Chemical Kinetics and Equilibrium - Review - MCAT General Chemistry Review

MCAT General Chemistry Review

Part I Review

Chapter 5: Chemical Kinetics and Equilibrium

It’s Friday night, and you’re on the first call of your pediatrics rotation. You’re a third-year medical student now, and you’re anxious to see some action. Hanging out in the call room near the ED of the children’s hospital, you get a page from the resident: Come to the emergency room, now, she says. They just brought in a kid with DKA. DKA, you know, stands for diabetic ketoacidosis and is a fairly common way for undiagnosed Type I diabetes to present. Entering the child’s room, the examination is already under way; he’s young—about 10 years old—conscious but agitated, and the most obvious sign—what you notice immediately—is his rapid, shallow breathing. You see he’s already receiving IV fluids and an insulin drip.

Later that evening, after the boy has stabilized, you and the resident are talking about diabetes and DKA. You remembered from your second-year lessons about endocrine pathophysiology that ketoacidosis can arise as a result of the body’s metabolism of fatty acids when insulin production finally shuts down in Type I diabetes. Because most of the cells of the human body can’t import glucose without the aid of insulin, the glucose accumulates in the plasma of the blood, producing hyperglycemia even as the cells of the body are in a state of glucose starvation. Fatty acids are metabolized into ketone bodies as an alternative energy source. Some of the ketones produced are ketoacids, and as the diabetic crisis continues and worsens, the concentration of these ketoacids increases, resulting in a plasma pH below 7.35 (metabolic acidosis). The combination of the acidosis, progressively severe dehydration due to the osmotic effect of glucose “spilling into” the urine, and other negative effects of the severe insulin depletion result in the host of signs and symptoms of diabetic ketoacidosis. You ask the resident why the boy was hyperventilating, and she takes a piece of paper and writes out the following:

H+ (aq) + HCO3- (aq) image H2CO3 (aq) image CO2 (g) + H2O (l)

It’s Le Châtelier’s principle, she deadpans, disappointed that you didn’t remember that. The respiratory system is trying to compensate for the metabolic acidosis; the increased breathing rate allows the patient to blow off more CO2, which causes the equilibrium to shift to the right. Hydrogen ions combine with bicarbonate ions to produce carbonic acid, which dissociates into CO2gas to replace the gas that’s being expelled from the lungs. Of course, the desired result is a decrease in the hydrogen ion concentration, which stabilizes the pH and keeps it from getting crazy low. It’s not perfect, but if you catch them soon enough, the pH hasn’t gone so low that they’ve essentially become a scrambled egg. You should know all of this by now.

You recognize the equation. In fact, you even remember studying it for your MCAT. What was that all about? Oh yeah, chemical equilibrium. Wow, chemistry really is essential for medical school!

This chapter focuses on two primary topics: chemical kinetics and chemical equilibrium. As the term suggests, chemical kinetics is the study of reaction rates, the effects of reaction conditions on these rates, and the mechanisms implied by such observations. Chemical equilibrium is a dynamic state of a chemical reaction at which the concentrations of reactants and products stabilize over time in a low-energy configuration. Pay particular attention to the concepts of chemical equilibrium, as we will return to them in our review of solutions, acid-base, and redox chemistry.

Chemical Kinetics


Reactions can be spontaneous or nonspontaneous; the change in Gibbs free energy determines whether or not a reaction will occur, by itself, without outside assistance (see Chapter 6, Thermochemistry). However, even if a reaction is spontaneous, this does not necessarily mean that it will run quickly. In fact, nearly every reaction that our very lives depend upon, while perhaps spontaneous, proceeds so slowly that without the aid of enzymes and other catalysts, we might not ever actually “see” the reaction occur over the course of an average human lifetime. In biology, we discuss the function of enzymes, which selectively enhance the rate of certain reactions (by a factor of 106 to 1014) over other thermodynamically feasible reaction pathways, thereby determining the course of cellular metabolism, the collection of all chemical reactions in a living cell. For now, however, let us review the topics of reaction mechanisms, rates, rate laws, and the factors that affect them.


Very rarely is the balanced reaction equation, with which we work to calculate limiting reactants and yields, an accurate representation of the actual steps involved in the chemical process from reactants to products. Many reactions proceed by more than one step, the series of which is known as themechanism of a reaction and the sum of which gives the overall reaction (the one that you, typically, are asked to balance). When you know the accepted mechanism of a reaction, this helps you explain the reaction’s rate, position of equilibrium, and thermodynamic characteristics (see Chapter 6). Consider this generic reaction:

Overall reaction: A2 + 2B image 2AB


Mechanisms are proposed pathways for a reaction that must coincide with rate data information from experimental observation. We will be studying mechanisms more in Organic Chemistry.

On its own, this equation seems to imply a mechanism in which two molecules of B collide with one molecule of A2 to form two molecules of AB. Suppose instead, however, that the reaction actually takes place in two steps:

Step 1: A2 + B image A2B (slow)
Step 2: A2B + B image 2AB (fast)

You’ll note that the two steps, taken together, give the overall (net) reaction. The molecule A2B, which does not appear in the overall reaction, is called an intermediate. Reaction intermediates are often difficult to detect, because they may be consumed almost immediately after they are formed, but a proposed mechanism that includes intermediates can be supported through kinetic experiments. One of the most important points for you to remember is that the slowest step in any proposed mechanism is called the rate-determining step, because it acts like a kinetic “bottleneck,” preventing the overall reaction from proceeding any faster than the slowest step. It holds up the entire process in much the same way that the overall rate of an assembly line production can only be as fast as the slowest step or slowest person (who will probably soon find himself out of a job).

