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22.4: Entropy

Difficulty Level: At Grade Created by: CK-12

Lesson Objectives

The student will:

• define entropy.
• calculate entropy.
• relate entropy to the tendency toward spontaneity.
• describe the factors that affect the increase or decrease in disorder.

• entropy

Introduction

In this section of the chapter, we explore the disorder of a system. Look at the diagrams of the two chessboards below.

At the start of a chess game, the chess pieces are all in place and orderly. After the game begins and you and your opponent have been playing for a while, there is more disorder in the positions of the chess pieces.

The same is true for reactions. Some reactions start out with more order than they end up with on the product side. Other reactions begin with a higher amount of disorder, but the products that form have a high amount of order. The study of the disorder of reactions is known as entropy and is the focus of this lesson.

The second law of thermodynamics states that the total entropy of the universe is continually increasing.

The Measure of the Disorder of a System

Entropy (S) is a measure of the disorder or randomness of a system. If there is more disorder in the system, there is more entropy. What does it mean to increase disorder? Consider the figure below.

The molecules in the gas have little or no attraction between the molecules, which means that there are many more possibilities of where each molecule can be found in space. As a result, the gas molecules have greater disorganization. Liquids have some disorganization. Although there is some attraction between the molecules, they can move somewhat more freely. In solids where there is great attraction between the molecules, there is much less possibility for the molecules to be unorganized. Therefore, \begin{align*}S_{\mathrm{gas}} > S_{\text{liquid}} > S_{\text{solid}}\end{align*} because there is more disorder or randomness in a gas than there is in a liquid than in a solid.

The most commonly used example to illustrate disorder is to use a deck of playing cards. If you buy a brand new deck of playing cards, all of the cards are in four suits lined up from Ace to King. This means the deck of cards has order. If, however, you were to take your new deck of cards and toss them up into the air, you now have disorder. If you pick up the mess and put the deck of cards back into order, this means you will once again put the cards into four suits lined up from Ace to King.

For a system, the change in entropy, \begin{align*}\triangle S\end{align*}, is measured by finding the difference between the entropy of the products and of the reactants, similar to how the change in enthalpy is calculated. Standard entropies are determined in the lab and published in standard entropy tables in the same manner as standard enthalpies. In fact, both standard enthalpy and standard entropy are often listed in the same table (Table below).

