# Entropy

• Define entropy.
• List factors that influence the entropy of a system.
• List situations that illustrate the concept of entropy.

What will it look like?

When the pieces of a jigsaw puzzle are dumped from the box, the pieces naturally hit the table in a very random pattern. In order to put the puzzle together, a great deal of work must be done to overcome the natural disorder of the pieces. The pieces need to be turned right-side up, then sorted by color or edge (some people like to put the border together first). Then comes the challenge of finding the exact spot of each piece of the puzzle in order to get the final organized picture.

### Entropy

There is a tendency in nature for systems to proceed toward a state of greater disorder or randomness. Entropy is a measure of the degree of randomness or disorder of a system. Entropy is an easy concept to understand when thinking about everyday situations. The entropy of a room that has been recently cleaned and organized is low. As time goes by, it likely will become more disordered and thus its entropy will increase (see Figure below ). The natural tendency of a system is for its entropy to increase.

Figure 20.1

The messy room on the right has more entropy than the highly ordered room on the left.

Chemical reactions also tend to proceed in such a way as to increase the total entropy of the system. How can you tell if a certain reaction shows an increase or a decrease in entropy? The molecular state of the reactants and products provide certain clues. The general cases below illustrate entropy at the molecular level.

1. For a given substance, the entropy of the liquid state is greater than the entropy of the solid state. Likewise, the entropy of the gas is greater than the entropy of the liquid. Therefore, entropy increases in processes in which solid or liquid reactants form gaseous products. Entropy also increases when solid reactants form liquid products.
2. Entropy increases when a substance is broken up into multiple parts. The process of dissolving increases entropy because the solute particles become separated from one another when a solution is formed.
3. Entropy increases as temperature increases. An increase in temperature means that the particles of the substance have greater kinetic energy. The faster moving particles have more disorder than particles that are moving more slowly at a lower temperature.
4. Entropy generally increases in reactions in which the total number of product molecules is greater than the total number of reactant molecules. An exception to this rule is when a gas is being produced from nongaseous reactants.

The examples below will serve to illustrate how the entropy change in a reaction can be predicted.

$text{Cl}_2(g) rightarrow text{Cl}_2(l)$

The entropy is decreasing because a gas is becoming a liquid.

$text{CaCO}_3(s) rightarrow text{CaO}(s)+text{CO}_ 2(g)$

The entropy is increasing because a gas is being produced and the number of molecules is increasing.

$text{N}_2(g)+3text{H}_2(g) rightarrow 2text{NH}_3(g)$

The entropy is decreasing because four total reactant molecules are forming two total product molecules. All are gases.

$text{AgNO}_3(aq)+text{NaCl}(aq) rightarrow text{NaNO}_3(aq)+text{AgCl}(s)$

The entropy is decreasing because a solid is formed from aqueous reactants.

$text{H}_2 (g)+text{Cl}_2(g) rightarrow 2text{HCl}(g)$

The entropy change is unknown (but likely not zero), because there are equal numbers of molecules on both sides of the equation and all are gases.

#### Summary

• Entropy is defined.
• Situations involving entropy changes are described.

#### Review

Questions

Indicate whether entropy increases or decreases in each of the following situations:

1. The jigsaw puzzle is taken apart and put back in the box.
2. Clothes are folded and put away.
3. ice  $rightarrow$ steam.
4. $text{Pb(NO}_3)_2(aq)+2text{KI}(aq) rightarrow text{PbI}_2(s)+2 text{KNO}_3$ .
• entropy: A measure of the degree of randomness or disorder of a system.

# Standard Entropy

• Perform change in entropy calculations involving 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  $S$ and the standard entropy of a substance is given by the symbol $S^{circ}$ , 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 $S^circ(text{J/K} cdot text{mol})$ 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  $(Delta S^circ)$ for a reaction. For example, the entropy change for the vaporization of water can be found as follows:

$Delta S^circ &=S^circ(text{H}_2text{O}(g)) - S^circ(text{H}_2text{O}(l)) \&=188.7 text{J/K} cdot text{mol} - 69.9 text{J/K} cdot text{mol}=118.8 text{J/K} cdot text{mol}$

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.

