## 5 Steps to a 5: AP Chemistry 2017 (2016)

### STEP __4__

### Review the Knowledge You Need to Score High

### CHAPTER 9

### Thermodynamics

**IN THIS CHAPTER**

**Summary: Thermodynamics** is the study of heat and its transformations.

**Thermochemistry** is the part of thermodynamics that deals with changes in heat that take place during chemical processes. We will be describing energy changes in this chapter. Energy can be of two types: kinetic or potential. **Kinetic energy** is energy of motion, while **potential energy** is stored energy. Energy can be converted from one form to another but, unless a nuclear reaction occurs, energy cannot be created or destroyed (Law of Conservation of Energy). We will discuss energy exchanges between a system and the surroundings. The **system** is that part of the universe that we are studying. It may be a beaker or it may be Earth. The **surroundings** are the rest of the universe.

The most common units of energy used in the study of thermodynamics are the joule and the calorie. The **joule (J)** is defined as:

1 J = 1 kg m^{2} /s^{2}

The **calorie** was originally defined as the amount of energy needed to raise the temperature of 1 g of water 1°C. Now it is defined in terms of its relationship to the joule:

1 cal = 4.184 J

It is important to realize that this is not the same calorie that is commonly associated with food and diets. That is the nutritional Calorie, Cal, which is really a kilocalorie (1 Cal = 1,000 cal).

**Keywords and Equations**

*S* ° = standard entropy

*G* ° = standard free energy

*c* = specific heat capacity

Δ*S* ° = Σ *S* ° products − Σ *S* ° reactants

*H* ° = standard enthalpy

*q* = heat

*C* _{p} = molar heat capacity at constant pressure

Gas constant, *R* = 8.31 J mol^{–1} K^{–1}

**Calorimetry**

**Calorimetry** is the laboratory technique used to measure the heat released or absorbed during a chemical or physical change. The quantity of heat absorbed or released during a chemical or physical change is represented as *q* and is proportional to the change in temperature of the system being studied. This system has what is called a **heat capacity** (*C* _{p} ), which is the quantity of heat needed to change the temperature 1 K. It has the form:

*C* _{p} = heat capacity = *q* /Δ*T*

Heat capacity most commonly has units of J/K. The **specific heat capacity** (**or specific heat** ) (**c** ) is the quantity of heat needed to raise the temperature of 1 g of a substance 1 K:

*c* = *q* /(*m* × Δ*T* ) or *q* = *cm* Δ*T* ,

where *m* is the mass of the substance.

The specific heat capacity commonly has units of J/g·K. Because of the original definition of the calorie, the specific heat capacity of water is 4.184 J/g·K. If the specific heat capacity, the mass, and the change of temperature are all known, the amount of energy absorbed can easily be calculated.

Another related quantity is the **molar heat capacity** ( ** C** ), the amount of heat needed to change the temperature of 1 mol of a substance by 1 K.

Calorimetry involves the use of a laboratory instrument called a calorimeter. Two types of calorimeter, a simple coffee-cup calorimeter and a more sophisticated bomb calorimeter, are shown in __Figure 9.1__ , on the next page. In both, a process is carried out with known amounts of substances and the change in temperature is measured.

The coffee-cup calorimeter can be used to measure the heat changes in reactions or processes that are open to the atmosphere: *q* _{p} , constant-pressure reactions. These might be reactions that occur in open beakers and the like. This type of calorimeter is also commonly used to measure the specific heats of solids. A known mass of solid is heated to a certain temperature and then is added to the calorimeter containing a known mass of water at a known temperature. The final temperature is then measured allowing us to calculate the Δ*T* . We know that the heat lost by the solid (the system) is equal to the heat gained by the surroundings (the water and calorimeter, although for simple coffee-cup calorimetry the heat gained by the calorimeter is small and is ignored):

−*q* _{solid} = *q* _{water}

**Figure 9.1 Two types of calorimeters.**

Substituting the mathematical relationship for *q* gives:

−(*c* _{solid} × *m* _{solid} × Δ*T* _{solid} ) = *c* _{water} × *m* _{water} × Δ*T* _{water}

This equation can then be solved for the specific heat capacity of the solid.

