Introduction to Enthalpy vs. Entropy

When you get to the thermodynamics section of AP® Chemistry, one of the hardest things to remember is the definition of enthalpy vs. entropy. This tutorial explains the theory behind enthalpy and entropy using the laws of thermodynamics. Then we will talk about the definition of enthalpy and how to calculate it using enthalpy practice questions. Once you have a firm hold on the definition of enthalpy, we will discuss entropy and look at entropy practice questions. Finally, we will revisit the topic of Gibbs free energy, of which you should already have a decent understanding, and how it relates to enthalpy vs. entropy.

Laws of Thermodynamics and Enthalpy vs. Entropy

There are three laws of thermodynamics:

1. The first law of thermodynamics is the theory of conservation of energy. It states that energy cannot be created or destroyed, only transferred between types of energy, such as heat and work.

2. The second law states that the entropy of the universe is always increasing. This means that the entropy of a closed or isolated system will always increase over time. This law explains that concentrated energy has a tendency to become dispersed energy in the form of random thermal motion.

3. This random motion, this entropy, is represented by S, and is a measure of the energy that is unable to do work. Its units are the change in energy (\Delta E \text{ per degree = Joules/Kelvin}). Any time that entropy is called random, disorganized, diffuse, dissipated, chaotic, etc., what is being indicated is the random molecular motion in a system that is unable to be harnessed to do work. As you know from general chemistry, increasing the temperature of a system increases its random motion, so it also increases its entropy.

{ S }_{ universe }={ \text{increases} } or \Delta { S }_{ universe }>0

Since we know that natural systems will all increase in entropy over time, we know that changes that cause an increase in total entropy will be spontaneous. We also know that there will never be a spontaneous decrease in entropy. Just like when you pour water on sand, it disperses randomly, so energy disperses randomly.

The third law states that the entropy of a perfect crystal (a crystal in which the molecules are in perfect alignment) approaches zero as the temperature approaches absolute zero.

Okay, so this law seems pointless, right? Actually, this law tells us that the entropy of a substance can be calculated at any temperature that is above 0 \text{ K}. It also explains how the entropy of a system can be higher than zero when the enthalpy and Gibbs free energy are both zero. This allows us to calculate an absolute entropy for molecules, not just a change in entropy.

Enthalpy vs. Entropy

Enthalpy is the heat energy in a closed system. Entropy is the energy in a closed system that is unavailable to do work

Enthalpy

The enthalpy of a reaction is the heat energy it contains. Enthalpy is represented by H in equations. The most important law when looking at enthalpy is Hess’s Law: the enthalpy change accompanying a chemical change is independent of the way in which the chemical change occurs. This means that if you convert reactants into products, the overall enthalpy change is the same whether you do it in one step or multiple steps.

This means that even with an intermediary compound, the change in enthalpy is equal to the sum of the change in enthalpy of the products minus the sum of the change in enthalpy of the reactants. This equation is Hess’s Law.

\Delta { H }^{ \circ }=\sum { { n\Delta H }_{ products }^{ \circ } } -\sum { { n\Delta H }_{ reactants }^{ \circ } }

The third law of thermodynamics allows us to calculate absolute entropy, but not absolute enthalpy. We cannot calculate absolute enthalpy values because enthalpy does not approach 0 as temperature approaches absolute zero. We can only calculate the enthalpy of a substance relative to other substances. So we use pre-calculated “enthalpy of formation” values to calculate the enthalpy change of a reaction and “absolute entropy” values to calculate entropy.

Example

Calculate the standard enthalpy of combustion for the following reaction using the enthalpy values below.

{ 2C }_{ 2 }{ H }_{ 5 }{ OH }_{ (l) } + 7{ O }_{ 2(g) } \rightarrow { 4CO }_{ 2(g) } + { 6H }_{ 2 }{ O }_{ (l) }

\Delta { H }^{ \circ } \text{ for } { CO }_{ 2(g) } = -393.5

\Delta { H }^{ \circ } \text{ for } { H }_{ 2 }{ O }_{ (g) } = -286

{ \Delta H }^{ \circ } \text{ for } { C }_{ 2 }{ H }_{ 5 }{ O }H_{ (l) } = -278

\Delta { H }^{ \circ } \text{ for } { O }_{ 2(g) } = 0

This problem can be solved in four easy steps.

