Thermochemistry - Review - MCAT General Chemistry Review

MCAT General Chemistry Review

Part I Review

Chapter 6: Thermochemistry

Back in the “bad old days,” before people became aware of the environmental havoc wreaked by Styrofoam, we used to drink our coffee from Styrofoam cups. Now, of course, coffee is usually served in paper cups consisting of some percentage of post-consumer recycled material, enveloped in a thin cardboard sleeve (also typically made from recycled material) for added insulation against the hot contents. In those rare instances when coffee is served in Styrofoam, the bearer of the cup may receive a few withering glances from those especially passionate about environmentalism and at the very least will probably feel a little guilty for using it.

While Styrofoam might be bad for the environment, it is certainly a good insulator. That’s why it was the material of choice for disposable coffee cups for so long. It keeps hot things (like coffee) hot and cool things (like your hands—relative to the coffee) cool. The layer of Styrofoam protects your hands from too quickly absorbing the energy from the hot coffee and getting burned. Paper coffee cups just aren’t as effective at insulation as Styrofoam cups are; hence, the need for the cardboard sleeve as an extra layer of protection.

Styrofoam cups are such good insulators that they can be used as holding containers for certain calorimetry experiments. “Coffee-cup calorimetry,” which uses Styrofoam cups to measure heats of solution and specific heats of metals and other materials, is low-tech, yet it can produce remarkably accurate results, as long as care has been taken to calibrate the calorimeter and to minimize heat loss through the top. The next time you are at your favorite overpriced coffee chain, standing at the milk and sugar station, think about what you are doing when you mix the cold milk into the hot coffee. What you have there in your hand is really a little science experiment. If you took the time to measure the masses and temperatures of the hot coffee and the cold milk before mixing them, measured the drink’s temperature after you had stirred them up, and then looked up the specific heats for water and milk, you would have enough information to calculate the amount of heat exchanged between the hot coffee and the cold milk. (And if you really wanted to be accurate, you would include in your calculation the mass of the little wooden stirrer and its specific heat as well!) Of course, if you actually used the milk and coffee station as your lab bench, you may be putting your life at risk in the hands of the angry, under- and overcaffeinated mob standing behind you impatiently waiting their turn.

This chapter will review the basic principles of thermochemistry, which is the study of the energy changes that accompany chemical and physical processes. Starting from the first law of thermodynamics, which states that energy is never created or destroyed but at most simply changed from one form to another, we will analyze the calculations that are done to quantify the various changes or exchanges in energy as a system moves from some initial state to a final state. As we go along, we will define what is meant by system and surroundings, state functions, heat, enthalpy, entropy, and Gibbs free energy.

Systems and Processes


For whatever reason, students seem to have some anxiety over what, exactly, constitutes a system and what, by exclusion from the system, constitute the surroundings or environment. Perhaps the problem isn’t so much with the definitions, which are fairly straightforward, but the way in which the boundary between the two can be “moved” to suit the interests of the experimenter or observer. Simply put, the system is the matter that is being observed. It’s the total amount of reactants and products, say, in a chemical reaction. It’s the amount of solute and solvent used to create a solution. It could even be the gas inside a balloon. Then the surroundings, or environment, are everything outside of what you’re looking at. However, the boundary between system and surroundings is not permanently fixed, and it can be moved. For example, you might consider the mass of coffee in your coffee cup to be the system and the cup containing it to be the environment. If this is the way you set your boundary, then you’re probably interested in determining, say, the amount of heat transferred from the hot coffee to the cooler coffee cup. Alternatively, you might define the system as the hot coffee and the cup and the environment as the air surrounding the coffee cup. If this is the way you’ve defined your boundary, then you’re probably interested in the heat exchange between the hot coffee/cup system and the cooler surrounding air. The boundary can be extended out farther and farther, until ultimately the entire mass of the universe is included in the system, at which point there are no surroundings. Where you place the boundary is really a decision based on what matter you’re interested in studying.

Systems can be further characterized by whether or not they can exchange heat and/ or matter with the surroundings. A system may be characterized as follows:

Isolated—The system cannot exchange energy (heat and work) or matter with the surroundings; for example, an insulated bomb calorimeter.

Closed—The system can exchange energy (heat and work) but not matter with the surroundings; for example, a steam radiator.

Open—The system can exchange both energy (again, heat and work) and matter with the surroundings; for example, a pot of boiling water.

When a system experiences a change in one or more of its properties (such as concentration of reactant or product, temperature, or pressure), we say that it undergoes a process. While processes, by definition, are associated with changes of state of systems, some processes are uniquely identified by some property that is constant throughout the process. For example, isothermal processes occur when the system’s temperature is constant. Constant temperature implies that the total internal energy of the system is constant throughout the process. Adiabatic processes occur when no heat is exchanged between the system and the environment; thus, the heat content of the system is constant throughout the process. Finally, isobaric processes occur when the pressure of the system is constant. Isothermal and isobaric processes are common, because it is usually easy to control temperature and pressure.

Processes themselves can be also classified as spontaneous or nonspontaneous. A spontaneous process is one that can occur by itself without having to be driven by energy from an outside source. Calculating the change in the Gibbs free energy (imageG) for a process, such as a chemical reaction, allows us to predict whether the process will be spontaneous or nonspontaneous. The same quantities that are used to calculate the change in the Gibbs free energy, imageH and imageS, can also tell us whether or not the process will be temperature-dependent; that is, spontaneous at some temperatures and nonspontaneous at others (see the section on Gibbs free energy).

Spontaneous reactions, as you’ll recall from our discussion of chemical kinetics and equilibrium in Chapter 5, will not necessarily happen quickly and may not go to completion. Many spontaneous reactions have very high activation energies and, therefore, rarely actually take place. For example, when was the last time you saw a match ignite itself spontaneously? Or for that matter, when was the last time you or a loved one spontaneously combusted? However, provide an amount of thermal energy (generated by the friction associated with “striking the match”) that equals or exceeds the energy of activation, and the match will light and burn spontaneously. Combustion, the combination of the chemical components of the match with molecular oxygen in the air, will not need any additional external energy input in order to proceed once the energy of activation has been supplied. Some spontaneous reactions proceed very slowly. The role of enzymes, biological catalysts, is to selectively enhance the rate of certain spontaneous but slow chemical reactions so that the biologically necessary products can be formed at a rate sufficient for sustaining life. As we learned in Chapter 5, some reactions do not go to completion but settle into a low-energy state called equilibrium. Spontaneous reactions may go to completion, but many simply reach equilibrium with dynamically stable concentrations of reactants and products.

States and State Functions


Every year in January, the president of the United States speaks to the nation in a State of the Union address. In the speech, the president gives an assessment of the parameters by which we can measure the relative well-being of the nation and its citizens. There is usually discussion of war or peace; economic indicators; domestic programs; and other measurements of the social, political, and economic status of the country. These indicators help the people understand where they “are” as a nation in comparison to where they were the year prior, but less so how they got to where they are now.