The constant-volume bomb calorimeter is used to measure the energy changes that occur during combustion reactions. A weighed sample of the substance being investigated is placed in the calorimeter, and compressed oxygen is added. The sample is ignited by a hot wire, and the temperature change of the calorimeter and a known mass of water is measured. The heat capacity of the calorimeter/water system is sometimes known.

For example, a 1.5886 g sample of glucose (C_{6} H_{12} O_{6} ) was ignited in a bomb calorimeter. The temperature increased by 3.682°C. The heat capacity of the calorimeter was 3.562kJ/°C, and the calorimeter contained 1.000 kg of water. Find the molar heat of reaction (i.e., kJ/mole) for:

C_{6} H_{12} O_{6} (s) + 6 O_{2} (g) → 6 CO_{2} (g) + 6 O(l)

Answer:

**Note:** The temperature increased so the reaction was exothermic (−)

→ −28.52 kJ

This is not molar (yet)

Thus:

**Laws of Thermodynamics**

The **First Law of Thermodynamics** states that the total energy of the universe is constant. This is simply the Law of Conservation of Energy. This can be mathematically stated as:

Δ*E* _{universe} = Δ*E* _{system} + Δ*E* _{surroundings} = 0

The Second Law of Thermodynamics involves a term called entropy. **Entropy** ( ** S** ) is related to the disorder of a system. The

**Second Law of Thermodynamics**states that all processes that occur spontaneously move in the direction of an increase in entropy of the universe (system + surroundings). Mathematically, this can be stated as:

Δ*S* _{universe} = Δ*S* _{system} + Δ*S* _{surroundings} > 0 for a spontaneous process

For a reversible process Δ*S* _{universe} = 0. The qualitative entropy change (increase or decrease of entropy) for a system can sometimes be determined using a few simple rules:

- Entropy increases when the number of molecules increases during a reaction.
- Entropy increases with an increase in temperature.
- Entropy increases when a gas is formed from a liquid or solid.
- Entropy increases when a liquid is formed from a solid.

Let us now look at some applications of these first two laws of thermodynamics.

**Products Minus Reactants**

**Enthalpies**

Many of the reactions that chemists study are reactions that occur at constant pressure. During the discussion of the coffee-cup calorimeter, the heat change at constant temperature was defined as *q* _{p} . Because this constant-pressure situation is so common in chemistry, a special thermodynamic term is used to describe this energy: enthalpy. The **enthalpy change** , Δ ** H** , is equal to the heat gained or lost by the system under constant-pressure conditions. The following sign conventions apply:

If Δ*H* > 0 the reaction is endothermic.

If Δ*H* < 0 the reaction is exothermic.

If a reaction is involved, Δ*H* is sometimes called Δ*H* _{reaction} . Δ*H* is often given in association with a particular reaction. For example, the enthalpy change associated with the formation of water from hydrogen and oxygen gases can be shown in this fashion:

2 H_{2} (g) + O_{2} (g) → 2 H_{2} O(g) Δ*H* = −483.6 kJ

The negative sign indicates that this reaction is exothermic. This value of Δ*H* is for the production of 2 mol of water. If 4 mol were produced, Δ*H* would be twice −483.6 kJ. The techniques developed in working reaction stoichiometry problems (see __Chapter 7__ , Stoichiometry) also apply here.

If the previous reaction for the formation of water were reversed, the sign of Δ*H* would be reversed. That would indicate that it would take 483.6 kJ of energy to decompose 2 mol of water. This would then become an endothermic process.

Δ*H* is dependent upon the state of matter. The enthalpy change would be different for the formation of liquid water instead of gaseous water.

Δ*H* can also indicate whether a reaction will be spontaneous. A negative (exothermic) value of Δ*H* is associated with a spontaneous reaction. However, in many reactions this is not the case. There is another factor to consider in predicting a reaction”s spontaneity. We will cover this other factor a little later in this chapter.