Step 1: Arrange the molecules and coefficients from the chemical reaction into the equation for calculating the change in entropy.

The n in the equation is the coefficient or number of moles of the molecule from the chemical reaction.

\Delta { H }^{ \circ } = [(4 \text{ mol})({ \Delta H }^{ \circ } \text{ for } { CO }_{ 2(g) }) + (6 \text{ mol})(\Delta { H }^{ \circ } \text{ for } { H }_{ 2 }{ O }_{ (g) })

- [(2 \text{ mol})({ \Delta H }^{ \circ } \text{ for } { C }_{ 2 }{ H }_{ 5 }{ O }H_{ (l) }) + (7 \text{ mol})(\Delta { H }^{ \circ } \text{ for } { O }_{ 2(g) })]

Step 2: Find the enthalpy of each molecule.

For this type of equation, the enthalpy values will be given to you either in a table on a separate piece of paper or with the question. In this case, the enthalpy values are given with the question.

Step 3: Place the enthalpy values in the equation.

\Delta { H }^{ \circ } = [(4 \text{ mol})(-393.5 \text{ kJ/mol}) + (6\text{ mol})(-286 \text{ kJ/mol})]

- [(2 \text{ mol})(-278 \text{ kJ/mol}) + (7 \text{ mol})(0 \text{ kJ/mol})]

Step 4: Solve the equation.

\Delta { H }^{ \circ } = [-1574 \text{ kJ} + (-1716\text{ kJ})] - [-556 \text{ kJ} + 0 \text{ kJ}]

\Delta { H }^{ \circ } = -3290 \text{ kJ} - (-556 \text{ kJ})

\Delta { H }^{ \circ } = -3290 \text{ kJ} + 556 \text{ kJ} = -2734 \text{ kJ}

Entropy

The entropy of a system is the amount of disorder it contains.

There are several ways to predict the entropy of a system:

1. Entropy increases as disorder or randomness increases, so the more random a system is, the higher its entropy.

2. When undergoing a phase change from solid to liquid and liquid to gas, the entropy of a substance always increases.

3. Except with carbonates, when a solid or liquid is dissolved in a solvent, the entropy of the substance increases. Carbonates actually increase the order of a system when dissolved in water.

4. Entropy increases when a gas molecule escapes from a solvent.

5. As molecular complexity increases (NaCl vs. MgCl_2), entropy generally increases due to the increase in moving electrons.

6. Reactions in which the moles of a particle increase often increase in entropy.

To calculate the standard change in entropy of a reaction, you subtract the sum of the entropies of the reactants from the sum of the entropies of the reactants.

\Delta { S }^{ \circ }=\sum { { n\Delta S }_{ products }^{ \circ } } -\sum { { n\Delta S }_{ reactants }^{ \circ } }

Also, entropy changes are different for condensation and vaporization. The change in entropy is calculated using the change in enthalpy and the temperature. The formulas necessary for each of these calculations are:

\Delta { S }_{ condensation } = \dfrac { { -n\Delta H }_{ vaporization } }{ { T }_{ boil } }

\Delta { S }_{ vaporization } = \dfrac { { n\Delta H }_{ vaporization } }{ { T }_{ boil } }

All of the formulas we have looked at so far have been used to calculate the enthalpy of the system. To calculate the enthalpy of the surroundings, the following formula must be used, which takes -1 times the change in enthalpy of the reaction divided by the temperature.

\Delta { S }_{ surr } = \dfrac { -\Delta { H }_{ rxn } }{ T }

Then, to calculate the total change in entropy of the universe, add the change in entropy of the reaction to the change in entropy of the surroundings, utilizing the following formula.

\Delta { S }_{ universe } =\Delta { S }_{ rxn } + \Delta { S }_{ surr }

Example Entropy Problems

Example 1

One mole of solid calcium carbonate decomposes to form solid calcium oxide and carbon dioxide gas. What is the standard entropy change of this reaction, given the absolute entropies for each compound below?