Reaction Rates


Reactions, unfortunately, do not come with handy built-in speedometers. We can’t just look at a dial or gauge and read the reaction rate. It takes a little more effort than that. To determine the rate at which a reaction proceeds, we must take measurements of concentrations of reactants and products and note their change over time.


If we consider a generic reaction, 2A + B image C, in which one mole of C is produced from every two moles of A and one mole of B, we can describe the rate of this reaction in terms of either the disappearance of reactants over time or the appearance of products over time. Because the reactants, by definition, are being consumed in the process of formation of the products, we place a minus sign in front of the rate expression in terms of reactants. For the previous reaction, the rate of the reaction with respect to A is -image[A]/imaget, with respect to B is -image[B]/imaget, and with respect to C is image[C]/imaget. You’ll notice that the stoichiometric coefficients for the reaction are not equal, and this tells you that the rates of change of concentrations are not equal. Because two moles of A are consumed for every mole of B consumed, rate -image[A] = 2 rate-image[B]. Furthermore, for every two moles of A consumed, only one mole of C is produced; thus, we can say that the rate-image[A] = 2 rateimage[C]. Based on the stoichoimetry, you can see that the rate of consumption of B is equal to the rate of production of C. To show a standard rate of reaction in which the rates with respect to all reaction species are equal, the rate of concentration change of each species should be divided by the species’ stoichiometric coefficient:


And for the general reaction aA + bB imagecC + dD:


Rate is expressed in the units of moles per liter per second (mol/L/s) or molarity per second (M/s).


Here’s something you can take to the bank (that is, to the MCAT testing center): In the Physical Sciences section of the MCAT, it is extremely unlikely that the test maker will give you a reaction equation that you can merely look at and write the correct rate law. Therefore, on the MCAT, whenever you are asked to determine the rate law for a reaction in which you are given the net equation, the first thing you will do is look for experimental data.

MCAT Expertise

Remember that the stoichiometric coefficients for the overall reaction will most likely be different from those for the rate-determining step and will, therefore, not be the same as the order of the reaction.

However, we’re getting ahead of ourselves, so let’s start with the basics. For nearly all forward, irreversible reactions, the rate is proportional to the product of the concentrations of the reactants, each raised to some power. For the general reaction

aA + bB imagecC + dD

the rate is proportional to [A]x[B]y. By including a proportionality constant, k, we can say that rate is determined according to the following equation:

rate = k[A]x[B]y

This expression is called the rate law for the general reaction shown here, where k is the reaction rate coefficient or rate constant. Rate is always measured in units of concentration over time; that is, molarity/second. The exponents x and y (or x, y, and z, if there are three reactants, etc.) are called the orders of the reaction: x is the order with respect to reactant A, and y is order with respect to reactant B. The overall order of the reaction is the sum of x + y (+ z...). These exponents may be integers, fractions, or zero and must be determined experimentally. The MCAT will focus almost entirely on zero-, first-, second-, and third-order reactions only. Furthermore, in most cases, the exponents will be integers.

Before we go any further in our consideration of rate laws, we must offer a few warnings about common traps that students often fall into. The first—and most common—is the mistaken assumption that the orders of a reaction are the same as the stoichiometric coefficients in the balanced overall equation. Pay close attention to this: The values of x and y usually aren’t the same as the stoichiometric coefficients . The orders of a reaction are usually must be determined experimentally. There are only two cases in which you can take stoichiometric coefficients as the orders of reaction. The first is when the reaction mechanism is a single step and the balanced “overall” reaction is reflective of the entire chemical process. The second is when the complete reaction mechanism is given and the rate-determining step is indicated. The stoichiometric coefficients on the reactant side of the rate-determining step are the orders of the reaction. Occasionally, even this can get a little complicated when the rate-determining step involves an intermediate as a reactant, in which case you must derive the intermediate molecule’s concentration by the law of mass action (that is, the equilibrium constant expression) for the step that produced it.

MCAT Expertise

Note that the exponents in the rate law are not equal to the stoichiometric coefficients, unless the reaction actually occurs via a single-step mechanism. Also, note that product concentrations never appear in a rate law. Don’t fall into the common trap of confusing the rate law with an equilibrium expression!

The second trap to be wary of is mistaking the equilibrium aspect of the law of mass action for the kinetic aspect. The equations for the two aspects do look similar, and if you’re not alert, you may mistake one for the other or use one when you should be using the other. The expression for equilibrium includes the concentrations of all the species in the reaction, both reactants and products. The expression for chemical kinetics, the rate law expression, includes only the reactants. Keq tells you where the reaction’s equilibrium position lies. The rate tells you how quickly the reaction will get there (that is, reach equilibrium).

The third trap we need to warn you about is regarding the rate constant, k. Technically speaking, it’s not a constant, because its particular value for any specific chemical reaction will depend on the activation energy for that reaction and the temperature at which the reaction takes place. However, for a specific reaction at a specific temperature, the rate coefficient is constant. For a reversible reaction, the Keq is equal to the ratio of the rate constant, k, for the forward reaction, divided by the rate constant, k-1, for the reverse reaction (see Chapter 6, Thermochemistry).

The fourth and final trap we need to warn you about is that the notion and principles of equilibrium apply to the system only at the end of the reaction; that is, the system has reached equilibrium. The reaction rate, while it theoretically can be measured at any time, is usually measured at or near the beginning of the reaction to minimize the effects of the reverse reaction.

