Entropy changes and their calculation

Even without knowing the actual values for the S^o of substances, it is possible to predict the sign of Delta S^o for a reaction or phase change. It is useful to remember that the entropy of a system will increase when:

1. A reaction in which a molecule is broken into two or more smaller molecules.

2. A reaction in which there is an increase in moles of gas.

3. A process in which a solid changes to a liquid or gas or a liquid changes to a gas.

Lets take the following example:

2 NaHCO₃ (s) ----> Na₂CO₃ (s) + H₂O (g) + CO₂ (g)

Looking at this reaction we see that two molecules of sodium bicarbonate (NaHCO₃) combine to form three molecules, Na₂CO₃, H₂O, and CO₂. Two of these molecules are gases. Therefore, rules 1 and 2 apply. The sodium bicarbonate is broken up into three smaller molecules and two of those are gases. We should suspect that the sign of [Delta]S^o will be positive. To confirm this, we can calculate the actual value of [Delta]S^o. Looking in table 18.1 (page 756) we find the following data:

S^o (NaHCO₃) = 102 J/K

S^o (Na₂CO₃) = 139 J/K

S^o (H₂O) = 188.7 J/K

S^o (CO₂) = 213.7 J/K

Before we proceed, it is important that we take into account the physical state of the substance for which we want to obtain is value of S^o In the case of water, there are two values in the table. We must choose the value of H₂O (g), which is the physical state of the water in this reaction. If we choose the value for H₂O (l) we will be calculating the wrong change in S^o, since that is not the state of the water in the chemical reaction.

We know that [Delta]S^o = [Sigma] S^o (products) - [Sigma] S^o (reactants); therefore:

[Delta]S^o = [(S^o Na₂CO₃) + (S^o H₂O) + S^o (CO₂)] - 2(S^o NaHCO₃), substituting we obtain:

[Delta]S^o = ([139+ 188.7 + 213.7] - 2[102]) J/K. When we do the math, we obtain a value of [Delta]S^o of

+337 J/K, which agrees with our prediction.

Of course, if the reaction is reversed, that is, we write it like:

Na₂CO₃ (s) + H₂O (g) + CO₂ (g) ---> 2 NaHCO₃ (s)

We can see that know three molecules are combinining to form two molecules and that of these molecules are solids, while we started with two gases and a solid. Obviously, the value of [Delta]S^o has to be negative. Since we calculated the value of [Delta]S^o for the opposite reaction, we know that when we reverse a chemical reaction, we must also reverse the sign of [Delta]S^o, which in this case will be -337 J/K.
You should practice problems like this in your book, in order to understand not only how to qualitatively assign the sign, but also to calculate the value of [Delta]S^o quantitatively.