Similarly, the state of a system is described by certain macroscopic properties of the system. These properties, or state functions, describe the system in an equilibrium state. They cannot describe the process of the system; that is, how the system got to its current equilibrium. They are useful only for comparing one equilibrium state to another. The pathway taken from one equilibrium state to another is described quantitatively by the process functions, the most important of which are mechanical work (W) and heat (Q).

The state functions include temperature (T), pressure (P), volume (V), density (image), internal energy (E or U), enthalpy (H), entropy (S), and Gibbs free energy (G). When the state of a system changes from one equilibrium to another, one or more of these state functions will change. In addition, while state functions are independent of the path (process) taken, they are not necessarily independent of one another. For example, Gibbs free energy is related to enthalpy, entropy, and temperature, as you will see.

Because systems can be in different equilibrium states at different temperatures and pressures, a set of standard conditions has been defined for measuring the enthalpy, entropy, and Gibbs free energy changes of a reaction. The standard conditions are defined as 25°C (298 K) and 1 atm. Don’t confuse standard conditions with standard temperature and pressure (STP), for which the temperature is 0°C (273 K) and 1 atm. This common MCAT Test Day mistake is easily made but also easily avoided. You’ll use standard conditions for thermodynamic problems of enthalpy or free energy, but you’ll use STP for ideal gas calculations.

MCAT Expertise

On the MCAT, be sure that you do not confuse standard conditions in thermodynamics with standard temperature and pressure (STP) in gas law calculations (see Chapter 7).

Under standard conditions, the most stable form of a substance is called the standard state of that substance. You should recognize the standard states for some elements and compounds commonly encountered on the test. For example, H2 (g), H2O (l), NaCl (s), O2 (g), and C (s) (graphite) are the most stable forms of these substances under standard conditions. Recognizing whether or not a substance is in its standard state is important for thermochemical calculations, such as heats of reactions and, in particular, the heat of formation. The changes in enthalpy, entropy, and free energy that occur when a reaction takes place under standard conditions are called the standard enthalpy, standard entropy, and standard free energy changes, respectively, and are symbolized by imageH°, imageS°, and imageG°.

The rest of this chapter will focus on the following state functions: enthalpy, entropy (energy dispersal), and free energy, with some concluding remarks on spontaneity.



Before we can examine the first of the four state functions that are the focus of this chapter, we must address the topic of heat, which is a source of some confusion for students. Perhaps the greatest barrier to a proper understanding of heat is the semantic conflation of the terms heat andtemperature. Many people use these terms interchangeably in everyday conversation. We might say, for example, that a midsummer afternoon was unbearably hot or that the temperature exceeded 100°F. Both convey the sense that the day was very, very warm, but what makes sense in everyday conversation needs to be clarified for a proper understanding of thermochemical principles. Temperature (T) is related to the average kinetic energy of the particles of the substance whose temperature is being measured. Temperature is the way that we scale how hot or cold something is (but not necessarily how hot or cold something feels to us: For reasons that will become clear in our discussion of specific heat, blacktop will feel hotter to our bare feet than a wooden boardwalk would, even if they are at the same temperature). We are familiar with a few temperature scales: Fahrenheit,Celsius, and Kelvin. The average kinetic energy of the particles in a substance is related to the thermal energy of the substance, but because we must also include consideration of how much substance is present to calculate total thermal energy content, the most we can say about temperature is that when a substance’s thermal energy increases, its temperature increases also. Nevertheless, we cannot say that something that is hot necessarily has greater thermal energy (in absolute terms) than a substance that is cold. For example, we might determine that a large amount of cool water has greatertotal heat content than a very small amount of very hot water.

Key Concept

Remember that heat and temperature are different. Heat is a specific form of energy that can enter or leave a system, while temperature is a measure of the average kinetic energy of the particles in a system.

Heat (Q) is the transfer of energy from one substance to another as a result of their difference in temperature. In fact, the zeroth law of thermodynamics implies that objects are in thermal equilibrium only when their temperatures are equal. Heat is therefore a process function, not a state function: We can quantify how much thermal energy is transferred between two or more objects as a result of their difference in temperatures by measuring the heat transferred.

The first law of thermodynamics states that the change in the total internal energy (imageU) of a system is equal to the amount of heat (thermal energy) transferred (Q) to the system minus the amount of work (W) (another form of energy transfer by the application of force through displacement) done by the system. This can be expressed mathematically as follows:

imageU = Q - W

Because heat and work are measured independently, we can assess the transfer of energy in the form of heat through any process regardless of the work done (or not done). Processes in which the system absorbs heat are called endothermic (+imageQ), while those processes in which the system releases heat are called exothermic (-imageQ). The unit of heat is the unit of energy: joule (J) or calorie (c) for which 1 c = 4.184 J.

Key Concept

Endothermic: positive imageH
Exothermic: negative imageH

You know how people who live in really hot desert climates, when asked, “How can you stand the heat?” often respond by pointing out, “But it’s a dry heat!” Well, there’s some truth to that. One of the most important ways that the body works to prevent overheating is through the production of sweat—that exocrine secretion of water, electrolytes, and urea. However, it’s not the production of sweat, per se, that is the cooling mechanism. It’s the evaporation of the sweat that helps cool the body. Evaporation (vaporization) from the liquid to gas phase is an endothermic process: Energy must be absorbed for the particles of the liquid to gain enough kinetic energy to escape into the gas phase. So the sweat that is excreted onto the skin must absorb energy in order to evaporate. Where does that necessary energy come from? It comes from the body itself. Hot, arid desert air has lowerpartial pressure of water vapor than humid tropical air, so sweat vaporizes more readily in the dry air than it does in the humid air. Although it might be hard to believe that any temperature in excess of 100°F could ever be considered comfortable, it probably is true that most people will feel more comfortable in “dry heat” than in “tropical heat.”

When substances of different temperatures are brought into thermal contact with each other (that is, some physical arrangement that allows for the transfer of heat energy), energy—in the form of heat—will transfer from the warmer substance to the cooler substance. When a substance undergoes a chemical reaction that is exothermic or endothermic, heat energy will be exchanged between the system and the environment. The process of measuring transferred heat is called calorimetry. Two basic types of calorimetry that you should know and understand for Test Day include constant-pressure calorimetry and constant-volume calorimetry. The coffee-cup calorimeter, introduced at the beginning of this lesson, is a low-tech example of a constant-pressure calorimeter, while the bomb calorimeter is an example of a constant-volume calorimeter. Constant-pressure and constant-volume are terms used to describe the conditions under which the heat changes are measured.