Experimental Determination of Rate Law

We’ve stated this a few times now, but it bears repeating: The values of k, x, and y in the rate law equation (rate = k[A]x[B]y) must usually be determined experimentally for a given reaction at a given temperature. Although rate laws can be quite complex and the orders of the reaction difficult to discern, the MCAT limits its coverage of this topic to fairly straightforward reaction mechanisms, experimental data, and rate laws.

The first step in determining a rate law for a specific reaction is to write out the generic rate law on the scratch material provided for you at the testing center. Then look for the necessary data. Typically, you’ll recognize a chart that includes initial concentrations of the reactants and the initial rates of product formation as a function of the reactant concentrations. Usually, the data for three or four trials are included in the chart.

Once you’ve located the data, the next step is to identify a pair of trials in which the concentration of one of the reactants is changed while the concentration of all other reactants remains constant. Under these conditions, any change in rate of product formation (if there is any) from the one trial to the other is due solely to the change in concentration of one reactant. Let’s imagine that compound A’s concentration is constant, while the concentration of B has been doubled. If the rate of the formation of product C has subsequently quadrupled, then you can say to yourself (using your in-the-head-voice, because nobody wants to hear you talk to yourself on Test Day!), “Doubling the concentration of B has resulted in a quadrupling of the production rate of C, so to determine the order of the reaction, y, with respect to reactant B, I need to calculate the power by which the number 2 must be raised to equal 4. Because 2y = 4, y = 2.” (Kaplan wants you to think exactly this way. When you read this sentence, exactly as we’ve written it, you will be analyzing the data in the correct manner and working through the correct process for generating the correct rate law.)

MCAT Expertise

Traditionally, the MCAT has loved rate problems. With practice, you’ll be able to do these quickly in your head with minimal paper-and-pencil calculations. Remember to look for pairs of reaction trials in which the concentration of only one species changes while the other(s) remain constant.

The next step is to repeat this process for the other reactant, using different data from a different pair of trials, always making sure that the concentration of only the reactant whose order you are trying to determine is changed from one trial to the other while the concentration of any other reactant remains the same. Once you’ve determined the orders of the reaction with respect to each reactant, you then can write the complete rate law, replacing the x and the y (and sometimes z) with actual numbers. To determine the value of the rate constant k, you will need to plug in actual values—you can use the data from any one of the trials; pick whichever trial has the most arithmetically convenient numbers—for the reactant concentrations and the product formation rate, once you know the values for the exponent for each reactant.

Example: Given the data below, find the rate law for the following reaction at 300 K.


Solution: First, look for two trials in which the concentrations of all but one of the substances are held constant.

a) In Trials 1 and 2, the concentration of A is kept constant, while the concentration of B is doubled. The rate increases by a factor of 8.1/2.0, approximately 4. Write down the rate expression of the two trials.

Trial 1: r1 = k[A]x [B]y = k(1.00)x (1.00)y

Trial 2: r2 = k[A]x [B]y = k(1.00)x (2.00)y

Divide the second equation by the first:


b) In Trials 2 and 3, the concentration of B is kept constant, while the concentration of A is doubled; the rate is increased by a factor of 15.9/8.1, approximately 2. The rate expressions of the two trials are as follows:

Trial 2: r2 = k(1.00)x (2.00)y Trial 3: r3 = k(2.00)x (2.00)y

Divide the second equation by the first,


The order of the reaction with respect to A is 1 and with respect to B is 2; the overall reaction order is 1 + 2 = 3.

To calculate k, substitute the values from any one of the above trials into the rate law; e.g.,

2.0 M/sec = k × 1.00 M × (1.00 M)2 k = 2.0 M-2 sec-1

Therefore, the rate law is r = 2.0 M-2 sec-1 [A][B]2.


We classify chemical reactions on the basis of kinetics into classes of reactions called zero-order, first-order, second-order, mixed-order, or higher-order reactions. We will continue to consider the generic reaction aA + bB imagecC + dD for this discussion.

Zero-Order Reactions

A zero-order reaction is one whose rate of formation of product C is independent of changes in concentrations of any of the reactants, A and B. These reactions have a constant reaction rate equal to the rate coefficient (rate constant) k. The rate law for a zero-order reaction is

rate = k[A]0[B]0 = k

where k has units of M·s-1. (Remember that any number raised to the zero power equals 1.) We will remind you that the rate constant itself is dependent upon temperature; thus, it is possible to change the rate for a zero-order reaction by changing the temperature. The only other way to change the rate of a zero-order reaction is by the addition of a catalyst, which lowers the energy of activation, thereby increasing the value of k.

MCAT Expertise

Temperature is the only factor that can change the rate of a zero-order reaction.

First-Order Reactions

A first-order reaction (order = 1) has a rate that is directly proportional to only one reactant, such that doubling the concentration of, say, reactant A results in a doubling of the rate of formation of product C. The rate law for a first-order reaction is

rate = k[A]1 or rate = k[B]1

where k has units of s-1. A classic example of a first-order reaction is the process of radioactive decay. From the rate law, in which the rate of decrease of the amount of a radioactive isotope A is proportional to the amount of A,


The concentration of radioactive substance A at any time t can expressed mathematically as

[At] = [Ao]e-kt

where [Ao] is the initial concentration of A, [At] is the concentration of A at time t, k is the rate constant, and t is time. It is important to recognize that a first-order rate law with a single reactant suggests that the reaction begins when the molecule undergoes a chemical change all by itself, without a chemical interaction and, usually, without a physical interaction with any other molecule.

Second-Order Reactions

A second-order reaction (order = 2) has a rate that is proportional either to the product of the concentrations of two reactants or to the square of the concentration of a single reactant (and zero-order with respect to any other reactant). The following rate laws all reflect second-order reactions:

rate = k[A]1[B]1 or rate = k[A]0[B]2 = k[B]2 or rate = k[A]2[B]0 = k[A]2

where k has units of M-1sec-1. It is important to recognize that a second-order rate law often suggests a physical collision between two reactant molecules, especially if the rate law is first-order with respect to each of the two reactants.