{ CaCO }_{ 3(s) }\rightarrow { CaO }_{ (s) }+{ CO }_{ 2(g) }

\Delta { S }^{ \circ }\text{ of }{ CaO }_{ (s) } = 39.8 \text{ J/mol K}

\Delta { S }^{ \circ }\text{ of }{ CO }_{ 2(g) } = 213.7 \text{ J/mol K}

\Delta { S }^{ \circ }\text{ of }{ CaCO }_{ 3(s) } = 92.9 \text{ J/mol K}

There are four simple steps you can follow to solve this type of problem.

Step 1: Arrange the molecules and coefficients from the chemical reaction into the equation for calculating the change in entropy.

The n in the equation is the coefficient or number of moles of the molecule from the chemical reaction.

\Delta { S }^{ \circ }=[(1 \text{ mol})({ \Delta S }^{ \circ } \text{ for } { CaO }_{ (s) })+(1 \text{ mol})(\Delta { S }^{ \circ } \text{ for } { CO }_{ 2(g) })] -[(1 \text{ mol})({ \Delta S }^{ \circ } \text{ for } { CaCO }_{ 3(s) })]

Step 2: Find the absolute entropy of each molecule in a table. For this type of equation, the entropy values will be given to you either on a separate piece of paper or with the question. In this case, the entropy values are given with the question.

Step 3: Place the absolute entropy values in the equation.

\Delta { S }^{ \circ } = [(1 \text{ mol})(39.8 \text{ J/mol K}) + (1 \text{ mol})(213.7\text{ J/mol K})] - [(1 \text{ mol})(92.9 \text{ J/mol K})]

Step 4: Solve the equation.

\Delta { S }^{ \circ } =(39.8 \text{ J/K}+213.7 \text{ J/K})-92.9 \text{ J/K}

\Delta { S }^{ \circ }=253.5 \text{ J/K} -92.9 \text{ J/K}

\Delta { S }^{ \circ }=160.6 \text{ J/K}

Example 2

What is the change in entropy when two moles of carbon monoxide condense at -191.4^\circ \text{C} ? The heat of vaporization of CO is 6.04 \text{ kJ/mol}.

Step 1: Make sure all variables have the correct units.

Temperature should be in Kelvin and change in enthalpy in \text{J/mol}.

n = 2 \text{ mol } CO

{ T }_{ boil }=(-191.4^{\circ}\text{C}+273.15)=81.75 \text{ K}

\Delta { H }_{ vaporization }=\dfrac { 6.04 \text{ kJ} }{ \text{mol} } \times \dfrac { 1000 \text{ J} }{ 1 \text{ kJ} } =6040 \text{ J/mol}

Step 2: Place the variables into the formula, including the number of moles of your molecule.

\Delta { S }_{ condensation }=\dfrac { { -n\Delta H }_{ vaporization } }{ { T }_{ boil } }

\Delta { S }_{ condensation }=\dfrac { { -(2 \text{ mol})(6040 \text{ J/mol} }) }{ { 81.75 \text{ K} } }

Step 3: Solve the equation.

\Delta { S }_{ condensation }=\dfrac { { -(2 \text{ mol})(6040 \text{ J/mol} }) }{ { 81.75 \text{ K} } } =\dfrac { -12080 \text{ J} }{ 81.75 \text{ K} } =-147.77 \text{ J/K}

Gibbs Free Energy, Entropy, and Enthalpy

Remember from learning about Gibbs free energy, that the change in free energy is equal to the sum of the free energy of the products minus the sum of the free energy of the reactants.

\Delta { G }^{ \circ }=\sum { { n\Delta G }_{ fproducts }^{ \circ } } -\sum { { n\Delta G }_{ freactants }^{ \circ } }

Notice that this is very close to the equation used to calculate \Delta H^\circ and \Delta S^\circ.

To review, let’s discuss the relationship between Gibbs free energy, entropy, and enthalpy.

Gibbs free energy is represented by G and is found with the following equation.