Enthalpies of reaction can be measured using a calorimeter. However, they can also be calculated in other ways. **Hess”s law** states that if a reaction occurs in a series of steps, then the enthalpy change for the overall reaction is simply the sum of the enthalpy changes of the individual steps. If, in adding the equations of the steps together, it is necessary to reverse one of the given reactions, then the sign of Δ*H* must also be reversed. Also, particular attention must be used if the reaction stoichiometry has to be adjusted. The value of an individual Δ*H* may need to be adjusted.

It doesn”t matter whether the steps used are the actual steps in the mechanism of the reaction, because Δ*H* _{reaction} (Δ*H* _{rxn} ) is a **state function** , a function that doesn”t depend on the pathway, but only on the initial and final states.

Let”s see how Hess”s law can be applied, given the following information:

find the enthalpy change for:

2C(s) + H_{2} (g) → C_{2} H_{2} (g)

Answer:

Enthalpies of reaction can also be calculated from individual enthalpies of formation (or heats of formation), Δ*H _{f} *, for the reactants and products. Because the temperature, pressure, and state of the substance will cause these enthalpies to vary, it is common to use a standard state convention. For gases, the standard state is 1 atm pressure. For a substance in an aqueous solution, the standard state is 1 molar concentration. And, for a pure substance (compound or element), the standard state is the most stable form at 1 atm pressure and 25°C. A degree symbol to the right of the

*H*indicates a standard state, Δ

*H*°. The

**standard enthalpy of formation**of a substance (Δ

*H*°) is the change in enthalpy when 1 mol of the substance is formed from its elements when all substances are in their standard states. These values are then tabulated and can be used in determining Δ

_{f}*H*°

_{rxn}.

Δ*H _{f} *° of an element in its standard state is zero.

Δ*H _{f} *°

_{rxn}can be determined from the tabulated Δ

*H*° of the individual reactants and products. It is the sum of the Δ

_{f}*H*° of the products minus the sum of the Δ

_{f}*H*° of the reactants:

_{f}Δ*H* °_{rxn} = Σ Δ*H _{f} *° products − Σ Δ

*H*° reactants

_{f}In using this equation be sure to consider the number of moles of each, because Δ*H _{f} *° for the individual compounds refer to the formation of 1 mol.

For example, let”s use standard enthalpies of formation to calculate Δ*H* _{rxn} for:

6 H_{2} O(g) + 4 NO(g)→ 5 O_{2} (g) + 4 NH_{3} (g)

Answer:

Using tabulated standard enthalpies of formation gives:

People commonly forget to subtract *all* the reactants from the products.

The values of Δ*H _{f} *° will be given to you on the AP exam, or you will be asked to stop before putting the numbers into the problem.

An alternative means of estimating the heat of reaction is to take the sum of the average bond energies of the reactant molecules and subtract the sum of the average bond energies of the product molecules.

**Entropies**

In much the same way as Δ*H* ° was determined, the **standard molar entropies** ( ** S** °) of elements and compounds can be tabulated. The standard molar entropy is the entropy associated with 1 mol of a substance in its standard state. Entropies are also tabulated, but unlike enthalpies, the entropies of elements are not zero. For a reaction, the standard entropy change is calculated in the same way as the enthalpies of reaction:

Δ*S* ° = Σ *S* ° products − Σ *S* ° reactants

Calculate Δ*S* ° for the following. If you do not have a table of *S* ° values, just set up the problems.