Standard Enthalpy of Formation and Standard Entropies for Some Selected Compounds
Name of Compound Formation Reaction Standard Enthalpy of Formation, \begin{align*}\triangle H_f^\circ\end{align*} (kJ/mol of product) Standard Entropy, \begin{align*}S^\circ, (\mathrm{J/mol}\cdot^\circ\mathrm{C})\end{align*}
aluminum oxide \begin{align*}2 \ \mathrm{Al}_{(s)} + \frac {3} {2} \mathrm{O}_{2(g)} \rightarrow \mathrm{Al}_2\mathrm{O}_{3(s)}\end{align*} \begin{align*}-1669.8\end{align*} \begin{align*}+51\end{align*}
ammonia \begin{align*}\frac {1} {2} \mathrm{N}_{2(g)} + \frac {3} {2} \mathrm{H}_{2(g)} \rightarrow \mathrm{NH}_{3(g)}\end{align*} \begin{align*}-46.1\end{align*} \begin{align*}+193\end{align*}
carbon dioxide \begin{align*}\mathrm{C}_{(s)} + \mathrm{O}_{2(g)} \rightarrow \mathrm{CO}_{2(g)}\end{align*} \begin{align*}-393.5\end{align*} \begin{align*}+214\end{align*}
carbon monoxide \begin{align*}\mathrm{C}_{(s)} + \frac {1} {2} \mathrm{O}_{2(g)} \rightarrow \mathrm{CO}_{(g)}\end{align*} \begin{align*}-110.5\end{align*} \begin{align*}+198\end{align*}
copper(I) oxide \begin{align*}\mathrm{Cu}_{(s)} + \frac {1} {2} \mathrm{O}_{2(g)} \rightarrow \mathrm{CuO}_{(s)}\end{align*} \begin{align*}-156\end{align*} \begin{align*}+43\end{align*}
iron(III) oxide \begin{align*}\mathrm{Fe}_{(s)} + \frac {3} {2} \mathrm{O}_{2(g)} \rightarrow \mathrm{Fe}_2\mathrm{O}_{3(s)}\end{align*} \begin{align*}-822.2\end{align*} \begin{align*}+90.\end{align*}
magnesium oxide \begin{align*}\mathrm{Mg}_{(s)} + \frac {1} {2} \mathrm{O}_{2(g)} \rightarrow \mathrm{MgO}_{(s)}\end{align*} \begin{align*}-602\end{align*} \begin{align*}+27\end{align*}
methane \begin{align*}\mathrm{C}_{(s)} + 2 \ \mathrm{H}_{2(g)} \rightarrow \mathrm{CH}_{4(g)}\end{align*} \begin{align*}-74.8\end{align*} \begin{align*}+188\end{align*}
nitrogen monoxoide \begin{align*}\frac {1} {2} \mathrm{N}_{2(g)} + \frac {1} {2} \mathrm{O}_{2(g)} \rightarrow \mathrm{NO}_{(g)}\end{align*} \begin{align*}+90.\end{align*} \begin{align*}+211\end{align*}
nitrogen dioxide \begin{align*}\frac {1} {2} N_{2(s) }+ O_{2(g)} \rightarrow NO_{2(g)}\end{align*} \begin{align*}+34\end{align*} \begin{align*}+240\end{align*}
sodium chloride \begin{align*}\mathrm{Na}_{(s)} + \frac {1} {2} \mathrm{Cl}_{2(g)} \rightarrow \mathrm{NaCl}_{(s)}\end{align*} \begin{align*}-411\end{align*} \begin{align*}+72\end{align*}
sulfur dioxide \begin{align*}\mathrm{S}_{(s)} + \mathrm{O}_{2(g)} \rightarrow \mathrm{SO}_{2(g)}\end{align*} \begin{align*}-297\end{align*} \begin{align*}+248\end{align*}
sulfur trioxide \begin{align*}\mathrm{S}_{(s)} + \frac {3} {2} \mathrm{O}_{2(g)} \rightarrow \mathrm{SO}_{3(g)}\end{align*} \begin{align*}-393.2\end{align*} \begin{align*}+257\end{align*}
water (gaseous) \begin{align*}\mathrm{H}_{2(g)} + \frac {1} {2} \mathrm{O}_{2(g)} \rightarrow \mathrm{H}_2\mathrm{O}_{(g)}\end{align*} \begin{align*}-241.8\end{align*} \begin{align*}+189\end{align*}
water (liquid) \begin{align*}\mathrm{H}_{2(g)} + \frac {1} {2} \mathrm{O}_{2(g)} \rightarrow \mathrm{H}_2\mathrm{O}_{(l)}\end{align*} \begin{align*}-285.8\end{align*} \begin{align*}+70\end{align*}.

It is important to note in the above table that the units for enthalpy of formation and entropy are not the same. Enthalpies of formation in the table have energy units in kilojoules while entropy has energy units in joules. If these values are entered into a formula together, one of them must be converted so that all energy units are the same.

Therefore, the change in entropy for a reaction \begin{align*}(\triangle S_{\mathrm{rxn}})\end{align*} can be found using the following equation:

\begin{align*}\triangle S_{\text{rxn}} = S_{\text{products}} - S_{\text{reactants}}\end{align*}

Example:

Given the following data, calculate the \begin{align*}\triangle S_{\mathrm{rxn}}\end{align*} for the following reaction:

\begin{align*}\mathrm{NH}_{3(g)} + \mathrm{HCl}_{(g)} \rightarrow \mathrm{NH}_4\mathrm{Cl}_{(s)} \ \ \ \ \ \triangle S_{\text{rxn}} = ?\end{align*}

Given that \begin{align*}S(\mathrm{NH}_{3(g)}) = 111.3 \ \mathrm{J/K}\cdot\mathrm{mol}\end{align*}, \begin{align*}S(\mathrm{HCl}_{(g)}) = 267.3 \ \mathrm{J/K} \cdot\mathrm{mol}\end{align*}, \begin{align*}S(\mathrm{NH}_4\mathrm{Cl}_{(s)}) = 94.56 \ \mathrm{J/K} \cdot\mathrm{mol}\end{align*}.