The heat (q) absorbed or released in a given process is calculated from this equation:

q = mcimageT

where m is the mass, c is the specific heat of the substance, and imageT is the change in temperature (in either Celsius or Kelvin). Specific heat is the amount of energy required to raise the temperature of one gram or kilogram of a substance by one degree Celsius or one unit Kelvin. Specific heat values will be provided for you on Test Day, but one value that you need to remember is the specific heat of H2O (l): one calorie per gram per Celsius degree (1 c/g°C). Speaking of easy to remember, you ought not have any problem remembering this equation, given that q = mcimageT looks like “qequals MCAT.”


Ever notice how the equation for heat spells out MCAT? Call it the Calorimetry Test if it helps you remember the equation!

Constant-Volume Calorimetry

To picture the setup of a constant-pressure calorimeter, just think of the coffee-cup calorimeter: an insulated container covered with a lid and filled with a solution in which a reaction or some physical process, such as dissolution, is occurring. The pressure, which is just the atmospheric pressure, remains constant throughout the process. The setup of a constant-volume calorimeter is perhaps a little less familiar to you or harder to picture. Furthermore, the term bomb calorimeter sounds rather ominous. Well, don’t worry—this isn’t some relic from the Cold War. Perhaps “bomb” is a bit misleading: A more accurately descriptive term is “decomposition vessel.” This better reflects what is actually taking place. A sample of matter, typically a hydrocarbon, is placed in the steel decomposition vessel, which is then filled with almost pure O2 gas. The decomposition vessel is then placed in an insulated container holding a known mass of water. The contents of the decomposition vessel are ignited by an electric ignition mechanism. The material combusts (burns) in the presence of the oxygen, and the heat that is evolved in the combustion is the heat of the reaction. Because W = image, no work is done in an isovolumic (imageV = 0) process, so Wcalorimeter = 0. Furthermore, because of the insulation, the whole calorimeter can be considered isolated from the rest of the universe, so we can identify the “system” as the sample plus oxygen and steel and the surroundings as the water. Because no heat is exchanged between the calorimeter and the rest of the universe, Qcalorimeter is 0. So imageUsystem + imageUsurroundings = imageUcalorimeter = Qcalorimeter-Wcalorimeter = 0. Therefore, imageUsystem = -imageUsurroundings. Because no work is done , qsystem = -qsurroundings, and msteelcsteelimageT+ moxygencoxygenimageT = -mwatercwaterimageT.

MCAT Expertise

We need a different formula to calculate q during phase changes, when imageT = 0. If we used q = mcimageT, we’d erroneously think q= 0.

Note that by using the layer of insulation to isolate the entire calorimeter from the rest of the universe, we’ve created an adiabatic process. This means that no heat is exchanged between the calorimeter and the rest of the universe, but it is exchanged between the steel decomposition vessel and the surrounding water. As the previous derivation shows, heat exchange between the system and its surroundings makes it possible for us to calculate the heat of the combustion.



Most reactions in the lab occur under constant pressure (at 1 atm) in closed thermodynamic systems. To express heat changes at constant pressure, chemists use the term enthalpy (H). Enthalpy is a state function, so we can calculate the change in enthalpy (imageH) for a system that has undergone a process—for example, a chemical reaction—by comparing the enthalpy of the final state to the enthalpy of the initial state, irrespective of the path taken. The change in enthalpy is equal to the heat transferred into or out of the system at constant pressure. To find the enthalpy change of a reaction, imageHrxn, you must subtract the enthalpy of the reactants from the enthalpy of the products:

imageHrxn = Hproducts-Hreactants

A positive imageHrxn corresponds to an endothermic process, and a negative imageHrxn corresponds to an exothermic process. It is not possible to measure enthalpy directly; only imageH can be measured, and only for certain fast and spontaneous processes. Several standard methods have been developed to calculate imageH for any process.

Standard Heat of Formation

The standard enthalpy of formation of a compound, imageH°f, is the enthalpy change that would occur if one mole of a compound in its standard state were formed directly from its elements in their respective standard states. Remember that standard state is the most stable physical state of an element or compound at 298 K and 1 atm. Note that imageH°f of an element in its standard state, by definition, is zero. The imageH°f ’s of most known substances are tabulated. You do not need to memorize these values, as they will be provided for you as necessary.

Standard Heat of Reaction

The standard heat of a reaction, imageH°rxn, is the hypothetical enthalpy change that would occur if the reaction were carried out under standard conditions. What this means is that all reactants must be in their standard states and all products must be in their standard states. This can be calculated by taking the difference between the sum of the standard heats of formation for the products and the sum of the standard heats of formation of the reactants:

imageH°rxn = Σ(imageH°f of products)-Σ(imageH°f of reactants)

Key Concept

State functions are path-independent. Always.

Hess’s Law

Enthalpy is a state function and is a property of the equilibrium state, so the pathway taken for a process is irrelevant to the change in enthalpy from one equilibrium state to another. As a consequence of this, Hess’s law states that enthalpy changes of reactions are additive. When thermochemical equations (chemical equations for which energy changes are known) are added to give the net equation for a reaction, the corresponding heats of reaction are also added to give the net heat of reaction. You can think of Hess’s law as being embodied in the enthalpy equations we’ve already introduced. For example, we can describe any reaction as the result of breaking down the reactants into their component elements, then forming the products from these elements. The enthalpy change for the reverse of any reaction has the same magnitude, but the opposite sign, as the enthalpy change for the forward reaction. Therefore,

imageH (reactants image elements) = - imageH (elements image reactants)

The imageHrxn written as

imageHrxn = imageH (reactants image elements) + imageH (elements image products)


imageHrxn = - imageH (elements image reactants) + imageH (elements image products) = imageH (elements image products) - imageH (elements image reactants)


imageH°rxn = Σ(imageH°f of products)- Σ(imageH°f of reactants)

Consider the following phase change:

Br2 (l) image Br2 (g) imageH°rxn = 31 kJ/mol

The enthalpy change for the phase change is called the heat of vaporization (imageH°vap). As long as the initial and final states exist at standard conditions, the imageH°rxn will always equal the imageH°vap, irrespective of the particular pathway that the process takes in vaporization. For example, it’s possible that Br2 (l) could first decompose to Br atoms, which then recombine to form Br2 (g), giving the following reaction mechanism. However, because the net reaction is the same as the one shown previously, the change in enthalpy will be the same.