**Note: These are thermochemical equations, so fractions are allowed.**

- H
_{2}(g) + ½ O_{2}(g) → H_{2}O(g) - H
_{2}(g) + ½ O_{2}(g) → H_{2}O(l) - CaCO
_{3}(s) + H_{2}SO_{4}(l) → CaSO_{4}(s) + H_{2}O(g) + CO_{2}(g)

Answers:

- H
_{2}O H_{2}O_{2}

188.7 J/mol K − [131.0 + 1/2(205.0)]J/mol K

= −44.8 J/mol K

- H
_{2}O H_{2}O_{2}

69.9 J/mol K − [131.0 + 1/2(205.0)]J/mol K

= −163.6 J/mol K

One of the goals of chemists is to be able to predict whether or not a reaction will be spontaneous. Some general guidelines for a spontaneous reaction have already been presented (negative Δ*H* and positive Δ*S* ), but neither is a reliable predictor by itself. Temperature also plays a part. A thermodynamic factor that takes into account the entropy, enthalpy, and temperature of the reaction should be the best indicator of spontaneity. This factor is called the Gibbs free energy.

**Gibbs Free Energy**

The **Gibbs free energy** ( ** G** ) is a thermodynamic function that combines the enthalpy, entropy, and temperature:

*G* = *H* − *TS* , where *T* is the Kelvin temperature

Like most thermodynamic functions, only the change in Gibbs free energy can be measured, so the relationship becomes:

Δ*G* = Δ*H* − *T* Δ*S*

Δ*G* is the best indicator chemists have as to whether or not a reaction is spontaneous:

- If Δ
*G*> 0, the reaction is not spontaneous; energy must be supplied to cause the reaction to occur. - If Δ
*G*< 0, the reaction is spontaneous. - If Δ
*G*= 0, the reaction is at equilibrium.

If there is a Δ*G* associated with a reaction and that reaction is then reversed, the sign of Δ*G* changes.

Just like with the enthalpy and entropy, the standard Gibbs free energy change, Δ*G* °, is calculated:

Δ*G* ° = Σ Δ*G _{f} *° products − Σ Δ

*G*° reactants

_{f}Δ*G _{f} *° of an element in its standard state is zero.

Δ*G* ° for a reaction may also be calculated by using the standard enthalpy and standard entropy of reaction:

Δ*G* ° = Δ*H* °_{rxn} − *T* Δ*S* °_{rxn}

Calculate Δ*G* ° for:

(If you do not have a table of Δ*G* ° values, just set up the problems.)

- 2 NH
_{4}Cl(s) + CaO(s) → CaCl_{2}(s) + H_{2}O(l) + 2 NH_{3}(g) - C
_{2}H_{4}(g) + H_{2}O(g) → C_{2}H_{5}OH(l) - Ca(s) + 2 H
_{2}SO_{4}(l) → CaSO_{4}(s) + SO_{2}(g) + 2 H_{2}O(l)

Answers:

**Thermodynamics and Equilibrium**

Thus far, we have considered only situations under standard conditions. But how do we cope with nonstandard conditions? The change in Gibbs free energy under nonstandard conditions is:

Δ*G* = Δ*G* ° + *RT* ln *Q* = Δ*G* ° + 2.303 log *Q*

*Q* is the activity quotient, products over reactants. This equation allows the calculation of Δ*G* in those situations in which the concentrations or pressures are not 1.

Using the previous concept, calculate Δ*G* for the following at 500.K:

Note that *Q* , when at equilibrium, becomes *K* . This equation gives us a way to calculate the equilibrium constant, *K* , from a knowledge of the standard Gibbs free energy of the reaction and the temperature.

If the system is at equilibrium, then Δ*G* = 0 and the equation above becomes:

Δ*G* ° = −*RT* ln *K* = −2.303 *RT* log *K*

For example, calculate Δ*G* ° for:

2O_{3} (g) 3O_{2} (g) *K* _{P} = 4.17 × 10^{14}

Note: ° = 298 K

Answer:

**Experiments**

The most common thermodynamic experiment is a calorimetry experiment. In this experiment, the heat of transition or heat of reaction is determined.

The experiment will require a balance to determine the mass of a sample and possibly a pipet to measure a volume, from which a mass may be calculated using the density. A calorimeter, usually a polystyrene (Styrofoam) cup, is needed to contain the reaction. Finally, a thermometer is required. Tables of heat capacities or specific heats may be provided.