$Delta S^circ=sum nS^circ(text{products}) - sum nS^circ (text{reactants})$

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 “ $n$ ” 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:

$& 2text{H}_2(g)+text{O}_2(g) rightarrow 2text{H}_2text{O}(l) \& Delta S^circ=2(69.9) - [2(131.0)+1(205.0)]=-327 text{J/K} cdot text{mol}$

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

http://www.science.uwaterloo.ca/~cchieh/cact/applychem/entropy.html

#### Review

Questions

1. When is the entropy of any material at its lowest?
2. In the reaction involving the formation of water from hydrogen and oxygen, why is the entropy value negative?
3. Why would diamond have a lower standard entropy value than graphite?
• standard entropy: The entropy of one mole of substance under standard conditions.

# Spontaneous and Nonspontaneous Reactions

• Define spontaneous reaction.
• Define nonspontaneous reaction.
• Give examples of spontaneous and nonspontaneous reactions.

Watch that nitro!

Nitroglycerin is tricky stuff. An active ingredient in dynamite (where it is stabilized), “raw” nitroglycerin is very unstable. Physical shock will cause the material to explode. The reaction is shown below:

$4text{C}_3text{H}_5(ONO_2)_3rightarrow 12text{CO}_2+10text{H}_2text{O}+6text{N}_2+text{O}_2$

The explosion of nitroglycerin releases large volumes of gases and is very exothermic.

### Spontaneous Reactions

Reactions are favorable when they result in a decrease in enthalpy and an increase in entropy of the system. When both of these conditions are met, the reaction occurs naturally. A spontaneous reaction is a reaction that favors the formation of products at the conditions under which the reaction is occurring. A roaring bonfire is an example of a spontaneous reaction, since it is exothermic (there is a decrease in the energy of the system as energy is released to the surroundings as heat). The products of a fire are composed partly of gases such as carbon dioxide and water vapor. The entropy of the system increases during a combustion reaction. The combination of energy decrease and entropy increase dictates that combustion reactions are spontaneous reactions.

Figure 20.2

Bonfire.

A nonspontaneous reaction is a reaction that does not favor the formation of products at the given set of conditions. In order for a reaction to be nonspontaneous, it must be endothermic, accompanied by a decrease in entropy, or both. Our atmosphere is composed primarily of a mixture of nitrogen and oxygen gases. One could write an equation showing these gases undergoing a chemical reaction to form nitrogen monoxide.

$text{N}_2(g)+text{O}_2(g)rightarrow 2text{NO}(g)$

Fortunately, this reaction is nonspontaneous at normal temperatures and pressures. It is a highly endothermic reaction with a slightly positive entropy change $(Delta text{S})$ . Nitrogen monoxide is capable of being produced at very high temperatures and has been observed to form as a result of lightning strikes.

One must be careful not to confuse the term spontaneous with the notion that a reaction occurs rapidly. A spontaneous reaction is one in which product formation is favored, even if the reaction is extremely slow. A piece of paper will not suddenly burst into flames, although its combustion is a spontaneous reaction. What is missing is the required activation energy to get the reaction started. If the paper were to be heated to a high enough temperature, it would begin to burn, at which point the reaction would proceed spontaneously until completion.

In a reversible reaction, one reaction direction may be favored over the other. Carbonic acid is present in carbonated beverages. It decomposes spontaneously to carbon dioxide and water according to the following reaction.

$text{H}_2text{CO}_3(aq)rightleftarrows text{CO}_2(g)+text{H}_2text{O}(l)$

If you were to start with pure carbonic acid in water and allow the system to come to equilibrium, more than 99% of the carbonic acid would be converted into carbon dioxide and water. The forward reaction is spontaneous because the products of the forward reaction are favored at equilibrium. In the reverse reaction, carbon dioxide and water are the reactants and carbonic acid is the product. When carbon dioxide is bubbled into water, less than 1% is converted to carbonic acid when the reaction reaches equilibrium. The reverse reaction, as written above, is not spontaneous.