Solution:

\begin{align*}\triangle S_{\text{rxn}} = S_{\text{products}} - S_{\text{reactants}}\end{align*}
\begin{align*}\triangle S_{\text{rxn}} = S(NH_4Cl_{(s)}) - [S(NH_{3(g)}) + S(HCl_{(g)})]\end{align*}
\begin{align*}\triangle S_{\text{rxn}} = (94.56\ \text{J/K} \cdot \;\mathrm{mol}) - (193\ \text{J/K} \cdot \text{mol} + 187\ \text{J/K} \cdot \text{mol})\end{align*}
\begin{align*}\triangle S_{\text{rxn}} = (94.56\ \text{J/K} \cdot \text{mol}) - (380\ \text{J/K} \cdot \text{mol})\end{align*}
\begin{align*}\triangle S_{\text{rxn}} = -285\ \text{J/K} \cdot \text{\text{mol}}\end{align*}

Therefore,

\begin{align*}\mathrm{NH}_{3(g)} + \mathrm{HCl}_{(g)} \rightarrow \mathrm{NH}_4\mathrm{Cl}_{(s)} \ \ \ \ \ \triangle S_{\text{rxn}} = -285 \ \text{J/K} \cdot\text{mol}\end{align*}.

What does it means when \begin{align*}\triangle S_{\mathrm{rxn}}\end{align*} is negative? In this system, two gases are coming together to form a solid, and the value of \begin{align*}\triangle S_{\mathrm{rxn}}\end{align*} is negative. We can make a conclusion that if the order of the system increases, then the change in entropy value \begin{align*}(\triangle S)\end{align*} will be negative. In other words, if a system goes from a state of high disorder (two moles of gas) to a state of low disorder (one mole of solid), the entropy change is negative. Let’s try another example.

Example:

For the reaction \begin{align*}\mathrm{CaCO}_{3(s)} \rightarrow \mathrm{CaO}_{(s)} + \mathrm{CO}_{2(g)}\end{align*} at \begin{align*}25^\circ\mathrm{C}\end{align*}, calculate the value of \begin{align*}\triangle S_{\mathrm{rxn}}\end{align*}.

Given: \begin{align*}S(\mathrm{CaCO}_{3(s)}) = 92.9 \ \mathrm{J/K}\cdot\mathrm{mol}\end{align*}; \begin{align*}S(\mathrm{CaO}_{(s)}) = 39.8 \ \mathrm{J/K} \cdot\mathrm{mol}\end{align*}; and \begin{align*}S(\mathrm{CO}_{2(g)}) = 213.6 \ \mathrm{J/K} \cdot\mathrm{mol}\end{align*}.

Solution:

\begin{align*}\triangle S_{\text{rxn}} = S_{\text{products}} - S_{\text{reactants}}\end{align*}
\begin{align*}\triangle S_{\text{rxn}} = [S(\mathrm{CaO}_{(s)}) + S(\mathrm{CO}_{2(g)})] - S(\mathrm{CaCO}_{3(s)})\end{align*}
\begin{align*}\triangle S_{\text{rxn}} = [39.8\ \text{J/K} \cdot\text{mol} + 213.6 \ \text{J/K} \cdot\text{mol}] - (92.9 \ \text{J/K} \cdot\text{mol})\end{align*}
\begin{align*}\triangle S_{\text{rxn}} = (253.4\ \text{J/K} \cdot\text{mol}) - (92.9\ \text{J/K} \cdot\text{mol})\end{align*}
\begin{align*}\triangle S_{\text{rxn}} = 160.5\ \text{J/K} \cdot\text{mol}\end{align*}

Therefore:

\begin{align*}\mathrm{CaCO}_{3(s)} \rightarrow \mathrm{CaO}_{(s)} + \mathrm{CO}_{2(g)} \ \ \ \ \ \triangle S_{rxn} = 160.5 \ \text{J/K} \cdot\text{mol}\end{align*}.