Mass and possible volume measurements, along with the initial and final temperatures, are needed. Remember: you *measure* the initial and final temperature so you can *calculate* the change in temperature.

After the temperature change is calculated, there are several ways to proceed. If the calorimeter contains water, the heat may be calculated by multiplying the specific heat of water by the mass of water by the temperature change. The heat capacity of the calorimeter may be calculated by dividing the heat by the temperature change. If a reaction is carried out in the same calorimeter, the heat from that reaction is the difference between the heat with and without a reaction.

Do not forget, if the temperature increases, the process is exothermic and the heat has a negative sign. The opposite is true if the temperature drops.

**Common Mistakes to Avoid**

- Be sure your units cancel giving you the unit desired in the final answer.
- Check your significant figures.
- Don”t mix energy units, joules, and calories.
- Watch your signs in all the thermodynamic calculations. They are extremely important.
- Don”t confuse enthalpy, Δ
*H*, and entropy, Δ*S*. - Pay close attention to the state of matter for your reactants and products, and choose the corresponding value for use in your calculated entropies and enthalpies.
- Remember:
**products minus reactants**. - Δ
*H*and Δ_{f}*G*are for 1 mol of substance. Use appropriate multipliers if needed._{f} - Δ
*G*and Δ_{f}*H*for an element in its standard state are zero._{f} - All temperatures are in kelvin.
- When using Δ
*G*° = Δ*H*°_{rxn}−*T*Δ*S*°_{rxn}, pay particular attention to your enthalpy and entropy units. Commonly, enthalpies will use kJ and entropies J.

** Review Questions**

Use these questions to review the content of this chapter and practice for the AP Chemistry exam. First are 16 multiple-choice questions similar to what you will encounter in Section I of the AP Chemistry exam. Following those is a multipart free-response question like the ones in Section II of the exam. To make these questions an even more authentic practice for the actual exam, time yourself following the instructions provided.

**Multiple-Choice Questions**

Answer the following questions in 20 minutes. You may not use a calculator. You may use the periodic table and the equation sheet at the back of this book.

** 1 .** Which of the following is the minimum energy required to initiate a reaction?

(A) free energy

(B) lattice energy

(C) kinetic energy

(D) activation energy

** 2 .** What is the minimum energy required to force a nonspontaneous reaction to occur?

(A) free energy

(B) lattice energy

(C) kinetic energy

(D) activation energy

** 3 .** The average _____________ is the same for any ideal gas at a given temperature.

(A) free energy

(B) lattice energy

(C) kinetic energy

(D) activation energy

** 4 .** What is the energy released when the gaseous ions combine to form an ionic solid?

(A) free energy

(B) lattice energy

(C) kinetic energy

(D) activation energy

** 5 .** Given the following information:

C(s) + O_{2} (g) → CO_{2} (g) Δ*H* = −393.5 kJ

H_{2} (g) + (1/2) O_{2} (g) → H_{2} O(l) Δ*H* = −285.8 kJ

C_{2} H_{2} (g) + (5/2) O_{2} (g) → 2 CO_{2} (g) + H_{2} O(l) Δ*H* = −1,299.8 kJ

Find the enthalpy change for:

2C(s) + H_{2} (g) → C_{2} H_{2} (g)

(A) 454.0 kJ

(B) −227.0 kJ

(C) 0.0 kJ

(D) 227.0 kJ

** 6 .** A sample of gallium metal is sealed inside a well-insulated, rigid container. The temperature inside the container is at the melting point of gallium metal. What can be said about the energy and the entropy of the system after equilibrium has been established? Assume the insulation prevents any energy change with the surroundings.

(A) The total energy increases. The total entropy will increase.

(B) The total energy is constant. The total entropy is constant.

(C) The total energy is constant. The total entropy will decrease.

(D) The total energy is constant. The total entropy will increase.

** 7 .** When ammonium chloride dissolves in water, the temperature drops. Which of the following conclusions may be related to this?