#### Summary

• Spontaneous and nonspontaneous reactions are defined.
• Examples of both types of reactions are given.

#### Practice

Questions

1. Why is system I a spontaneous reaction?
2. Why is system II not spontaneous?
3. Why is system III spontaneous?

#### Review

Questions

1. Why is a combustion reaction spontaneous?
2. Is NO formation spontaneous at room temperature?
3. How do we know that the equilibrium between carbonic acid and CO 2 goes strongly to the right?
• nonspontaneous reaction: A reaction that does not favor the formation of products at the given set of conditions.
• spontaneous reaction:  A reaction that favors the formation of products at the conditions under which the reaction is occurring.

# Free Energy

• Define free energy.
• Describe how changes in  $Delta H$ and  $Delta S$ affect $Delta G$ .

All aboard!

The steam engine pictured above is slowly going out of style, but is still a picturesque part of the modern railroad. The water in a boiler is heated by a fire (usually fueled by coal) and turned to steam. This steam then pushes the pistons that drive the wheels of the train. It is the pressure created by the steam which allows work to be done in moving the train.

### Free Energy

Many chemical reactions and physical processes release energy that can be used to do other things. When the fuel in a car is burned, some of the released energy is used to power the vehicle. Free energy is energy that is available to do work. Spontaneous reactions release free energy as they proceed. Recall that the determining factors for spontaneity of a reaction are the enthalpy and entropy changes that occur for the system. The free energy change of a reaction is a mathematical combination of the enthalpy change and the entropy change.

$Delta G^circ=Delta H^circ - T Delta S^circ$

The symbol for free energy is $G$ , in honor of American scientist Josiah Gibbs (1839-1903), who made many contributions to thermodynamics. The change in Gibbs free energy is equal to the change in enthalpy minus the mathematical product of the change in entropy multiplied by the Kelvin temperature. Each thermodynamic quantity in the equation is for substances in their standard states. The usual units for $Delta H$ is kJ/mol, while $Delta S$ is often reported in J/K • mol. It is necessary to change the units for $Delta S$ to kJ/K • mol, so that the calculation of $Delta G$ is in kJ/mol.

A spontaneous reaction is one that releases free energy, and so the sign of $Delta G$ must be negative. Since both $Delta H$ and $Delta S$ can be either positive or negative, depending on the characteristics of the particular reaction, there are four different general outcomes for $Delta G$ and these are outlined in the Table below :

 $Delta H$ $Delta S$ $Delta G$ − value (exothermic) + value (disordering) always negative + value (endothermic) + value (disordering) negative at higher temperatures − value (exothermic) − value (ordering) negative at lower temperatures + value (endothermic) − value (ordering) never negative

Keep in mind that the temperature in the Gibbs free energy equation is the Kelvin temperature and so can only be positive. When $Delta H$ is negative and $Delta S$ is positive, the sign of $Delta G$ will always be negative, and the reaction will be spontaneous at all temperatures. This corresponds to both driving forces being in favor of product formation. When $Delta H$ is positive and $Delta S$ is negative, the sign of $Delta G$ will always be positive, and the reaction can never be spontaneous. This corresponds to both driving forces working against product formation.

When one driving force favors the reaction, but the other does not, it is the temperature that determines the sign of $Delta G$ . Consider first an endothermic reaction (positive $Delta H$ ) that also displays an increase in entropy (positive $Delta S$ ). It is the entropy term that favors the reaction. Therefore, as the temperature increases, the $T Delta S$ term in the Gibbs free energy equation will begin to predominate and $Delta G$ will become negative. A common example of a process which falls into this category is the melting of ice. At a relatively low temperature (below 273 K), the melting is not spontaneous because the positive $Delta H$ term “outweighs” the $T Delta S$ term. When the temperature rises above 273 K, the process becomes spontaneous because the larger $T$ value has tipped the sign of $Delta G$ over to being negative.

