191.5 + 391.8]

   S = 384.6 – 583.3 = –198.7 J/mol•K

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