(A) Ammonium chloride is more soluble in hot water.

(B) Ammonium chloride produces an ideal solution in water.

(C) The heat of solution for ammonium chloride is exothermic.

(D) Ammonium chloride has a low lattice energy.

** 8 .** Choose the reaction expected to have the greatest increase in entropy.

(A) H_{2} O(g) → H_{2} O(l)

(B) 2 KClO_{3} (s) → 2 KCl(s) + 3 O_{2} (g)

(C) Ca(s) + H_{2} (g) → CaH_{2} (s)

(D) N_{2} (g) + 3H_{2} (g) → 2 NH_{3} (g)

** 9 .** Under standard conditions, calcium metal reacts readily with chlorine gas. What conclusions may be drawn from the fact?

(A) *K* _{eq} < 1 and Δ*G* ° > 0

(B) *K* _{eq} > 1 and Δ*G* ° = 0

(C) *K* _{eq} < 1 and Δ*G* ° < 0

(D) *K* _{eq} > 1 and Δ*G* ° < 0

** 10 .** Which of the following combinations is true when sodium chloride melts?

(A) Δ*H* > 0 and Δ*S* > 0

(B) Δ*H* = 0 and Δ*S* > 0

(C) Δ*H* > 0 and Δ*S* < 0

(D) Δ*H* < 0 and Δ*S* < 0

** 11 .** Which of the following reactions have a negative entropy change?

(A) 2 C_{2} H_{6} (g) + 7 O_{2} (g) → 4 CO_{2} (g) + 6 H_{2} O(g)

(B) 2 NH_{3} (g) → N_{2} (g) + 3 H_{2} (g)

(C) CaCl_{2} (s) → Ca(s) + Cl_{2} (g)

(D) 2 H_{2} (g) + O_{2} (g) → 2 H_{2} O(l)

** 12 .** A certain reaction is nonspontaneous under standard conditions, but becomes spontaneous at higher temperatures. What conclusions may be drawn under standard conditions?

(A) Δ*H* < 0, Δ*S* > 0, and Δ*G* > 0

(B) Δ*H* > 0, Δ*S* < 0, and Δ*G* > 0

(C) Δ*H* > 0, Δ*S* > 0, and Δ*G* > 0

(D) Δ*H* < 0, Δ*S* < 0, and Δ*G* > 0

** 13 .** 2 H

_{2}(g) + O

_{2}(g) → 2 H

_{2}O(g)

From the table below, determine the enthalpy change for the above reaction.

(A) 0 kJ

(B) 485 kJ

(C) −485 kJ

(D) 464 kJ

** 14 .** Which of the following reactions would be accompanied by the greatest decrease in entropy?

(A) N_{2} (g) + 3 H_{2} (g) → 2 NH_{3} (g)

(B) C(s) + O_{2} (g) → CO_{2} (g)

(C) 2 H_{2} (g) + O_{2} (g) → 2 H_{2} O(g)

(D) 2 Na(s) + Cl_{2} (g) → 2 NaCl(s)

** 15 .** CO(g) + 2 H

_{2}(g) → CH

_{3}OH(g) Δ

*H*= −91kJ

Determine Δ*H* for the above reaction if CH_{3} OH(l) were formed in the above reaction instead of CH_{3} OH(g). The Δ*H* of vaporization for CH_{3} OH is 37 kJ/mol.

(A) −128 kJ

(B) −54 kJ

(C) +128 kJ

(D) +54 kJ

** 16 .** A solution is prepared by dissolving solid ammonium nitrate, NH

_{4}NO

_{3}, in water. The initial temperature of the water was 25°C, but after the solid had dissolved, the temperature had fallen to 20°C. What conclusions may be made about Δ

*H*and Δ

*S*?

(A) Δ*H* < 0 Δ*S* > 0

(B) Δ*H* > 0 Δ*S* > 0

(C) Δ*H* > 0 Δ*S* < 0

(D) Δ*H* < 0 Δ*S* < 0

**Answers and Explanations for the Multiple-Choice Questions**

** 1 . D** —You may wish to review the Kinetics chapter if you have forgotten what the activation energy is.

** 2 . A** —The free energy is the minimum energy required for a nonspontaneous reaction and the maximum energy available for a spontaneous reaction.

