20.2: Standard Entropy
How much energy is available?
As scientists explore energy supplies, geothermal sources look very appealing. The natural geysers that exist in some parts of the world could possibly be harnessed to provide power for many purposes. The change in energy content and the release of energy caused by steam condensing to liquid can help fill some of our growing energy needs.
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 \begin{align*}S\end{align*} and the standard entropy of a substance is given by the symbol \begin{align*}S^{\circ}\end{align*}, 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 in the Table below:
Substance | \begin{align*}S^\circ(\text{J/K} \cdot \text{mol})\end{align*} |
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 knowledge of the absolute entropies of substances allows us to calculate the entropy change \begin{align*}(\Delta S^\circ)\end{align*} for a reaction. For example, the entropy change for the vaporization of water can be found as follows:
\begin{align*}\Delta S^\circ &=S^\circ(\text{H}_2\text{O}(g)) - S^\circ(\text{H}_2\text{O}(l)) \\ &=188.7 \ \text{J/K} \cdot \text{mol} - 69.9 \ \text{J/K} \cdot \text{mol}=118.8 \ \text{J/K} \cdot \text{mol}\end{align*}
The entropy change for the vaporization of water is positive because the gas state has higher entropy than the liquid state.
In general, the entropy change for a reaction can be determined if the standard entropies of each substance are known. The equation below can be applied.
\begin{align*}\Delta S^\circ=\sum nS^\circ(\text{products}) - \sum nS^\circ (\text{reactants})\end{align*}
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 “\begin{align*}n\end{align*}” signifies that each entropy must first be multiplied by its coefficient in the balanced equation. The entropy change for the formation of liquid water from gaseous hydrogen and oxygen can be calculated using this equation:
\begin{align*}& 2\text{H}_2(g)+\text{O}_2(g) \rightarrow 2\text{H}_2\text{O}(l) \\ & \Delta S^\circ=2(69.9) - [2(131.0)+1(205.0)]=-327 \ \text{J/K} \cdot \text{mol}\end{align*}
The entropy change for this reaction is highly negative because three gaseous molecules are being converted into two liquid molecules. According to the drive towards higher entropy, the formation of water from hydrogen and oxygen is an unfavorable reaction. In this case, the reaction is highly exothermic and the drive towards a decrease in energy allows the reaction to occur.
Summary
- Calculations of change in entropy using standard entropy are described.
Practice
Read the material at the link below and do the problems:
http://www.science.uwaterloo.ca/~cchieh/cact/applychem/entropy.html
Review
Questions
- When is the entropy of any material at its lowest?
- In the reaction involving the formation of water from hydrogen and oxygen, why is the entropy value negative?
- Why would diamond have a lower standard entropy value than graphite?
Image Attributions
- Perform change in entropy calculations involving standard entropy.
Concept Nodes:
- standard entropy: The entropy of one mole of substance under standard conditions.
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