Br2 (l) image Br2 (g) imageH = (31 kJ/mol)(1 mol) = 31 kJ

Example: Given the following thermochemical equations:


C3H8 (g) + 5 O2 (g) image 3 CO2 (g) + 4 H2O (l)

imageHa = -2,220.1 kJ


C (graphite) + O2 (g) image CO2 (g)

imageHb = -393.5 kJ


H2 (g) + 1/2 O2 (g) image H2O (l)

imageHc = -285.8 kJ

Calculate imageH for this reaction:

d) 3 C (graphite) + 4 H2 (g) image C3H8 (g)

Solution: Equations a, b, and c must be combined to obtain equation d. Since equation d contains only C, H2, and C3H8, we must eliminate O2, CO2, and H2O from the first three equations. Equation a is reversed to get C3H8 on the product side (this gives equation e). Next, equation b is multiplied by 3 (this gives equation f) and c by 4 (this gives equation g). The following addition is done to obtain the required equation d: 3b + 4c + e.

e) 3 CO2 (g) + 4 H2O (l) image C3H8 (g) + 5 O2 (g) imageHe = 2,220.1 kJ

f) 3 × [C ( graphite) + O2 ( g) image CO2 ( g)] imageHf = 3 × -393.5 kJ

g) 4 × [H2 ( g) + 1/2 O2 (g) image H2O (l )] imageHg = 4 × -285.8 kJ

3 C ( graphite) + 4 H2 (g) image C3 H8 ( g) imageHd = -103.6 kJ where imageHd = imageHe + imageHf + imageHg.

MCAT Expertise

When doing a problem like this on the MCAT, make sure to switch signs when you reverse the equation. Also, make sure to multiply by the correct stoichiometric coefficients when performing your calculations.

It is important to realize that Hess’s law applies to any state function, including entropy and Gibbs free energy.

Bond Dissociation Energy

Hess’s law can also be expressed in terms of bond enthalpies, also called bond dissociation energies, when these are given. Bond dissociation energy is the average energy that is required to break a particular type of bond between atoms in the gas phase (remember, bond dissociation is an endothermic process). Bond dissociation energy is given as kJ/mol of bonds broken. For example, the bond enthalpy of the double bond in the diatomic oxygen molecule O2 is 498 kJ/mol (of double bonds broken). The tabulated bond enthalpies for bonds found in compounds other than diatomic molecules are the average of the bond energies for the bonds in many different compounds. For example, the C–H bond enthalpy (415 kJ/mol) that you will find listed in a table of bond enthalpies is averaged from measurements of the individual C–H bond enthalpies of thousands of different organic compounds. Please note that bond formation, the opposite of bond breaking, has the same magnitude of energy but is negative rather than positive; that is, energy is released when bonds are formed. The enthalpy change associated with a reaction is given by

imageHrxn = ΣimageHbonds broken + ΣimageHbonds formed = total energy absorbed - total energy released

For example,

H2 (g) image 2H (g) imageH = 436 kJ/mol

Key Concept

Because it takes energy to pull two atoms apart, bond breakage is always endothermic. The reverse process, bond formation, must always be exothermic.

In this decomposition reaction, diatomic hydrogen gas is cleaved to produce mono-atomic hydrogen gas. For each mole of H2 cleaved, 436 kJ of energy is absorbed by the system in order to overcome the bonding force. Since energy is absorbed, the bond-breaking reaction is endothermic.

If you have a hard time remembering whether bond formation or dissociation is endothermic, think about bonds as if they were two bar magnets stuck together. You have to exert pulling forces (invest energy) to pull apart two bar magnets (endothermic). On the other hand, the two bar magnets, once separated, “want” to come back together because they exert an attractive force between their opposite poles. Allowing them to stick together reduces their potential energy (exothermic).

Example: Calculate the enthalpy change for the following reaction:

C(s) + 2 H2(g) image CH4(g) imageH = ?

Bond dissociation energies of H–H and C–H bonds are 436 kJ/mol and 415 kJ/mol, respectively.

imageHf of C( g) = 715 kJ/mol

Solution: CH4 is formed from free elements in their standard states (C in solid and H2 in gaseous state).

Thus here, imageHrxn = imageHf.

The reaction can be written in three steps:

a) C (s) image C ( g)


b) 2 [H2 (g) image 2 H (g)]


c) C (g) + 4 H (g) image CH4 (g)


and imageHf = [imageH1 + 2 imageH2] + [imageH3].

imageH1 = imageHf C ( g) = 715 kJ/mol

imageH2 is the energy required to break the H–H bond of one mole of H2. So,

imageH2 = bond energy of H2 = 436 kJ/mol

imageH3 is the energy released when 4 C–H bonds are formed. So,

imageH3 = -(4 × bond energy of C–H) = -(4 × 415 kJ/mol) = -1,660 kJ/mol

(Note: Because energy is released when bonds are formed, imageH3 is negative.)


imageHrxn = imageHf = [715 + 2(436)] - (1,660) kJ/mol = -73 kJ/mol

MCAT Expertise

With practice, you’ll become accustomed to the patterns and shortcuts that will make things easier on Test Day.

Key Concept

The larger the alkane reactant, the more numerous the combustion products.

Heats of Combustion

One more type of standard enthalpy change that you should be aware of for the MCAT is the standard heat of combustion, imageH°comb. Because measurements of enthalpy change require a reaction to be spontaneous and fast, combustion reactions are the ideal process for such measurements. Most combustion reactions presented on the MCAT occur in the presence of O2 in the atmosphere, but keep in mind that there are other combustion reactions in which oxygen is not the oxidant. Diatomic fluorine, for example, can be used as an oxidant, and hydrogen gas will combust with chlorine gas to form gaseous hydrochloric acid and, in the process, will evolve a large amount of heat and light characteristic of combustion reactions. The reactions listed in the C3H8 (g) example shown earlier are combustion reactions with O2 (g) as the oxidant. Therefore, the enthalpy change listed for each of the three reactions is the imageHcomb for each of the reactions.



Many, many students are genuinely perplexed by the concept of entropy. Enthalpy makes, perhaps, intuitive sense, especially when the energy change from reactants to products is large, fast, and dramatic (as in combustion reactions involving explosions). Entropy seems to be less intuitive. Except that it isn’t. In fact, our understanding of what constitutes normal life experience and even the passage of time is based, fundamentally, upon the property of entropy. Consider, for example, how “normal” each of the following seems to you: hot tea cools down, frozen drinks melt, iron rusts, buildings crumble, balloons deflate, living things die, and so on.

Key Concept

Entropy changes that accompany phase changes (see Chapter 8) can be easily estimated, at least qualitatively. For example, freezing is accompanied by a decrease in entropy, as the relatively disordered liquid becomes a well-ordered solid. Meanwhile, boiling is accompanied by a large increase in entropy, as the liquid becomes a much more disordered gas. For any substance, sublimation will be the phase transition with the greatest entropy change.