Higher-Order Reactions

Fortunately there are very few—almost zero—reactions in which a single-reaction step involves a termolecular process; in other words, there are almost no elementary processes whose rate is third-order with respect to a single reactant. This is because it is almost impossible to get three particles to collide simultaneously. Any order higher than 3 is virtually unknown.

Mixed-Order Reactions

Mixed-order reactions sometimes refer to noninteger orders (fractions) and in other cases to reactions whose order varies over the course of the reaction. Fractions are more specifically described as broken-order, and in recent times, the term mixed-order has come to refer solely to reactions whose order changes over time. Knowing those two definitions will probably be enough for you on Test Day.

An example of a mixed-order reaction is a catalyzed reaction whose rate law is given by


where A is the single reactant and E the catalyst. (The overall reaction and its mechanism are beyond the relevance and scope of the MCAT, and the derivation of this rate law is even more unnecessary!) The result of the large value for [A] at the beginning of the reaction is that k3[A] >> k2, and the reaction will appear to be first-order; at the end of the reaction, k2 >> k3[A] because [A] will have a low value, making the reaction appear to be second-order. While the MCAT will not ask you to derive a rate expression for a mixed-order reaction, you are responsible for being able to recognize how the rate order changes as the specific reactant concentration changes.


It’s one thing to say “A2 reacts with B2 to form 2AB”; it’s quite another to be able to describe, as precisely as possible, the actual interactions that occur between A2 and B2 to produce AB at some rate. Various theories have been proposed to explain the events that are taking place at the atomic level through the process of a reaction.

Collision Theory of Chemical Kinetics

For a reaction to occur, molecules must collide with each other in much the same way that for children to enjoy themselves, there usually must be some rough physical contact involved; that is, of course, until someone’s eye gets poked out. The collision theory of chemical kinetics states that the rate of a reaction is proportional to the number of collisions per second between the reacting molecules.

The theory suggests, however, that not all collisions result in a chemical reaction. An effective collision (that is, one that leads to the formation of products) occurs only if the molecules collide with each other in correct orientation and sufficient energy to break the existing bonds and form new ones. Not every punch that a professional boxer throws is going to result in a knockout, only the ones that have the right angle and energy. The minimum energy of collision necessary for a reaction to take place is called the activation energy, Ea or the energy barrier. Only a fraction of colliding particles have enough kinetic energy to exceed the activation energy. This means that only a fraction of all collisions are effective. The rate of a reaction can therefore be expressed as follows:

rate = fZ

where Z is the total number of collisions occurring per second and f is the fraction of collisions that are effective.

Transition State Theory

The discussion of this theory introduces us to some terms that will be more fully defined and explored in the next chapter, Thermochemistry, so if some of these concepts are not clear to you, be patient, because you’ll have another opportunity to consider them more carefully soon enough.

When molecules collide with sufficient energy at least equal to the activation energy, they form a transition state in which the old bonds are weakened and the new bonds begin to form. The transition state then dissociates into products, and the new bonds are fully formed. For the reaction A2+ B2 = 2AB, the change along the reaction coordinate, which is a measure of the extent to which the reaction has progressed from reactants to products, can be represented as shown in Figure 5.1.


Figure 5.1

Key Concept

Relative to reactants and products, transition states have the highest energy. So they are only theoretical structures and cannot be isolated. Nevertheless, we can still use the proposed structures to understand better the reactions in which they are involved.

The transition state, also called the activated complex, has greater energy than either the reactants or the products and is denoted by the symbol ‡. An amount of energy at least equal to the activation energy is required to bring the reactants to this energy level. Once an activated complex is formed, it can either dissociate into the products or revert to reactants without any additional energy input. Transition states are distinguished from reaction intermediates in that, existing as they do at energy maxima, transition states exist on a continuum rather than having distinct identities and finite lifetimes.

A potential energy diagram illustrates the relationship between the activation energy, the heats of reaction, and the potential energy of the system. The most important features to recognize in such diagrams are the relative energies of all of the products and reactants. The enthalpy change of the reaction (imageH) is the difference between the potential energy of the products and the potential energy of the reactants (see Chapter 6). A negative enthalpy change indicates an exothermic reaction (one in which heat is given off), and a positive enthalpy indicates an endothermic reaction (one in which heat is absorbed). The activated complex, the transition state, exists at the top of the energy barrier. The difference in potential energies between the activated complex and the reactants is the activation energy of the forward reaction; the difference potential in energies between the activated complex and the products is the activation energy of the reverse reaction.

Key Concept

-imageH = exothermic = heat given off +imageH = endothermic = heat absorbed

For example, consider the formation of HCl from H2 and Cl2. The overall reaction is

H2 (g) + Cl2 (g) image 2 HCl (g)

Figure 5.2 shows that the reaction is exothermic. The potential energy of the products is less than the potential energy of the reactants; heat is evolved, and the enthalpy change of the reaction is negative.


Figure 5.2

MCAT Expertise

Kinetics and thermodynamics should be considered separately. Note that the potential energy of the product can be raised or lowered, thereby changing the value of imageH without affecting the value of forward Ea.


If you imagine a group of children playing together in a rough-and-tumble game of tag, we can identify certain conditions that would result in a more vigorous, more fun, more effective game. We can use this analogy to remember the factors that can affect chemical reaction rates.

