### Entropy changes and their calculation

Even without knowing the actual values for the So of substances, it is possible to predict the sign of Delta So 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]So will be positive. To confirm this, we can calculate the actual value of [Delta]So. Looking in table 18.1 (page 756) we find the following data:

So (NaHCO3) = 102 J/K

So (Na2CO3) = 139 J/K

So (H2O) = 188.7 J/K

So (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 So 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 So, since that is not the state of the water in the chemical reaction.

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

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

[Delta]So = ([139+ 188.7 + 213.7] - 2[102]) J/K. When we do the math, we obtain a value of [Delta]So 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]So has to be negative. Since we calculated the value of [Delta]So for the opposite reaction, we know that when we reverse a chemical reaction, we must also reverse the sign of [Delta]So, 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]So quantitatively.

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