** 3 . C** —This is a basic postulate of kinetic molecular theory.

** 4 . B** —This is the reverse of the lattice energy definition.

__5__ . D

Simple rounding to the nearest 100 kJ gives 200 kJ.

** 6 . D** —The system is insulated and no work can be done on or by the system (rigid container); thus, the energy is constant. At the melting point, some of the gallium will spontaneously melt; changing a solid to a liquid increases the entropy.

** 7 . A** —The process is endothermic (the ammonium chloride is absorbing heat to cool the water). Endothermic processes are “helped” by higher temperatures. Answer C and possibly D would give an increase in temperature. There is insufficient information about answer B.

** 8 . B** —The reaction showing the greatest increase in the number of moles of gas will show the greatest entropy increase. If no gases are present, then the greatest increase in the number of moles of liquid would yield the greatest increase.

** 9 . D** —If the reaction occurs readily, it must be spontaneous. Spontaneous reactions require Δ

*G*° < 0. A negative free energy leads to a large

*K*(>1).

** 10 . A** —Heat is required to melt something (Δ

*H*> 0). A transformation from a solid to a liquid gives an increase in entropy (Δ

*S*> 0).

** 11 . D** —This equation has an overall decrease in the amount of gas (high entropy) present. The other answers produce more gas (increases entropy).

** 12 . C** —Nonspontaneous means that Δ

*G*> 0. Since the reaction becomes spontaneous, the sign must change. Recalling: Δ

*G*= Δ

*H*–

*T*Δ

*S*. The sign change at higher temperature means that the entropy term (with Δ

*S*> 0) must become more negative than the enthalpy term (Δ

*H*> 0).

** 13 . C** —[2(436 kJ) + 499 kJ] – {2[2(464 kJ)]} = –485 kJ

** 14 . A** —The reaction that produces the most gas will have the greatest increase in entropy; the one losing the most gas would have the greatest decrease.

__15__ .

** 16 . B** —Dissolving almost always has: Δ

*S*> 0. A decrease in temperature means the process has: Δ

*H*> 0 (the system is absorbing energy from the surroundings).

**Free-Response Question**

You have 10 minutes to answer the following question. You may use a calculator and the tables in the back of the book.

**Question 1**

Xe(g) + 3F_{2} (g) XeF_{6} (g)

Under standard conditions, the enthalpy change for the reaction going from left to right (forward reaction) is Δ*H* ° = −294 kJ.

(a) Is the value of Δ*S* °, for the above reaction, positive or negative? Justify your conclusion.

(b) The above reaction is spontaneous under standard conditions. Predict what will happen to Δ*G* for this reaction as the temperature is increased. Justify your prediction.

(c) Will the value of *K* remain the same, increase, or decrease as the temperature increases? Justify your prediction.

(d) Show how the temperature at which the reaction changes from spontaneous to nonspontaneous can be predicted. What additional information is necessary?

**Answer and Explanation for the Free-Response Question**

(a) The value is negative. The decrease in the number of moles of gas during the reaction means there is a decrease in entropy.

Give yourself 1 point if you predicted this. Give yourself 1 point for discussing the number of moles of gas. You may get this point even if you did not get the first point.

(b) Recalling Δ*G* = Δ*H* – *T* Δ*S* (this equation is given on the equation page of the AP exam).

The value of Δ*G* will increase (become less negative).

Give yourself 1 point for this answer if it is obvious that increasing means less negative.

In general, both Δ*H* and Δ*S* are relatively constant with respect to small temperature changes. As the temperature increases, the value of the entropy term, *T* Δ*S* , becomes more negative. The negative sign in front of this term leads to a positive contribution. The value of Δ*G* will first become less negative (more positive), and eventually the value will be positive (no longer spontaneous).