Reactant Concentrations

A group of children playing tag will probably have more fun the greater the number of children playing: The participation of more children equals more opportunities to chase, trip, and tag each other. The greater the concentrations of the reactants, the greater the number of effective collisions per unit time. Therefore, the reaction rate will increase for all but zero-order reactions. For reactions occurring in the gaseous state, the partial pressures of the gas reactants serve as a measure of concentration (see Chapter 7, The Gas Phase).

MCAT Expertise

We saw earlier in this chapter that increasing reactant concentrations might increase the reaction rate, but be aware of each reactants order before making this assumption.


You can imagine that children’s enthusiasm for playing tag outside increases as the temperature warms from the bone-chilling snowstorms of winter to the pleasant sun of June. For nearly all reactions, the reaction rate will increase as the temperature increases. Because the temperature of a substance is a measure of the particles’ average kinetic energy, increasing the temperature increases the average kinetic energy of the molecules. Consequently, the proportion of molecules having energies greater than Ea (thus capable of undergoing reaction) increases with higher temperature. You’ll often hear that raising the temperature of a system by 10°C will result in an approximate doubling of the reaction rate. You have to be careful with this approximation, as it is generally true for biological systems but not so for many other systems. (Don’t forget also: If the temperature gets too high, a catalyst may denature—and then the reaction rate plummets!)


The rate at which a reaction takes place may also be affected by the medium in which it takes place. Just as children playing tag would prefer to play on a grassy field, but another group of children wanting to get a game of ice hockey going would be looking for an ice rink, some molecules are more likely to react with each other in aqueous environments, while others are more likely to react in a nonaqueous solvent, such as DMSO (dimethylsulfoxide) or ethanol. Furthermore, the physical state of the medium (liquid, solid, or gas) can also have a significant effect. Generally, polar solvents are preferred because their molecular dipole tends to polarize the bonds of the reactants, thereby lengthening and weakening them, which permits the reaction to occur faster.


Catalysts are substances that increase reaction rate without themselves being consumed in the reaction. Catalysts interact with the reactants, either by adsorption or through the formation of intermediates, and stabilize them so as to reduce the energy of activation necessary for the reaction to proceed. While many catalysts, including all enzymes, chemically interact with the reactants, upon formation of the products, they return to their original chemical state. They may increase the frequency of collisions between the reactants; change the relative orientation of the reactants, making a higher percentage of the collisions effective; donate electron density to the reactants; or reduce intramolecular bonding within reactant molecules. In homogeneous catalysis, the catalyst is in the same phase (solid, liquid, gas) as the reactants. In heterogeneous catalysis, the catalyst is in a distinct phase. Figure 5.3 compares the energy profiles of catalyzed and uncatalyzed reactions.


Figure 5.3


Enzymes are essential to most biological processes.

If you look closely at the energy profiles in the figure, you’ll notice that the only effect of the catalyst is the decrease in the energies of activation, Ea, for both the forward and reverse reactions. The presence of the catalyst has no impact on the potential energies of the reactants or the products or the difference between them. This means that catalysts change only the rate of reactions, and in fact, they change the forward rate and the reverse rate by the same factor. Consequently, they have no impact whatsoever on the equilibrium position or the measure of Keq. Of course, you have to remember that as useful as catalysts are in biological and nonbiological systems, catalysts are not miracle workers: They will not transform a nonspontaneous reaction into a spontaneous one; they only make spontaneous reactions go more quickly toward equilibrium.



We’ve been dancing around this term for the past couple of pages now. We warned you not to confuse the chemical equilibrium expression for the rate expression. We stressed that catalysts make reactions go faster toward their equilibrium position but that they can’t actually change the equilibrium position or alter the value of Keq. Well, now we’re really going to get into it—and you’d better start paying attention, because the principles and concepts that are the focus of the rest of this chapter will direct our discussion in the upcoming chapters about some of the most important general chemistry topics for the MCAT: solutions, acids and bases, and redox reactions.



Equilibrium, like biological homeostasis, is a dynamic process that seeks to find balance in all systems. We can use this concept to our advantage on the MCAT in all four of the basic sciences. Equilibria are dynamic, meaning that they do undergo change but their net change will be zero.

So far, in our discussion of reaction rates, we have been assuming that the reactions were irreversible; that is, the reaction proceeds in one direction only, the reaction goes to completion, and the amount of product formed is the maximum as determined by the amount of limiting reactant present.Reversible reactions are those in which the reaction can proceed in one of two ways: forward and reverse. (From the perspective of the direction in which the overall reaction is written, the forward reaction is the one that goes from “reactants” on the left to “products” on the right.) Reversible reactions usually do not proceed to completion because (by definition) the products can react together to re-form the reactants. When the reaction system is closed and no products or reactants are removed or added, the system will eventually “settle” into a state in which the rate of the forward reaction equals the rate of the reverse reaction and the concentrations of the products and reactants are constant. In this dynamic equilibrium state, the forward and reverse reactions are occurring—they haven’t stopped, as would be the case in a static equilibrium—but they are going at the same rate; thus, there is no net change in the concentrations of the products or reactants. Consider the following generic reversible reaction:

A image B

At equilibrium, the concentrations of A and B are constant (though not necessarily equal), and the reactions A image B and B image A continue to occur at equal rates.

Equilibrium can be thought of as a balance between the two reactions (forward and reverse). Better still, equilibrium should be understood on the basis of entropy, which is the measure of the distribution of energy throughout a system or between a system and its environment. For a reversible reaction at a given temperature, the reaction will reach equilibrium when the system’s entropy—or energy distribution—is at a maximum and the Gibbs free energy of the system is at a minimum.