In this problem, the system becomes more disordered since a solid and a gas are the products. The value of the entropy change for the reaction is positive. We can conclude that if the disorder of the reaction increases, then the change in entropy value \begin{align*}(\triangle S)\end{align*} will be positive. In other words, if a system goes from a state of low disorder to a state of high disorder, the entropy change is positive.

Entropy and Spontaneous Reactions

A reaction will tend to be spontaneous if the reaction moves from a state of low disorder to a state of high disorder. Look at the example below.

\begin{align*}\mathrm{C}_{(s)} + \mathrm{H}_2\mathrm{O}_{(g)} + \text{energy} \rightarrow \mathrm{H}_{2(g)} + \mathrm{CO}_{(g)}\end{align*}

In the equation above, the enthalpy is positive \begin{align*}(\triangle H > 0)\end{align*}. We also know that in the reaction there are two moles of reactants (one of which is a solid) and two moles of products (both of which are gases). Gases have the highest entropies because they have higher disorder than solids or liquids. Therefore this system is one in which moves from a state of low disorder to a state of high disorder. It will tend to be spontaneous. Chemical reactions are driven by the combination of a tendency toward minimum enthalpy (lowest potential energy) and a tendency toward maximum entropy (greatest disorder).

Generalizations for Determining Entropy

Less Organized Phases Contain More Entropy

There are a number of factors that affect the disorder of a system. For instance, when a liquid is formed from a solid or a gas is formed from a liquid, the disorder increases.

\begin{align*}\mathrm{Na}_{(s)} \rightarrow \mathrm{Na}_{(l)} \ \ \ \ \ \triangle S = 51.4\ \text{J/K} \times \ \text{mol}\end{align*}
\begin{align*}\mathrm{C}_2\mathrm{H}_5\mathrm{OH}_{(l)} \rightarrow \mathrm{C}_2\mathrm{H}_5\mathrm{OH}_{(g)} \ \ \ \ \ \triangle S = 122.0\ \text{J/K} \cdot \text{mol}\end{align*}

We can also predict that disorder increases when a solid or liquid dissolves in water. The equation below shows solid ammonium chloride dissolving in water. Notice the value of \begin{align*}\triangle S\end{align*} is positive, indicating the increase in disorder. Entropy increases in this system since the particles are no longer held in place by their electrostatic attractions and are able to move more freely in solution.

\begin{align*}\mathrm{NH}_4\mathrm{Cl}_{(s)} \rightarrow \mathrm{NH}^{+}_{4(aq)} + \mathrm{Cl}^-_{(aq)} \ \ \ \ \ \triangle S = 74.7\ \text{J/K} \cdot\text{mol}\end{align*}

The system therefore increases in disorder, and the entropy is positive. A gas, however, decreases in disorder when it is dissolved in water. The gas molecules have less possible positions. As a result, the disorder decreases and the sign of \begin{align*}\triangle S\end{align*} would be negative.

Generally, large molecules have a larger entropy than smaller molecules of simpler structures. For example, when we consider methane (\begin{align*}\mathrm{CH}_4\end{align*}), ethane (\begin{align*}\mathrm{C}_2\mathrm{H}_6\end{align*}), and propane (\begin{align*}\mathrm{C}_3\mathrm{H}_8\end{align*}), we can see that the relationships between their entropies will be that \begin{align*}S_{\mathrm{propane}} > S_{\mathrm{ethane}} > S_{\mathrm{methane}}\end{align*}. Propane, with more atoms, has more possibilities of rotating and twisting around than ethane and therefore greater entropy.

Increase in Number of Particles Increases Entropy

The equation below represents water being produced from its elements.

\begin{align*}2\ \mathrm{H}_{2(g)} + \mathrm{O}_{2(g)} \rightarrow 2 \ \mathrm{H}_2\mathrm{O}_{(l)} \ \ \ \ \ \triangle S = -326.0\ \text{J/K} \times \text{mol}\end{align*}

Notice that there are three moles of reactants combining to produce two moles of products. With three moles of gas reactants, the higher entropy is on the reactant side of the chemical equation. The negative value of entropy means the disorder for the system is decreasing. Since the number of particles is decreasing, the particles are going from a gas state to a liquid state, and the disorder is decreasing, \begin{align*}\triangle\end{align*} \begin{align*}S \le 0\end{align*}.