{ G }^{\circ}={ H }^{\circ}-{ TS }^{\circ}

Since we are looking at the changes in energy, we need to account for that in our equation:

\Delta { G }^{\circ}={\Delta H }^{\circ}-{\Delta TS }^{\circ}

When using this equation, we almost always are calculating a change during constant temperature:

\Delta { G }^{\circ}={ \Delta H }^{\circ}-{ T\Delta S }^{\circ}

…and we also almost always calculate changes in standard state conditions:

\Delta { G }^{\circ}={ \Delta H }^{\circ}-{ T\Delta S }^{\circ}

Using this equation, we can now calculate the change in free energy from the change in enthalpy and entropy values, as demonstrated in this example.

Example

In the decomposition of calcium carbonate, what is the change in free energy?

We have already calculated the change in entropy in the previous example. Using a similar equation, we can calculate the change in enthalpy.

\Delta { H }^{ \circ }=\sum { { n\Delta H }_{ products }^{ \circ } } -\sum { { n\Delta H }_{ reactants }^{ \circ } }

\Delta { H }^{ \circ }=[(1 \text{ mol})({ \Delta H }^{ \circ }\text{ for }{ CaO }_{ (s) })+(1 \text{ mol})(\Delta { H }^{ \circ }\text{ for }{ CO }_{ 2(g) })]-[(1 \text{ mol})({ \Delta H }^{ \circ }\text{ for }{ CaCO }_{ 3(s) })]

\Delta { H }^{ \circ }=[(1 \text{ mol})(-635.3 \text{ kJ/mol})+(1 \text{ mol})(-393.5 \text{ kJ/mol})]-[(1 \text{ mol})(-1207.0 \text{ kJ/mol})]

\Delta { H }^{ \circ }=[(-635.3 \text{kJ/mol})+(-393.5 \text{ kJ/mol})]-[(-1207.0 \text{ kJ/mol})]

\Delta { H }^{ \circ }=-758.8 \text{ kJ/mol}+1207.0 \text{ kJ/mol}=178.2 \text{ kJ}

Though entropy is recorded in \text{J/K}, free energy’s standard units are \text{kJ}. Always make sure to convert entropy from \text{J} to \text{kJ} before calculating Gibbs free energy!

{ \Delta S }^{ \circ }=\dfrac { 160.6 \text{ J} }{ \text{K} } \times \dfrac { 1 \text{ kJ} }{ 1000 \text{ J} } =0.1606

Now we can use the new equation relating the two to Gibbs free energy. Thermodynamic standard temperature (298.15 \text{ K or } 25^\circ \text{C}) is used for T.

\Delta { G }^{\circ}={ \Delta H }^{\circ}-{ T\Delta S }^{\circ}=178.2 \text{ kJ}-(298.15 \text{ K} \times 0.1606 \text{ kJ})=130.3 \text{ kJ}

Wrapping Up Enthalpy vs. Entropy

The definition of enthalpy vs. entropy should be much clearer now. If you have any other questions or comments about enthalpy vs. entropy, please let us know in the comments below! If you would like more practice with enthalpy vs. entropy problems, we have included more practice problems and their answers below.

Enthalpy Practice Problems

1. The standard enthalpy of formation of AgNO_{3(s)} is -123.02 \text{ kJ/mol}. What is the standard enthalpy of formation of AgNO_{2(s)}, given the following reaction?

{ AgNO }_{ 3(s) }\rightarrow { AgNO }_{ 2(s) }+\dfrac { 1 }{ 2 } { O }_{ 2(g) }

2. Given the following standard enthalpies of formation, calculate the heat of combustion per mole of gaseous water formed during the combustion of ethane gas.

{ 2C }_{ 2 }{ H }_{ 6(l) }+{ 7O }_{ 2(g) }\rightarrow { 4CO }_{ 2(g) }+{ 6H }_{ 2 }{ O }_{ (g) }

• C_2H_{6(g)}: -84.68
• O_{2(g)}: 0
• CO_{2(g)}: -393.5
• H_2O_{(g)}: -241.8

Entropy Practice Problems

1. Which of these substances should have a higher entropy? Answer without calculating and assume that there is one mole of each substance at 25^\circ \text{C} and 1 \text{ bar}.