For a generic reversible reaction aA + bB imagecC + dD, the law of mass action states that if the system is at equilibrium at a given temperature, then the following ratio is constant:


The law of mass action is actually related to the expressions for the rates of the forward and reverse reactions. Consider the following one-step reversible reaction:

2A image B + C

Because the reaction occurs in one step, the rates of the forward and reverse reactions are given by

ratef = kf[A]2, and rater = kr[B][C]

When ratef = rater, the system is in equilibrium. Because the rates are equal, we can set the rate expressions for the forward and reverse reactions equal to each other:


Because kf and kr are both constants, we can define a new constant Kc, where Kc is called the equilibrium constant and the subscript c indicates that it is in terms of concentration. (When dealing with gases, the equilibrium constant is referred to a Kp, and the subscript p indicates that it is in terms of pressure.) For dilute solutions, Kc and Keq are used interchangeably. The new equation can thus be written


Key Concept

For most purposes, you will not need to distinguish between different K values. For dilute solutions, Keqimage Kc and both are calculated in units of concentration.

While the forward and the reverse reaction rates are equal at equilibrium, the concentrations of the reactants and products are not usually equal. This means that the forward and reverse reaction rate constants, kf and kr, are not usually equal. The ratio of kf to kr is Kc (Keq).


When a reaction occurs by more than one step, the equilibrium constant for the overall reaction is found by multiplying together the equilibrium constants for each step of the reaction. When you do this, the equilibrium constant for the overall reaction is equal to the concentrations of the products divided by the concentrations of the reactants in the overall reaction, each concentration term raised to the stoichiometric coefficient for the respective species. The forward and reverse rate constants for the nth step are designated kn and k-n, respectively. For example, if the reaction

aA + bB imagecC + dD . . .

occurs in three steps, then


Example: What is the expression for the equilibrium constant for the following reaction?

3 H2 (g) + N2 (g) image 2 NH3 (g)

Solution: Kc = 129

MCAT Expertise

Remember our earlier warning about confusion between reaction coefficients and rate laws? Well, here the coefficients are equal to the exponents in the equilibrium expression. On Test Day, if the reaction is balanced, then the equilibrium expression should just about write itself on your scratch paper.


The law of mass action defines the position of equilibrium by stating that the ratio of the product of the concentrations of the products, each raised to their respective stoichiometric coefficients, to the product of the concentrations of the reactants, each raised to their respective stoichiometric coefficients, is constant. However, equilibrium is a state that is achieved only through time. Depending on the actual rates of the forward and reverse reactions, equilibrium might be achieved in minutes or years. What can we use as a kind of “timer” that tells us how far along in the process toward equilibrium the reaction has reached? We can take another measurement of concentrations called the reaction quotient, Qc. At any point in time of a reaction, we can measure the concentrations of all the reactants and products and calculate the reaction quotient according to the following equation:


You should be struck by how similar this looks to the equation for Keq. It’s the same form, but the information it provides is quite different. While the concentrations used for the law of mass action are equilibrium (constant) concentrations, when calculating a value of Qc for a reaction, the concentrations of the reactants and products may not be constant. In fact, if Qc changes over time because the concentrations of reaction species are changing, the reaction by definition is not at the equilibrium state. Thus, the utility of Qc is not the value itself but rather the comparison that can be made of the calculated Qc at any given moment in time of the reaction to the known Keq for the reaction at a given temperature. For any reaction, if

Qc < Keq, then the reaction has not yet reached equilibrium.

Qc > Keq, then the reaction has exceeded equilibrium.

Qc = Keq, then the reaction is in dynamic equilibrium.

Any reaction that has not yet reached the equilibrium state, as indicated by Qc < Keq , will continue spontaneously in the forward direction (that is, consuming reactants to form products) until the equilibrium ratio of reactants and products is reached. Any reaction in the equilibrium state will continue to react in the forward and reverse direction, but the reaction rates for the forward and reverse reactions will be equal and the concentrations of the reactants and products will be constant, such that Qc = Keq. Once a reaction is at equilibrium, any further “movement” either in the forward direction (resulting in an increase in products) or in the reverse direction (resulting in the re-formation of reactants) will be nonspontaneous. In Chapter 6, Thermochemistry, we’ll review methods of introducing changes in systems at equilibrium and discuss how those systems respond in terms of enthalpy, spontaneity, and entropy.


Remember the following characteristics of the law of mass action and the equilibrium constant expression:

• The concentrations of pure solids and pure liquids (the solvent, really) do not appear in the equilibrium constant expression, because for the purposes of the MCAT, their concentrations do not change in the course of the reaction. (Actually, the concentrations of any pure solids and/or pure liquids are included in the expression, but we assign them a value of 1 and “hide” the actual concentrations by incorporating them into the equilibrium constant.

• Keq is characteristic of a particular reaction at a given temperature: The equilibrium constant is temperature dependent.

• Generally, the larger the value of Keq, the farther to the right we’ll find the equilibrium and the more complete the reaction.

• If the equilibrium constant for a reaction written in one direction is Keq, the equilibrium constant for the reaction written in reverse is 1/Keq.

Le Châtelier’s Principle


We can think of few other examples of a scientist’s legacy being so tarnished by such rampant mispronunciation of her or his name. While we can all be thankful that the MCAT is not an oral examination with penalties for mispronunciation, let’s just clear it up once and for all: Henry Louis’s surname is pronounced image. The principle that bears his name, and for which he is most famous, states that a system to which a “stress” is applied tends to shift so as to relieve the applied stress. No matter what the particular form the stress takes (e.g., change in concentration of one component or another, change in pressure, or change in temperature), the effect of the stress is to cause the reaction to move temporarily out of its equilibrium state, either because the concentrations or partial pressures of the system are no longer in the equilibrium ratio or because the equilibrium ratio has actually changed as a result of a change in the temperature of the system. The reaction then responds by reacting in whichever direction (that is, either forward or reverse) that results in a re-establishment of the equilibrium state.