When the reaction is exothermic (negative $Delta H$ ) but undergoes a decrease in entropy (negative $Delta S$ ), it is the enthalpy term that favors the reaction. In this case, a spontaneous reaction is dependent upon the $T Delta S$ term being small relative to the $Delta H$ term, so that $Delta G$ is negative. The freezing of water is an example of this type of process. It is spontaneous only at a relatively low temperature. Above 273 K, the larger $T Delta S$ value causes the sign of $Delta G$ to be positive, and freezing does not occur.

#### Summary

• Free energy is defined.
• Relationships between enthalpy, entropy, and free energy are described.

#### Practice

Questions

Click on the image above for more content

2. What happens to the total energy when the ball rolls down the slide?
3. How does $H$ change in a spontaneous reaction?
4. How does $S$ change in a spontaneous reaction?

#### Review

Questions

1. What do spontaneous reactions do?
2. What are the units for $Delta H$ ?
3. What are the units for $Delta S$ ?
• free energy: Energy that is available to do work.

# Calculating Free Energy Change (ΔG°)

• Perform free energy calculations using enthalpy, entropy, and temperature values.

Time for dessert!

When you are baking something, you heat the oven to the temperature indicated in the recipe. Then you mix all the ingredients, put them in the proper baking dish, and place them in the oven for a specified amount of time. If you had mixed the ingredients and left them out at room temperature, not much would change. The materials need to be heated to a given temperature for a set time in order for the ingredients to react with one another and produce a delicious final product.

### Calculating Free Energy  $(Delta G^circ)$

The free energy change of a reaction can be calculated using the following expression:

$Delta G^circ=Delta H^circ - TDelta S^circ$

where  $Delta G = text{free energy change (kJ/mol)}$

$Delta H = text{change in enthalpy (kJ/mol)}$
$Delta S = text{change in entropy (J/K} cdot text{mol)}$
$T = text{temperature (Kelvin)}$

Note that all values are for substances in their standard state. In performing calculations, it is necessary to change the units for $Delta S$  to kJ/K • mol, so that the calculation of $Delta G$  is in kJ/mol.

#### Sample Problem: Gibbs Free Energy

Methane gas reacts with water vapor to produce a mixture of carbon monoxide and hydrogen according to the balanced equation below.

$text{CH}_4(g)+text{H}_2text{O}(g) rightarrow text{CO}(g)+3text{H}_2(g)$

The  $Delta H^circ$ for the reaction is +206.1 kJ/mol, while the  $Delta S^circ$ is +215 J/K • mol . Calculate the  $Delta G^circ$ at 25°C and determine if the reaction is spontaneous at that temperature.

Step 1: List the known values and plan the problem.

Known

• $Delta H^circ =206.1 text{kJ/mol}$
• $Delta S^circ = 215 text{J/K}cdot text{mol}=0.215 text{kJ/K}cdot text{mol}$
• $T =25^circ text{C}=298 text{K}$

Unknown

• $Delta G^circ =? text{kJ/mol}$

Prior to substitution into the Gibbs free energy equation, the entropy change is converted to  kJ/K • mol and the temperature to Kelvins.

Step 2: Solve.

$Delta G^circ =Delta H^circ - TDelta S^circ = 206.1 text{kJ/mol} - 298 text{K}(0.215 text{kJ/K}cdot text{mol})=+142.0 text{kJ/mol}$

The resulting positive value of  $Delta G$ indicates that the reaction is not spontaneous at 25°C.

The unfavorable driving force of increasing enthalpy outweighed the favorable increase in entropy. The reaction will be spontaneous only at some elevated temperature.

Available values for enthalpy and entropy changes are generally measured at the standard conditions of 25°C and 1 atm pressure. The values are slightly temperature dependent and so we must use caution when calculating specific  $Delta G$ values at temperatures other than 25°C. However, since the values for  $Delta H$ and  $Delta S$ do not change a great deal, the tabulated values can safely be used when making general predictions about the spontaneity of a reaction at various temperatures.