These examples have a common denominator: In all of them, energy of some form or another is going from being localized or concentrated to being spread out or dispersed. The thermal energy in the hot tea is spreading out to the cooler air that surrounds it. The thermal energy in the warmer air is spreading out to the cooler frozen drink. The chemical energy in the bonds of elemental iron and oxygen is released and dispersed as a result of the formation of the more stable (lower-energy) bonds of iron oxide (rust). The potential energy of the building is released and dispersed in the form of light, sound, and heat (motional energy) of the ground and air as the building crumbles and falls. The motional energy of the pressurized air is released to the surrounding atmosphere as the balloon deflates. The chemical energy of all the molecules and atoms in living flesh is released into the environment during the process of death and decay.

This is the second law of thermodynamics: Energy spontaneously disperses from being localized to becoming spread out if it is not hindered from doing so. Pay attention to this: The usual way of thinking about entropy as “disorder” must not be taken too literally, a trap that many students fall into. Be very careful in thinking about entropy as disorder. The old analogy between a messy (disordered) room and entropy is arguably deficient and may not only hinder understanding but actually increase confusion.

Entropy, then, according to statistical mechanics, is the measure of the spontaneous dispersal of energy at a specific temperature: how much energy is spread out, or how widely spread out energy becomes, in a process. The equation for calculating the change in entropy is


where imageS is the change in entropy, Qrev is the heat that is gained or lost in a reversible process (a process that proceeds with infinitesimal changes in the system), and T is the temperature in Kelvin. The units of entropy are usually kJ/mol • K.When energy is distributed into a system at a given temperature, its entropy increases. When energy is distributed out of a system at a given temperature, its entropy decreases.

Notice that the second law states that energy will spontaneously disperse; it does not say that energy can never be localized or concentrated. However, the concentration of energy will rarely, if ever, happen spontaneously in a closed system (there is an exceedingly small but measurable chance that it could, but this is beyond the scope of the MCAT). Work usually must be done to concentrate energy. For example, refrigerators move thermal energy against a temperature gradient (that is, they cause heat to be transferred from “cool” to “warm”), thereby “concentrating” energy outside of the system in the surroundings. Nevertheless, refrigerators consume a lot of energy (they do a lot of work) to accomplish this movement of energy against the temperature gradient.


Figure 6.1

The second law has been described as “time’s arrow,” because there is a unidirectional limitation on the movement of energy by which we recognize “before and after” or “new and old.” For example, you would instantly recognize whether a video recording of an explosion was running forward or backward. This is what is meant by the phrase “time’s arrow.” Another way of understanding this is to say that energy in a closed system will spontaneously spread out and entropy will increase if it is not hindered from doing so. Remember that a system can be variably defined to include ultimately the entire universe; in fact, the second law ultimately claims that the entropy of the universe is increasing. That is to say, energy concentrations at any and all locations in the universe are in the process of becoming distributed and spread out.

imageSuni verse = imageSsystem + imageS surroundings > 0

Entropy is a state function, so a change in entropy from one equilibrium state to another is pathway-independent and only depends upon the difference in entropies of the final and initial states:

imageS = Sfinal-Sinitial

The standard entropy change for a reaction, imageS°rxn, is calculated using the standard entropies of reactants and products:

imageS°rxn = ΣimageS°products - ΣimageS°reactants

Gibbs Free Energy


The final state function that we will examine in this chapter is Gibbs free energy, G. Now, you can be sure that Josiah Gibbs isn’t giving it away for free. Actually, given that the man died way back in 1903, it’s unlikely that he’s got much more energy to give away, anyway. This state function is a combination of the two that we’ve just examined: enthalpy and entropy. The change in Gibbs free energy, imageG, is a measure of the change in the enthalpy and the change in entropy as a system undergoes a process, and it indicates whether a reaction is spontaneous or nonspontaneous. The change in the free energy is the maximum amount of energy released by a process, occurring at constant temperature and pressure, that is available to perform useful work. The change in Gibbs free energy is defined as follows:

imageG = imageH-image

where T is the temperature in Kelvin and image represents the total amount of energy that is absorbed by a system when its entropy increases reversibly.


image Get High Test Scores!

A helpful visual aid for conceptualizing Gibbs free energy is to think of it as a valley between two hills. Just as a ball would tend to roll down the hill into the valley and eventually come to rest at the lowest point in the valley (as long as nothing prevents it from doing so), any system, including chemical reactions, will move in whatever direction results in a reduction of the free energy of the system. The bottom of the valley, the lowest point, represents equilibrium, and the sides of the hill represent the various points in the pathway toward or away from equilibrium. Movement toward the equilibrium position is associated with a decrease in Gibbs free energy (-imageG) and is spontaneous, while movement away from the equilibrium position is associated with an increase in Gibbs free energy (+imageG) and is nonspontaneous. You wouldn’t quite believe your eyes if, all of a sudden, the ball at the bottom of the valley began to roll up the hill with no assistance. Once at the energy minimum state, the position of equilibrium (the bottom of the valley), the system will resist any changes to its state, and the change in free energy is zero for all systems at equilibrium. To summarize:

1. If imageG is negative, the reaction is spontaneous.

2. If imageG is positive, the reaction is nonspontaneous.

3. If imageG is zero, the system is in a state of equilibrium; thus imageH = image.

Key Concept

Recall that thermodynamics and kinetics are separate topics. When a reaction is thermodynamically spontaneous, it has no bearing on how fast it goes. It only means that it will proceed eventually without external energy input.

Because the temperature in Gibbs free energy is in units of Kelvin, it is always positive. Therefore, the effects of the signs on imageH and imageS and the effect of temperature on the spontaneity of a process can be summarized as follows:






spontaneous at all temperatures



nonspontaneous at all temperatures



spontaneous only at high temperatures



spontaneous only at low temperatures

Key Concept

imageG is temperature-dependent when imageH and imageS have the same sign.

Phase changes are examples of temperature-dependent processes. The phase changes of water should be familiar to you. Have you ever wondered why water doesn’t boil at, say, 20°C instead of 100°C? Well, first of all, be thankful that it doesn’t because 20°C is room temperature. It’d be hard to keep a glass of water, not to mention you, around for very long. (Water makes up almost 99 percent of the total number of molecules in the human body. Excluding adipose tissue, the human body is about 75 percent water by mass!) So we all know that water boils at 100°C, but why? The answer lies in phase transformation:

H2O (l) image H2O (g) imageHvap = 40.65 kJ/mol

When water boils, hydrogen bonds (H-bonds) are broken, and the water molecules gain sufficient potential energy to escape into the gas phase. Thus, boiling (vaporization) is an endothermic process, and imageH is positive. As thermal energy is transferred to the water molecules, energy is distributed through the molecules entering the gas phase, and entropy is positive and image is positive. Both imageH and image are positive, so the reaction will be spontaneous only if image is greater than imageH, giving a negative imageG. These conditions are met only when the temperature of the system is greater than 373 K (100°C). Below 100°C, the free energy change is positive, and boiling is nonspontaneous; the water remains a liquid. At 100°C, imageH = image and imageG = 0; equilibrium between the liquid and gas phases is established in such a way that the water’s vapor pressure equals the ambient pressure. This is the definition of the boiling point: the temperature at which the vapor pressure equals the ambient pressure.