Give yourself 1 point for the Δ*G* = Δ*H* – *T* Δ*S* argument even if you did not get the first point.

(c) Recalling Δ*G* = –*RT* ln *K* (this equation is given on the equation page of the AP exam).

The value of *K* will decrease.

You get 1 point for this answer.

As the value of Δ*G* increases (see part **b** ), the value of *K* will decrease.

You get 1 point for using Δ*G* = –*RT* ln *K* in your discussion.

If you got the justification for part **b** wrong, and you used the same argument here, you will not be penalized twice. You still get your point.

(d) Recalling Δ*G* = Δ*H* – *T* Δ*S*

Rearranging this equation to: *T* = (Δ*G* – Δ*H* )/Δ*S* will allow the temperature to be estimated.

This is worth 1 point. To do the calculation, the value of Δ*S* is necessary. Give yourself 1 point for this.

There are a total of 8 points possible. All of the mathematical relations presented in the answers are in the exam booklet.

Note that many students lose points on a question like this because their answers are too long. Keep your answers short and to the point, even if it appears that you have multiple pages available.

** Rapid Review**

- Thermodynamics is the study of heat and its transformations.
- Kinetic energy is energy of motion, while potential energy is stored energy.
- The common units of energy are the joule, J, and the calorie, cal.
- A calorimeter is used to measure the heat released or absorbed during a chemical or physical change. Know how a calorimeter works.
- The specific heat capacity is the amount of heat needed to change the temperature of 1 gram of a substance by 1 K, while the molar heat capacity is the heat capacity per mole.
- The heat lost by the system in calorimetry is equal to the heat gained by the surroundings.
- The specific heat (
*c*) of a solid can be calculated by: −(*c*_{solid}×*m*_{solid}× Δ*T*_{solid}) =*c*_{water}×*m*_{water}× Δ*T*_{water}or by*γ*=*cm*Δ*T*. - The First Law of Thermodynamics states that the total energy of the universe is constant. (Energy is neither created nor destroyed.)
- The Second Law of Thermodynamics states that all spontaneous processes move in a way that increases the entropy (disorder) of the universe.
- The enthalpy change, Δ
*H*, is equal to the heat lost or gained by the system under constant pressure conditions. - Δ
*H*values are associated with a specific reaction. If that reaction is reversed, the sign of Δ*H*changes. If one has to use a multiplier on the reaction, it must also be applied to the Δ*H*value. - The standard enthalpy of formation of a compound, Δ
*H*°, is the enthalpy change when 1 mol of the substance is formed from its elements and all substances are in their standard states._{f} - The standard enthalpy of formation of an element in its standard state is zero.
- Δ
*H*°_{rxn}= Σ Δ*H*° products − Σ Δ_{f}*H*° reactants. Know how to apply this equation._{f} - Δ
*H*°_{rxn}is usually negative for a spontaneous reaction. - Δ
*S*° = Σ*S*° products − Σ*S*° reactants. Know how to apply this equation. - Δ
*S*° is usually positive for a spontaneous reaction. - The Gibbs free energy is a thermodynamic quantity that relates the enthalpy and entropy, and is the best indicator for whether or not a reaction is spontaneous.
- If Δ
*G*° > 0 the reaction is not spontaneous; if Δ*G*° < 0, the reaction is spontaneous; and if Δ*G*° = 0, the reaction is at equilibrium. - Δ
*G*° = Σ Δ*G*° products − Σ Δ_{f}*G*° reactants. Know how to apply this equation._{f} - Δ
*G*° = Δ*H*°_{rxn}−*T*Δ*S*°_{rxn}. Know how to apply this equation. - For a system not at equilibrium: Δ
*G*= Δ*G*° +*RT*ln*Q*= Δ*G*° + 2.303*RT*log*Q*. Know how to apply this equation. - For a system at equilibrium: Δ
*G*° = −*RT*ln*K*= −2.303*RT*log*K*. Know how to apply this equation to calculate equilibrium constants.