MCAT Expertise

Le Châtelier’s principle applies to a wide variety of systems and, as such, appears as a fundamental concept in both MCAT science sections.


When you add or remove reactants or products from a reaction in equilibrium, you are causing the reaction to be no longer at its energy minimum state. Metaphorically, you have pushed the reaction, like a ball, up a little ways along the slope of the energy hills on either side of the energy valley. The actual effect is that with the change in concentration of one or another of the chemical species, you have caused the system to have a ratio of products to reactants that is not equal to the equilibrium ratio. In other words, changing the concentration of either a reactant or a product results in Qc ≠ Keq. By adding reactant or removing product, you have created a situation in which Qc < Keq, and the reaction will spontaneously move in the forward direction, increasing the value of Qc until Qc = Keq. By removing reactant or adding product, you have created a situation in which Qc > Keq, and the reaction will spontaneously react in the reverse direction, thereby decreasing the value of Qc until once again Qc = Keq. A simple way to remember this is: The system will always react in the direction away from the added species or toward the removed species.


Remember this equation:

CO2 + H2O image HCO3- + H+

In the tissues, there is a lot of CO2, and the reaction shifts to the right. In the lungs, CO2 is lost, and the reaction shifts to the left. Note that blowing off CO2 (hyperventilation) is used as a mechanism of dealing with acidosis (excess H+).

We often take advantage of this particular tendency of chemical reactions in order to improve the yield of chemical reactions. For example, where possible in the industrial production of chemicals, products of reversible reactions are removed as they are formed so as to prevent the reactions from ever reaching their equilibrium states. The reaction will continue to go in the forward direction, producing more and more product (assuming continual replenishment of reactants as they are consumed in the reaction). You could also drive a reaction forward by starting with higher concentrations of reactants. This will lead to an increase in the absolute quantities of products formed, but the reaction would still eventually reach its equilibrium state, unless product was removed as it formed.


Because liquids and solids are essentially incompressible, only chemical reactions that involve at least one gas species will be affected by changes to the system’s volume and pressure. When you compress a system, its volume decreases, and the total pressure increases. The increase in the total pressure is associated with an increase in the partial pressures of all the gases in the system, and this results in the system no longer being in the equilibrium state, such that Qp ≠ Keq. The system will move either forward or in reverse but always toward whichever side has the lower total number of moles of gas. This result is a consequence of the ideal gas law, which tells us that there is a direct relationship between the number of moles of gas and the pressure of the gas. If you increase the pressure on a system, it will respond by decreasing the pressure by means of decreasing the number of gas moles present. (In this case, the volume of the system was decreased and then held constant while the system returned to its equilibrium state.) When you expand a system, its volume increases, and the total pressure and partial pressures decrease. The system is no longer in its equilibrium state and will react in the direction of the side with the greater number of moles of gas.

Consider the following reaction:

N2 (g) + 3H2image 2NH3 (g)

The left side of the reaction has a total of four moles of gas molecules, whereas the right side has only two moles. When the pressure of this system is increased, the system will react in the direction that produces fewer moles of gas. In this case, that direction is to the right: More ammonia will form. However, if the pressure is decreased, the system will react in the direction that produces more moles of gas; the favored reaction will be the reverse, and more reactants will re-form.


Le Châtelier’s principle tells us that changing the temperature of a system will also cause the system to react in a particular way so as to “return” to its equilibrium state, but we have to be careful here, because unlike the effect of changing concentrations or pressures, the result of changing temperature is not simply a change in the reaction quotient, Qc or Qp, but a change in Keq. The change in temperature doesn’t cause the concentrations or partial pressures of the reactants and products to change immediately, so the Q immediately after the temperature change is the same before the temperature change. Before the temperature change, the system was at equilibrium, and Q was equal to Keq; now after the temperature change, Keq is a different value (because it depends on temperature), so Q ≠ Keq. The system has to move in whichever direction allows it to reach its new equilibrium state at the new temperature. That direction is determined by the enthalpy of the reaction (see Chapter 6, Thermochemistry). You can think of heat as a reactant if the reaction is endothermic (+imageH) and as a product if the reaction is exothermic (-imageH). Thinking about heat as a reactant or product allows you to apply the principle that we discussed in regards to concentration changes to temperature changes. For example, consider the following exothermic reaction:

A image B + heat (-imageH)

If we placed this system in an ice bath, its temperature would decrease, driving the reaction to the right to replace the heat lost. Conversely, if the system were placed in a boiling water bath, the reaction would shift to the left due to the increased “concentration” of heat. All of this is to say that on Test Day, when you are asked to predict the direction a reaction would go in response to a change in temperature, you must look for the enthalpy change for the reaction, which will be given somewhere in the passage, figure, or question stem.

Key Concept

The reaction

A (aq) + 2 B ( g) image C (g) + heat

Will shift to the right if...

Will shift to the left if...

A or B added

C added

C removed

A or B removed

pressure increased or volume reduced

volume increased or pressure reduced

temperature reduced

temperature increased


We’ve discussed some very important concepts and principles in this chapter related to the studies of reaction rates and chemical equilibria. We began with a consideration of chemical reactions and the mechanisms that illustrate the possible individual steps necessary to transform reactants into products. We demonstrated the way to derive a reaction’s rate law through the analysis of experimental data, and we looked at the factors that can affect the rates of chemical reactions. The second part of this chapter focused on the law of mass action and the significance of the equilibrium state of a chemical reaction. With our understanding of the significance of Keq and Q, we are able to predict the direction that a reaction will go in response to various stresses—concentration, pressure, or temperature changes—that might be applied to a system.