It is very important to remember that the rate of a reaction depends on the activation energy Ea, not the imageG. Spontaneous reactions may be fast or slow. Sometimes a reversible reaction may produce two products that differ both in their stability, as measured by the change in the Gibbs free energy associated with their production, and in their kinetics, as measured by their respective energies of activation. Sometimes the thermodynamically more stable product will have the slower kinetics due to higher activation energy. In this situation, we talk about kinetic versus thermodynamic reaction control. For a period of time after the reaction begins, the “dominant” product—that is, the major product—will be the one that is produced more quickly as a result of its lower energy of activation. The reaction can be said to be under kinetic control at this time. Given enough time, however, and assuming a reversible process, the dominant product will be the thermodynamically more stable product as a result of its lower free energy value. The reaction can then be said to be under thermodynamic control. Eventually, the reaction will reach its equilibrium, as defined by its Keq expression.

A quick illustration may help to make this distinction clearer. Cats are famous for their ability to find the warmest spot in a house to take a nap. (Frankly, we find the word nap a bit lacking in its descriptive power as a reference to the 16+ hours per day that the typical house cat sleeps!) Imagine a cat wandering through a house on a sunny but cold winter day. She scurries up the staircase from the basement to the first floor and discovers a luscious patch of sunlight and basks in the warmth. She knows that there’s something even better up the next flight of stairs on the second floor: a roaring fire in the big brick fireplace. But she’s so tired and the sunlight is warm enough for now. So she lies down and sleeps for a little while. Eventually, though, the thought of all that crackling heat is too much to resist. So, she gets up, stretches, scurries up the second flight of stairs (Oh—so much energy to get up these stairs! she complains.) and nestles down in front of the fireplace. Sighing the sigh of pure contentment, she resolves never to leave this spot.

Standard Gibbs Free Energy

By now, it shouldn’t surprise you to learn that the free energy change of reactions can be measured under standard state conditions to yield the standard free energy, imageG°rxn. For standard free energy determinations, the concentrations of any solutions in the reaction are 1 M. The standard free energy of formation of a compound, imageG°f , is the free energy change that occurs when 1 mole of a compound in its standard state is produced from its respective elements in their standard states under standard state conditions. The standard free energy of formation for any element in its most stable form under standard state conditions (and therefore already in its standard state) is, by definition, zero. The standard free energy of a reaction, imageG°rxn, is the free energy change that occurs when that reaction is carried out under standard state conditions; that is, when the reactants in their standard states are converted to the products in their standard states, at standard conditions of temperature (298 K) and pressure (1 atm). For example, under standard state conditions, conversion of carbon in the form of diamond to carbon in the form of graphite is spontaneous (because graphite is the standard state for carbon). However, the reaction rate is so slow that the conversion is never actually observed. You can imagine the distress caused to brides around the world if suddenly all those very expensive diamonds turned into the form of carbon used in pencils.

imageG°rxn = Σ(imageG°f of products) - Σ(imageG°f of reactants)

MCAT Expertise

To make things easy for Test Day: Note the similarity of this equation to Hess’s law. Almost any state function could be substituted for imageG here.

Free Energy, Keqand Q

We can derive the standard free energy change for a reaction from the equilibrium constant Keq for this reaction:

imageG°rxn = -RT ln Keq

where Keq is the equilibrium constant, R is the gas constant, and T is the temperature in K. This is a very valuable equation to know and understand for the MCAT, as it allows you to make not only quantitative evaluations of the free energy change of a reaction that goes from standard state concentrations of reactants to equilibrium concentrations of reactants and products, but also qualitative assessment of the spontaneity of the reaction. The greater the value of Keq is, the more positive the value of its natural log. The more positive the natural log, the more negative the standard free energy change. The more negative the standard free energy change, the more spontaneous the reaction.

Once a reaction begins, however, the standard state conditions (i.e., 1 M solutions) no longer hold. The value of the equilibrium constant must be replaced with another number that is reflective of where the reaction is in its path toward equilibrium. To determine the free energy change for a reaction that is in progress, we relate imageGrxn (not imageG°rxn) to the reaction quotient Q. We reviewed Q in Chapter 5, Chemical Kinetics and Equilibrium. As a reminder,


Key Concept

Note that the right side of this equation is the same as that for Keq. The use of the Q rather than K indicates that the system is not at equilibrium.

And the relation between imageGrxn and Q is given as follows:

imageGrxn = imageG°rxn + RT lnQ

where R is the gas constant and T is the temperature in Kelvin. This can be simplified to


This reaction is very useful for quick qualitative assessments of Test Day problems in which you are asked to determine the spontaneity of a reaction at some point in the reaction process. By calculating the value of the reaction quotient and then comparing that value to the known equilibrium constant for the reaction, you will be able to predict whether the free energy change for the reaction is positive (giving a nonspontaneous reaction) or negative (giving a spontaneous reaction). That is, if the ratio of Q/Keq is less than one (Q < Keq) , then the natural log will be negative and the free energy change will be negative, so the reaction will spontaneously proceed forward until equilibrium is reached. If the ratio of Q/Keq is greater than one (Q > Keq) , then the natural log will be positive, and the free energy change will be positive. In that case, the reaction will spontaneously move in the reverse direction until equilibrium is reached. Of course, if the ratio is equal to one, then that means that the reaction quotient is equal to the equilibrium constant; the reaction is at equilibrium, and by definition, the free energy change is zero (the natural log of 1 is 0).


We began our discussion of thermochemistry with a review of different ways in which we characterize systems (isolated, closed, etc.) and processes (adiabatic, isothermal, etc.). We then further classified systems according to their state functions: system properties such as volume, pressure, temperature, enthalpy, entropy, and Gibbs free energy that describe the equilibrium state. We defined enthalpy as the heat content of the system and the change in enthalpy as the change in heat content of the system from one equilibrium state to another. Enthalpy is characterized as the energy found in the intermolecular interactions and in the bonds of the compounds in the system. We explored the various ways Hess’s law can be applied to calculate the total enthalpy change for a series of reactions. Moving on to entropy, we described this property as a measure of the degree to which energy in a system becomes spread out through a process. There is danger in thinking too literally about entropy as “disorder,” as a system’s entropy may be increasing even if there is no observable change in the system’s macroscopic disorder (e.g., ice warming from -10°C to -5°C). Gibbs free energy combines enthalpy and entropy considerations, and the change in Gibbs function determines whether a process will be spontaneous or nonspontaneous. When the change in Gibbs function is negative, the process is spontaneous, but when the change in Gibbs function is positive, the process is nonspontaneous.