#### Summary

• Calculations of free energy changes are described.

#### Practice

Questions

Watch the video at the link below and answer the following questions:

Click on the image above for more content

1. Why is  $Delta H$ negative in this example?
2. What would happen if you forgot to change the sign of the  $TDelta S$ value in the first calculation?
3. What indicates that the reaction is spontaneous?

#### Review

Questions

1. What would happen to  $Delta H$ if you forgot to change the units for  $Delta S$ to kJ/K • mol ?
2. What are standard conditions for enthalpy and entropy changes?
3. At what temperature would the reaction become spontaneous?

# Temperature and Free Energy

• Describe the effect of temperature on $Delta G$ .

How is steel produced?

Iron ore (Fe 2 O 3 ) and coke (an impure form of carbon) are heated together to make iron and carbon dioxide. The reaction is non-spontaneous at room temperature, but becomes spontaneous at temperature above 842 K. The iron can then be treated with small amounts of other materials to make a variety of steel products.

### Temperature and Free Energy

Consider the reversible reaction in which calcium carbonate decomposes into calcium oxide and carbon dioxide gas. The production of CaO (called quicklime) has been an important reaction for centuries.

$text{CaCO}_3(s) rightleftarrows text{CaO}(s) +text{CO}_2(g)$

The  $Delta H^circ$ for the reaction is 177.8 kJ/mol, while the  $Delta S^circ$ is 160.5 J/K • mol. The reaction is endothermic with an increase in entropy due to the production of a gas. We can first calculate the  $Delta G^circ$ at 25°C in order to determine if the reaction is spontaneous at room temperature.

$Delta G^circ=Delta H^circ -T Delta S^circ=177.8 text{kJ}/ text{mol} - 298 K (0.1605 text{kJ} / text{K} cdot text{mol})=130.0 text{kJ} / text{mol}$

Since the  $Delta G^circ$ is a large positive quantity, the reaction strongly favors the reactants and very little products would be formed. In order to determine a temperature at which  $Delta G^circ$ will become negative, we can first solve the equation for the temperature when  $Delta G^circ$ is equal to zero.

$0 &=Delta H^circ - T Delta S^circ \T &=frac{Delta H^circ}{Delta S^circ}=frac{177.8 text{kJ} / text{mol}}{0.1605 text{kJ} / text{K} cdot text{mol}}=1108 text{ K}=835^circ text{ C}$

So at any temperature higher than 835°C, the value of  $Delta G^circ$ will be negative and the decomposition reaction will be spontaneous.

Figure 20.3

This lime kiln in Cornwall was used to produce quicklime (calcium oxide), an important ingredient in mortar and cement.

Recall that the assumption that  $Delta H^circ$ and  $Delta S^circ$ are independent of temperature means that the temperature at which the sign of  $Delta G^circ$ switches from being positive to negative (835°C) is an approximation. It is also important to point out that one should not assume that absolutely no products are formed below 835°C and that at that temperature decomposition suddenly begins. Rather, at lower temperatures, the amount of products formed is simply not great enough to say that the products are favored. When this reaction is performed, the amount of products can be detected by monitoring the pressure of the CO 2 gas that is produced. Above about 700°C, measurable amounts of CO 2 are produced. The pressure of CO 2 at equilibrium gradually increases with increasing temperature. Above 835°C, the pressure of CO 2 at equilibrium begins to exceed 1 atm, the standard-state pressure. This is an indication that the products of the reaction are now favored above that temperature. When quicklime is manufactured, the CO 2 is constantly removed from the reaction mixture as it is produced. This causes the reaction to be driven towards the products according to LeChâtelier’s principle.

#### Summary

• The influence of temperature on free energy is described.

#### Practice

Questions

1. Is an exothermic reaction where entropy increases spontaneous or nonspontaneous?
2. Give an example of a process where a process proceeds spontaneously only below a certain temperature.
3. If  $Delta H > 0$ and $Delta S < 0$ , can the reaction ever be spontaneous?