For the reaction below, the reverse is happening.

\begin{align*}2 \ \mathrm{H}_2\mathrm{O}_{2(l)} \rightarrow 2 \ \mathrm{H}_2\mathrm{O}_{2(l)} + \mathrm{O}_{2(g)} \ \ \ \ \ \triangle S = 125.6\ \text{J/K} \times \text{mol}\end{align*}

In the equation above, there are two moles of reactant forming three moles of products. Looking at this equation, we see that these two moles of reactant are in the liquid state and that they form products in both the liquid and gas state. An increase in the number of particles and the formation of the gas product means an increase in disorder. Therefore, \begin{align*}\triangle S > 0\end{align*}. The driving force for this reaction is the oxygen gas being produced.

Example:

Predict whether the entropy will be positive or negative for each of the following:

1. \begin{align*}\mathrm{Br}_{2(g)} \rightarrow \mathrm{Br}_{2(l)}\end{align*}
2. \begin{align*}\mathrm{Ca(OH)}_{2(s)}\rightarrow \mathrm{Ca}^{2+}{_{(aq)}} + 2 \ \mathrm{OH}^-{_{(aq)}}\end{align*}
3. \begin{align*}\mathrm{CO}_{(g)} + 3 \ \mathrm{H}_{2(g)} \rightarrow \mathrm{CH}_{4(g)} + \mathrm{H}_2\mathrm{O}{_{(g)}}\end{align*}

Solution:

1. Going from a gas to a liquid decreases the disorder of the system, therefore \begin{align*}\triangle S < 0\end{align*}.
2. Disorder increases when a solid or liquid dissolves in water, therefore \begin{align*}\triangle S > 0\end{align*}.
3. There are four moles of reactants forming two moles of products. A decrease in the number of particles means a decrease in disorder and therefore \begin{align*}\triangle S < 0\end{align*}.

Higher Temperature Favors More Entropy

When the temperature of a substance is increased, the molecules have greater average kinetic energy and therefore move around with greater velocity. The greater velocity of the particles means they collide more often and with greater force. The greater force of the collisions cause the molecules to spread further apart. Consequently, increasing the temperature of a system favors the tendency toward greater randomness or maximum entropy.

Lesson Summary

• Entropy (S) is a measure of the disorder of a system. If the order of the reaction increases, then the change in entropy value \begin{align*}(\triangle S)\end{align*} will be negative. If the order of the reaction decreases, then the change in entropy value \begin{align*}(\triangle S)\end{align*} will be positive.
• A system that goes from a state of low disorder to a state of high disorder will tend to be spontaneous. The disorder of a system increases if a liquid is formed from a solid, a gas is formed from a liquid, and when a solid or liquid dissolves in water.
• A gas decreases in disorder when it dissolved in water.