Great job! This is a difficult material, but your focus and attention to these topics will pay off in points on Test Day. The changes in energy that accompany chemical processes are tested heavily on the MCAT.



image Systems can be characterized as isolated (no heat or matter exchange), closed (no matter exchange), or open (both heat and matter exchange possible).

image Some processes are identified by some constant property of the system: isothermal (constant internal energy/temperature), adiabatic (no heat exchange), and isobaric (constant pressure).

image State functions are physical properties of a system that describe the equilibrium state; as such, changes in state functions are pathway-independent. Some examples of state functions are temperature, volume, pressure, internal energy, enthalpy, entropy, and Gibbs free energy.

image Standard state of a substance is the most stable form (phase) of the substance under standard state conditions (298 K and 1 atm). Standard enthalpy, standard entropy, and standard free energy changes are measured under standard state conditions.

image Heat and temperature are not the same thing. Temperature is a scale related to the average kinetic energy of the molecules in a substance. Heat is the transfer of energy that results from two objects at different temperatures being put in thermal contact with each other. Heat energy transferred from one substance to another is measured by the methods of calorimetry.

image Enthalpy is a measure of the potential energy of a system found in the intermolecular interactions and bonds of molecules. Hess’s law states that the total change in the potential energy of a system is equal to the changes in potential energies of the individual steps (reactions) of the process.

image Entropy is a measure of the degree to which energy has been spread out through a system or between a system and its surroundings. It is a ratio of heat transferred per unit Kelvin. Systems reach maximum possible entropy (maximum possible energy dispersal) only at equilibrium.

image Gibbs free energy is a calculation involving both enthalpy and entropy values for a system. The change in Gibbs function determines whether a process is spontaneous or nonspontaneous. When the change in Gibbs function is negative, the process is spontaneous, but when the change in Gibbs function is positive, the process is nonspontaneous.

image Temperature-dependent processes are those that change between spontaneous and nonspontaneous at a certain temperature. For example, water freezes spontaneously only at temperatures below 273 K and boils spontaneously only at temperatures above 373 K.

image The larger and more positive the value of the equilibrium constant is, the more spontaneous a reaction will be (that is, the more negative the change in Gibbs function for the system as it moves from its initial position to its equilibrium).



Practice Questions

1. Consider the cooling of an ideal gas in a closed system. This process is illustrated in the pressure-volume graph shown in Figure 1.


Figure 1

Based on the information in Figure 1, the process is

A. adiabatic.

B. isobaric.

C. isothermal.

D. Two of the above

2. A reaction has a positive entropy and enthalpy. What can be inferred about the progress of this reaction from this informationimage

A. The reaction is spontaneous.

B. The reaction is nonspontaneous.

C. The reaction is at equilibrium.

D. There is not enough information to determine if the reaction is spontaneous.

3. Pure sodium metal spontaneously combusts upon contact with room temperature water. What is true about the equilibrium constant of this combustion reaction at 25°Cimage

A. Keq < 1.

B. Keq > 1.

C. Keq = 1.

D. There is not enough information to determine the equilibrium constant.

4. Which of the following processes has the most exothermic heat of reactionimage

A. Combustion of ethane

B. Combustion of propane

C. Combustion of n-butane

D. Combustion of isobutane

5. Methanol reacts with acetic acid to form methyl acetate and water, as shown in the following table in the presence of an acid catalyst.

CH3OH (image) + CH3COOH (aq) image CH3COOCH3 (aq) + H2O (image)

Type of Bond

Bond Disassociation Energy (kJ/mol)





C = O






What is the heat of formation of methyl acetate in kJ/molimage

A. 0 kJ/mol

B. 464 kJ/mol

C. 824 kJ/mol

D. 1,288 kJ/mol

6. At standard temperature and pressure, a chemical process is at equilibrium. What is the free energy of reaction (imageG) for this process?

A. imageG > 0.

B. imageG < 0.

C. imageG = 0.

D. There is not enough information to determine the free energy of reaction.

7. For a certain chemical process, imageG° = -4.955 kJ/mol. What is the equilibrium constant Keq for this reaction?

A. Keq = 0.13

B. Keq = 7.4

C. Keq = 8.9

D. Keq = 100

8. Consider the chemical reaction in the vessel depicted in the following diagram.


A. The reaction is spontaneous.

B. The reaction is nonspontaneous.

C. The reaction is at equilibrium.

D. There is not enough information to determine if the reaction is spontaneous.

Suppose imageGrxn° = -2,000 kJ/mol for a chemical reaction. At 300 K, what is the change in Gibb’s free energy, imageGimage

A. imageG = -2,000 kJ/mol + (300 K) (8.314 Jmol-1K-1)ln(Q).

B. imageG = -2,000 kJ/mol - (300 K) (8.314 Jmol-1K-1)ln(Q).

C. imageG = -2,000 kJ/mol + (300 K) (8.314 Jmol-1K-1)log(Q).

D. imageG = -2,000 kJ/mol - (300 K) (8.314 Jmol-1K-1)log(Q).

10. An ideal gas undergoes a reversible expansion at constant pressure. Which of the following terms could describe this expansion?

I. Adiabatic

II. Isothermal

III. Isobaric

A. I only

B. I and II only

C. I and III only

D. I, II, and III

11. A chemical reaction has a negative enthalpy and a negative entropy. Which of the following terms describes the energy of this reaction?

A. Exothermic

B. Endothermic

C. Endergonic

D. Exergonic

12. Consider the chemical reaction in the vessel pictured in the following figure.


What can we say about the entropy of this reaction?

A. imageS > 0

B. imageS < 0

C. imageS = 0

D. There is not enough information in the picture to determine imageS.

13. Which of the following statements is true of a spontaneous process?

A. imageG > 0 and Keq > Q

B. imageG > 0 and Keq < Q

C. imageG < 0 and Keq > Q

D. imageG < 0 and Keq < Q

14. Which of the following devices would be the most appropriate to use to measure the heat capacity of a liquidimage

A. Thermometer

B. Calorimeter

C. Barometer

D. Volumetric flask

15. Which of the following equations does not state a law of thermodynamics?

A. imageE system + imageE surroundings = imageE universe.

B. imageS system + imageS surroundings = imageS universe.

C. imageHsystem + imageHsurroundings = imageHuniverse.

D. Suniverse = 0 at T = 0 K.

16. A reaction coordinate for a chemical reaction is displayed in the graph below.


Which of the following terms describes the energy of this reaction?

A. Endothermic

B. Exothermic

C. Endergonic

D. Exergonic

Small Group Questions

1. Why is energy required to break bonds? Why is it released when bonds are formed?

2. Why is imageH so much easier to determine than instantaneous H (enthalpy)? What standard methods have been developed to help scientists find H through calculations?