#### Review

Questions

1. If you increased the pressure of CO 2 in the quicklime reaction, what would happen to the equilibrium?
2. Why do we calculate the situation where  $Delta G$ is zero?
3. At temperatures below 835°C, is any product formed?

# Changes of State and Free Energy

• Describe how a change of state influences free energy.

#### How are energy and changes of state related?

Energy in a body of water can be gained or lost depending on conditions.  When water is heated above a certain temperature steam is generated.  The increase in heat energy creates a higher level of disorder in the water molecules as they boil off and leave the liquid.

### Changes of State and Free Energy

At the temperature at which a change of state occurs, the two states are in equilibrium with one another. For an ice-water system, equilibrium takes place at 0°C, so  $Delta G^circ$ is equal to 0 at that temperature. The heat of fusion of water is known to be equal to 6.01 kJ/mol, and so the Gibbs free energy equation can be solved for the entropy change that occurs during the melting of ice. The symbol  $Delta S_{text{fus}}$ represents the entropy change during the melting process, while  $T_{text{f}}$ is the freezing point of water.

$Delta G & = 0 = Delta H - T Delta S\Delta S_{text{fus}} & = frac{Delta H_{text{fus}}}{T_{text{f}}} = frac{6.01 text{ kJ/mol}}{273 text{ K}} = 0.0220 text{ kJ/K} cdot text{mol} = 22.0 text{ J/K} cdot text{mol}$

The entropy change is positive as the solid state changes into the liquid state. If the transition went from the liquid to the solid state, the numerical value for  $Delta S$ would be the same, but the sign would be reversed since we are going from a less ordered to a more ordered situation.

A similar calculation can be performed for the vaporization of liquid to gas. In this case we would use the molar heat of vaporization. This value would be 40.79 kJ/mol. The  $Delta S_{text{vap}}$ would then be as follows:

$Delta S = frac{40.79 text{ kJ/mol}}{373 text{ K}} = 0.1094 text{ kJ/K} cdot text{mol} = 109.4 text{ J/K} cdot text{mol}$

The value is positive, again reflecting the increase in disorder going from liquid to vapor. Condensation from vapor to liquid would give a negative value for $Delta S$ .

#### Summary

• Calculations are shown for determining entropy changes at transition temperatures (ice → water or water → vapor and reverse).

#### Practice

Questions

http://www.everyscience.com/Chemistry/Physical/Entropy/f.1311.php

1. Is the transfer of heat reversible or irreversible at the transition temperature?
2. If the phase transition is exothermic, is the entropy change positive or negative?
3. What is Trouton’s Rule?

#### Review

Questions

1. What precautions need to be taken in selecting a value for $Delta H$ ?
2. Why is temperature selection important?
3. Why would the entropy of vaporization be so much larger than the entropy of fusion?

# Calculations of Free Energy and Keq

• Describe the relationship between  $Delta G^circ$ and  $K_{eq}$ .
• Perform calculations involving these two parameters.

What are these formations called when they point down?

Formation of stalactites (pointing down) and stalagmites (pointing up) is a complex process. Solutions of minerals drip down and absorb carbon dioxide as water flows through the cave. Calcium carbonate dissolves in this liquid and redeposits on the rock as the carbon dioxide is dissipated into the environment.

### Equilibrium Constant and $Delta G$

At equilibrium the  $Delta G$ for a reversible reaction is equal to zero. $K_{eq}$ relates the concentrations of all substances in the reaction at equilibrium. Therefore we can write (through a more advanced treatment of thermodynamics) the following equation:

$Delta G^circ=-RT ln K_{eq}$

The variable  $R$ is the ideal gas constant (8.314 J/K • mol),  $T$ is the Kelvin temperature, and  $ln K_{eq}$ is the natural logarithm of the equilibrium constant.