Review Questions

1. Define entropy.
2. Give an everyday example of entropy.
3. Which of the following examples will result in an increase in entropy?
1. \begin{align*}\mathrm{H}_2\mathrm{O}_{(l)} \rightarrow \mathrm{H}_2\mathrm{O}_{(s)}\end{align*}
2. \begin{align*}(\mathrm{NH}_4)_{2}\mathrm{SO}_{4(s)} \rightarrow 2 \ \mathrm{NH}^{+}_{4(aq)} + \mathrm{SO}^{2-}_{4(aq)}\end{align*}
3. \begin{align*}\mathrm{H}_2\mathrm{O}_{(g)} \rightarrow \mathrm{H}_2\mathrm{O}_{(l)}\end{align*}
4. \begin{align*}\mathrm{Ag}^+_{(aq)} + \mathrm{Cl}^-_{(aq)} \rightarrow \mathrm{AgCl}_{(s)}\end{align*}
4. Which of the following would have the greatest entropy?
1. \begin{align*}\mathrm{CO}_{2(s)}\end{align*}
2. \begin{align*}\mathrm{H}_2\mathrm{O}_{(g)}\end{align*}
3. \begin{align*}\mathrm{CCl}_{4(l)}\end{align*}
4. \begin{align*}\mathrm{CHCl}_{3(l)}\end{align*}
5. From the following equations, select those that tend to be spontaneous. \begin{align*}\\ &\text{i.} \ \mathrm{N}_{2(g)} + \mathrm{O}_{2(g)} \rightarrow \mathrm{N}_2\mathrm{O}_{5(g)} + \mathrm{heat}\\ &\text{ii.} \ \mathrm{H}_2\mathrm{O}_{(s)} + \mathrm{heat} \rightarrow \mathrm{H}_2\mathrm{O}_{(l)}\\ &\text{iii.} \ \mathrm{N}_{2(g)} + \mathrm{O}_{2(g)} \rightarrow 2 \ \mathrm{NO}_{(g)} \ \ \ \triangle H = 180.6 \ \mathrm{kJ/mol}\\ &\text{iv.} \ \mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_{6(s)} + 6 \ \mathrm{O}_{2(g)} \rightarrow 6 \ \mathrm{CO}_{2(g)} + 6 \ \mathrm{H}_2\mathrm{O}_{(l)} \ \ \ \triangle H = -2802 \ \mathrm{kJ/mol}\\ &\text{v.} \ \mathrm{CaCl}_{2(s)} \rightarrow \mathrm{Ca}^{2+}_{(aq)} + 2 \ \mathrm{Cl}^-_{(aq)} + \mathrm{heat}\end{align*}
1. i, ii, and v
2. i and iii
3. iii and iv
4. i, iv, and v
6. Calculate the entropy of the following reactions. Use the data from Table below.
1. \begin{align*}\mathrm{CH}_3\mathrm{CH}_2\mathrm{OH}_{(l)} + 3 \ \mathrm{O}_{2(g)} \rightarrow 2 \ \mathrm{CO}_{2(g)} + 3 \ \mathrm{H}_2\mathrm{O}_{(l)}\end{align*}
2. \begin{align*}2 \ \mathrm{AsF}_{3(l)} \rightarrow 2 \ \mathrm{As}_{(s)} + 3 \ \mathrm{F}_{2(g)}\end{align*}
Compound \begin{align*}\triangle S\end{align*} Compound \begin{align*}\triangle S\end{align*}
\begin{align*}\mathrm{CH}_3\mathrm{CH}_2\mathrm{OH}_{(l)}\end{align*} \begin{align*}213.\ \mathrm{J/K} \times \mathrm{mol}\end{align*} \begin{align*}\mathrm{As}_{(s)}\end{align*} \begin{align*}35.1 \ \mathrm{J/K} \times \mathrm{mol}\end{align*}
\begin{align*}\mathrm{O}_{2(g)}\end{align*} \begin{align*}69.9 \ \mathrm{J/K} \times \mathrm{mol}\end{align*} \begin{align*}\mathrm{F}_{2(g)}\end{align*} \begin{align*}202.7 \ \mathrm{J/K} \times \mathrm{mol}\end{align*}
\begin{align*}\mathrm{CO}_{2(g)}\end{align*} \begin{align*}160.7 \ \mathrm{J/K} \times \mathrm{mol}\end{align*} \begin{align*}\mathrm{AsF}_{3(l)}\end{align*} \begin{align*}181.2 \ \mathrm{J/K} \times \mathrm{mol}\end{align*}
\begin{align*}\mathrm{H}_2\mathrm{O}_{(l)}\end{align*} \begin{align*}205.0 \ \mathrm{J/K} \times \mathrm{mol}\end{align*}
1. Predict whether the entropy will be positive or negative for each of the following:
1. \begin{align*}\mathrm{CO}_{2(s)} \rightarrow \mathrm{CO}_{2(g)}\end{align*}
2. \begin{align*}\mathrm{C}_{12}\mathrm{H}_{22}\mathrm{O}_{11(s)} \rightarrow \mathrm{C}_{12}\mathrm{H}_{22}\mathrm{O}_{11(aq)}\end{align*}
3. \begin{align*}2 \ \mathrm{NO}_{(g)} + \mathrm{O}_{2(g)} \rightarrow 2 \ \mathrm{NO}_{2(g)}\end{align*}

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