1. Hg(l) or CO(g)

2. CH_3OH(l) or CH_3CH_2OH(l)

3. KI(s) or CaS(s)

2. Predict the sign of the entropy change for the following reaction and use the given entropy values to calculate the change in enthalpy of the reaction at 25^\circ \text{C}.

{ 2H }_{ 2(g) }+{ O }_{ 2(g) }\rightarrow { 2H }_{ 2 }{ O }_{ (g) }

\Delta { S }^{ \circ } \text{ of } { H }_{ 2 }{ O }_{ (g) }=188.8 \text{ J/mol K}

\Delta { S }^{ \circ } \text{ of }{ H }_{ 2 }{ (g) }=130.7 \text{ J/mol K}

\Delta { S }^{ \circ } \text{ of }{ O }_{ 2 }{ (g) }=205.1 \text{ J/mol K}

Combined Practice Problems

1. Hydrogen and oxygen react to form water vapor with a \Delta S_{rxn}= -88.99 \text{ J/K}. What are the entropy changes of the surroundings and the universe at 25^\circ \text{C}?

2. Calculate the free energy change for the following reaction at 25^\circ \text{C} from the change in enthalpy of the reaction and the change in entropy of the reaction. (Remember to check the enthalpy vs. entropy section if you need clarification.)

{ CO }_{ (g) }+\dfrac { 1 }{ 2 } {O }_{ 2(g) }\rightarrow { CO }_{ 2(g) }

Enthalpy Practice Problems Key

1. The standard enthalpy of formation of AgNO_{3(s)} is -123.02 \text{ kJ/mol}. What is the standard enthalpy of formation of AgNO_{2(s)}, given the following reaction?

{ AgNO }_{ 3(s) }\rightarrow { AgNO }_{ 2(s) }+\dfrac { 1 }{ 2 } { O }_{ 2(g) }

a. -44.35 \text{ kJ}

2. Given the following standard enthalpies of formation, calculate the heat of combustion per mole of gaseous water formed during the combustion of ethane gas.

a. -345 \text{ kJ/mol}

Entropy Practice Problems Answer Key

1. Which of these substances should have a higher entropy? Answer without calculating and assume that there is one mole of each substance at 25^\circ \text{C} and 1 \text{ bar}.

a. Since the motion of molecules in a gas are more random than the motion of molecules in a liquid, carbon monoxide gas has a higher entropy than liquid mercury.

b. More complex molecules have more ways of distributing their energy (electrons) at a given temperature, so the larger molecule (liquid ethanol, CH_3CH_2OH(l)) has a higher entropy than the liquid methanol.

c. There is more of an attraction between the ions in calcium sulfide than in potassium iodide. As the attraction between ions decreases, the entropy will increase, because movement between ions will be easier. Therefore, potassium iodide will have a higher entropy.

2. Predict the sign of the entropy change for the following reaction and use the given entropy values to calculate the change in enthalpy of the reaction at 25^circ \text{C}.

a. -88.9 \text{ J/K}

Combined Practice Problems Answer Key

1. Hydrogen and oxygen react to form water vapor with a \Delta S_{rxn}= -88.99 \text{ J/K}. What are the entropy changes of the surroundings and the universe at 25^\circ \text{C}?

a. \Delta H_{rxn} = -483.6 \text{ kJ}

b. \Delta S_{rxn} = 1.62 \text{ kJ/K}

c. \Delta S_{univ} = 1.53 \text{ kJ/K}

d. Since the \Delta S_{univ} is greater than 0, the reaction is spontaneous.

2. Calculate the free energy change for the following reaction at 25^\circ \text{C} from the change in enthalpy of the reaction and the change in entropy of the reaction. (Remember to check the enthalpy vs. entropy section if you need clarification.)

{ CO }_{ (g) }+\dfrac { 1 }{ 2 } {O }_{ 2(g) }\rightarrow { CO }_{ 2(g) }

a. \Delta H_{rxn} = -283.0 \text{ kJ/mol}

b. \Delta S_{rxn} = -86.6 \text{ J/mol K}

c. \Delta G_{rxn} = -257 \text{ kJ/mol}

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