When  $K_{eq}$ is large, the products of the reaction are favored and the negative sign in the equation means that the  $Delta G^circ$ is negative. When  $K_{eq}$ is small, the reactants of the reaction are favored. The natural logarithm of a number less than one is negative and so the sign of $Delta G^circ$ is positive. The Table below summarizes the relationship of  $Delta G^circ$ to $K_{eq}$ :

 $K_{eq}$ $ln K_{eq}$ $Delta G^circ$ Description >1 positive negative Products are favored at equilibrium. 1 0 0 Reactants and products are equally favored. <1 negative positive Reactants are favored at equilibrium.

Knowledge of either the standard free energy change or the equilibrium constant for a reaction allows for the calculation of the other. The following two sample problems illustrate each case.

#### Sample Problem: Gibbs Free Energy and the Equilibrium Constant

The formation of nitrogen monoxide from nitrogen and oxygen gases is a reaction that strongly favors the reactants at 25°C.

$text{N}_2(g)+text{O}_2(g) rightleftarrows 2text{NO}(g)$

The actual concentrations of each gas would be difficult to measure, and so the  $K_{eq}$ for the reaction can more easily calculated from the $Delta G^circ$ , which is equal to 173.4 kJ/mol.

Step 1: List the known values and plan the problem.

Known

• $Delta G^circ=+173.4 text{kJ} / text{mol}$
• $R=8.314 text{J} / text{K} cdot text{mol}$
• $T=25^circ text{C}=298 text{K}$

Unknown

• $K_{eq}=?$

In order to make the units agree, the value of $Delta G^circ$ will need to be converted to J/mol (173,400 J/mol). To solve for $K_{eq}$ , the inverse of the natural logarithm, $e^x$ , will be used.

Step 2: Solve .

$Delta G^circ &=-RT ln K_{eq} \ln K_{eq}&=frac{-Delta G^circ}{RT} \K_{eq} &=e^{frac{-Delta G^circ}{RT}}=e^{frac{-173, 400 text{J} / text{mol}}{8.314 text{J} / text{K} cdot text{mol}(298 text{K})}}=4.0 times 10^{-31}$

The large positive free energy change leads to a $K_{eq}$ value that is extremely small. Both lead to the conclusion that the reactants are highly favored and very few product molecules are present at equilibrium.

#### Sample Problem: Free Energy from $K_{sp}$

The solubility product constant  $(K_{sp})$ of lead(II) iodide is 1.4 × 10 -8 at 25°C. Calculate  $Delta G^circ$ for the dissociation of lead(II) iodide in water.

$text{PbI}_2(s) rightleftarrows text{Pb}^{2+}(aq) + 2text{I}^-(aq)$

Step 1: List the known values and plan the problem .

Known

• $K_{eq}=K_{sp}=1.4 times 10^{-8}$
• $R=8.314 text{J} / text{K} cdot text{mol}$
• $T=25^circ text{C}=298 text{K}$

Unknown

• $Delta G^circ=? text{kJ} / text{mol}$

The equation relating  $Delta G^circ$ to  $K_{eq}$ can be solved directly.

Step 2: Solve.

$Delta G^circ&=-RT ln K_{eq} \&=-8.314 text{J} / text{K} cdot text{mol}(298 text{K}) ln (1.4 times 10^{-8}) \&=45,000 text{J} / text{mol} \&=45 text{kJ} / text{mol}$

The large, positive  $Delta G^circ$ indicates that the solid lead(II) iodide is nearly insoluble and so very little of the solid is dissociated at equilibrium.

#### Summary

• The relationship between  $Delta G$ and  $K_{eq}$ is described.
• Calculations involving these two parameters are shown.

#### Practice

Questions

1. What is the difference between $Delta G$ and $Delta G^circ$ ?
2. At equilibrium, why does the equation between free energy and equilibrium constant reduce to $Delta G^circ = -RT ln K_{eq}$ ?
3. What other equilibrium units could we use?

#### Review

Questions

1. When  $K_{eq}$ is large, what will be the sign of $Delta G$ ?
2. When  $K_{eq}$ is small, are reactants or products favored?
3. What does  $R$ stand for?