3. What is the difference between exergonic and exothermic? What is the difference between endergonic and endothermic?

Explanations to Practice Questions

1. A

The process is adiabatic. Adiabatic describes any thermodynamic transformation that does not involve a heat transfer (Q). An adiabatic process can be either reversible or irreversible, though for MCAT purposes, adiabatic expansions or contractions refer to volume changes in a closed system without experimentally significant losses or gains of heat. The internal energy of the system changes (U = - image, where U = Q - W, W = image, and Q = 0). (B) is incorrect because isobaric processes occur at constant pressures. (C) is incorrect because an isothermal process requires a heat transfer at constant temperature, and there is no heat transfer along an adiabatic curve. Once it becomes clear that the process is adiabatic, that rules out (D).

2. D

There is not enough information in the problem to determine whether or not the reaction is nonspontaneous. Start with imageG = imageH-image. If the signs of enthalpy and entropy are the same, the reaction is temperature-dependent. If the signs of these terms are different, we can find the sign of imageGwithout using temperature. The most common thermochemistry questions on the MCAT test your ability to manipulate and interpret the imageG = imageH-image equation. Memorize it.

3. B

Sodium oxidizes easily at standard conditions. No calculation is necessary here. There is enough information in the problem to predict the equilibrium constant, eliminating (D). If Keq <1, the reverse reaction is favored, indicating that the forward reaction is nonspontaneous. If Keq = 1, the reaction is at equilibrium. If Keq >1, the forward reaction proceeds spontaneously. The question states that the sample spontaneously combusts at room temperature (i.e., 25°C). The answer is (B), Keq >1.

4. C

Combustion involves the reaction of a hydrocarbon with oxygen to produce carbon dioxide and water. Longer hydrocarbon chains yield greater amounts of combustion products and release more heat in the process (i.e., the reaction is more exothermic). Isobutane combusts less easily than n-butane because of its branched structure.

5. A

At first glance, this might seem like a math-heavy problem, but it really doesn’t require any calculations at all. We just have to keep track of which bonds are broken and which bonds are formed. Remember, breaking bonds requires energy, while forming bonds releases energy.

Nucleophile: methanol’s oxygen CH3O–H Electrophile: acetic acid’s carbonyl carbon CH3COOH

Through nucleophilic attack and leaving group separation, substitution occurs so that CH3O takes the place of acetic acid’s OH to create methyl acetate. A proton transfer occurs between methanol and acetic acid’s leaving group, OH.

Bonds broken:

1 O–H (-464 kJ/mol): Proton leaves methanol.

1 C–O (-360 kJ/mol): Between the carbonyl carbon and the hydroxyl oxygen in acetic acid

Bonds formed:

1 C–O (+360 kJ/mol): Between the carbonyl carbon and the attacking oxygen

1 O–H (+464 kJ/mol): Between the leaving group oxygen and the transferred proton

The sum of all our bonding events is 0 kJ. We can reason through this intuitively. If one O–H bond is broken and another is made, the two values will cancel each other out, and the net energy change must be 0 kJ. Similarly, if one C–O bond is broken and another is made, the net energy change will also be 0 kJ.

6. C

Standard temperature and pressure indicates 0°C and 1 atm. Gibbs free energy is temperature-dependent. If a reaction is at equilibrium, imageG = 0. (C) is the correct answer.

7. B

The correct answer is (B), using Keq = e-imageG°/RT from imageG° = -RT ln(Keq). Use e = 2.7. R is the gas constant, 8.314 Jmol-1 K-1, and T = 298 K, because of the standard-state sign.

Convert -4.955 kJ to -4,955 J and then round -4,955 to -5,000. Let’s also round 298 K to 300 K. The exponent’s denominator will be 8.314 × 300, which we can estimate as 2,500.


So we’re left with e(-5,000/-2,500) = Keq, or e2 = Keq.

Plugging in 2.7 for e, we can calculate 2.7 (2) = Keq. Because 22 is 4 and 32 is 9, we can assume that we want an answer choice somewhere in between, which eliminates (A) and (D). Let’s use some logic to choose between (B) and (C). Obviously, 8.9 is very close to 9, so we can assume that its square root is very, very close to 3 (it’s actually around 2.984). The answer choice should be a bit smaller, so (B), 7.4, is correct.

8. D

There is not enough information available to determine the energy of this reaction. Only its entropy is obvious.

9. A

This problem asks you to calculate the free energy of reaction at nonstandard conditions, which you can do using the equation imageG = imageG° + RT ln(Q). (R is the gas constant, 8.314 Jmol-1K-1).

10. C

There is not enough information to deduce anything about reaction temperature, which eliminates (B) and (D). Adiabatic and isothermal processes are necessarily opposite because adiabatic processes do not involve heat transfers. A reaction at constant pressure can be either adiabatic (no heat transfer to change volume) or isobaric (constant pressure, as the word roots imply).

11. A

This question requires interpreting the equation imageG = imageH - image. Endergonic (C) indicates a nonspontaneous reaction, and an exergonic reaction (D) indicates a spontaneous reaction. In contrast, endothermic and exothermic reactions suggest the sign of the enthalpy of the reaction. The problem does not provide enough information to determine the free energy of this temperature-dependent reaction. (A) is the correct answer.

12. A

Disorder in the vessel increases over the course of the reaction, so (A) is correct. While we cannot make a quantitative determination of entropy from the picture, we can estimate the relative amount of disorder from the beginning to the end of the reaction in the vessel.

13. C

For a process to occur spontaneously, Q must be less than Keq and will therefore have a tendency to move in the direction toward equilibrium. A spontaneous reaction’s free energy is negative by convention. (C) is the correct answer. (A), (B), and (D) are opposites.

14. B

A calorimeter measures specific heat or heat capacity. Though calorimeters often incorporate thermometers, the thermometer itself only tracks heat transfers, not the specific heat value itself, so (A) is incorrect. (C) is irrelevant; barometers measure changes in pressure. (D) is also incorrect, as volumetric flasks measure liquid quantities, not the heat capacity of the liquid.

15. C

Memorize the laws of thermodynamics prior to Test Day; they may be stated in several different forms. You should know the equation and how to rephrase it into a sentence or two. The first law often confuses students—keep in mind that it refers to the overall energy of the universe, not to the enthalpy of the universe, even though enthalpy is usually substituted for this term in introductory college chemistry experiments. The third law (absolute zero) is not an equation, so keep that in mind as you work.

16. A

Eliminate (C) and (D), which describe the free energy of reaction and cannot be determined from this graph. If the heat of formation of the products is greater than that of the reactants, the reaction is endothermic. We can determine this information by their relative magnitude on the graph. An exothermic graph would reflect products with a lower enthalpy than that of the reactants.