You’ve worked through another chapter and reviewed some very important topics for Test Day. We cannot stress enough how much all of your hard work and dedication will pay off in points on the MCAT. There’s more to learn and review, but you have already made great progress and will continue to do so. Take pride in the work you are doing, and have confidence in yourself and in your preparation with Kaplan.



image Reaction mechanisms propose a series of steps, the sum of which gives the overall reaction that explains the chemical processes in the transformation of reactants into products. The slowest step in a reaction mechanism is the rate-determining step, and it limits the maximum rate at which the reaction can proceed.

image Reaction rates can be measured in terms of the rate of disappearance of reactant or the appearance of product, as measured by changes in their respective concentrations.

image The generic rate law is rate = k[A]x[B]y, and a particular reaction’s actual rate law usually must be determined by analyzing experimental rate data that relate concentrations of reactants to rates of product formation.

image For a reaction to occur, molecules of reactants must collide with each other at the proper angle and with an amount of kinetic energy at least as great as the energy maximum of the transition state, known as the energy of activation, Ea.

image Reaction rates can be increased by increasing reactant concentrations (except for zero-order reactions), increasing the temperature, changing the medium, or adding a catalyst.

image Reversible chemical reactions will eventually “settle” into an energy minimum state (for which there is maximum entropy) called equilibrium. This is a dynamic equilibrium in which the concentrations of reactants and products are constant because the rates of the forward and reverse reactions are equal.

image The law of mass action gives the equilibrium constant (Keq) expression. It states that at equilibrium, the ratio of products to reactants will be constant. At equilibrium, the ratio of the forward rate to the reverse rate will be equal to one; however, the ratio of the equilibrium concentrations of products to reactants will usually not be equal to one.

image The reaction quotient, Qc, is a calculated value involving reactant and product concentrations at any time within a reaction. Comparison of the calculated Qc value to the known Keq value will tell you “where” the reaction is with respect to its equilibrium state.

image Pure solids and pure liquids are not included in the law of mass action, and Keq is temperature dependent.

image Le Châtelier’s principle states that a chemical system that experiences a stress (changes to concentration, pressure, or temperature) will react in whichever direction results in the re-establishment of the equilibrium state.



Practice Questions

1. In a third-order reaction involving two reactants and two products, doubling the concentration of the first reactant causes the rate to increase by a factor of 2. What will happen to the rate of this reaction if the concentration of the second reactant is cut in half?

A. It will increase by a factor of 2.

B. It will increase by a factor of 4.

C. It will decrease by a factor of 2.

D. It will decrease by a factor of 4.

2. In a certain equilibrium process, the activation energy of the forward reaction is greater than the activation energy of the reverse reaction. This reaction is

A. endothermic.

B. exothermic.

C. spontaneous.

D. nonspontaneous.

3. Carbonated beverages are produced by dissolving carbon dioxide in water to produce carbonic acid

CO2 (g) + H2O(l) image H2CO3 (aq)

When a bottle containing carbonated water is opened, the taste of the beverage gradually changes until all of the carbonation is lost. Which of the following statements best explains this phenomenon?

A. The change in pressure and volume causes the reaction to shift to the left, thereby decreasing the amount of aqueous carbonic acid.

B. The change in pressure and volume causes the reaction to shift to the right, thereby decreasing the amount of gaseous carbon dioxide.

C. Carbonic acid reacts with environmental oxygen and nitrogen.

D. Carbon dioxide reacts with environmental oxygen and nitrogen.

4. A certain chemical reaction is endothermic. It occurs spontaneously. Which of the following must be true for this reaction?

I. imageH > 0

II. imageG < 0

III. imageS > 0

A. I only

B. I and II only

C. II and III only

D. I, II, and III

5. A certain ionic salt, A3B, has a molar solubility of 10 M at a certain temperature. What is the Ksp of this salt at the same temperature?

A. 104

B. 3 × 104

C. 2.7 × 105

D. 8.1 × 105

6. Acetic acid dissociates in solution according to the following equation:

CH3COOH image CH 3COO- + H+

If sodium acetate is added to a solution of acetic acid in excess water, which of the following effects would be observed in the solution?

A. Decreased pH

B. Increased pH

C. Decreased pKa

D. Increased pKa

7. A certain chemical reaction follows the following rate law:

Rate = k [NO2] [Br2]

Which of the following statements describe(s) the kinetics of this reaction?

I. The reaction is a second-order reaction.

II. The amount of NO2 consumed is equal to the amount of Br2 consumed.

III. The rate will not be affected by the addition of a compound other than NO2 and Br2.

A. I only

B. I and II only

C. I and III only

D. III only

8. The following data shown in the table were collected for the combustion of the theoretical compound XH4:

XH4 + 2 O2image XO2 + 2 H2O


What is the rate law for the reaction described here?

A. Rate = k [XH4] [O2]

B. Rate = k [XH4] [O2]2

C. Rate = k [XH4]2 [O2]

D. Rate = k [XH4]2 [O2]2

9. Which of the following actions does NOT affect the equilibrium of a reaction?

A. Adding/subtracting heat

B. Adding/removing a catalyst

C. Increasing/decreasing concentration of reactants

D. Increasing/decreasing volume of reactants

10. In a sealed 1 L container, 1 mol of nitrogen gas reacts with 3 mol of hydrogen gas to form 0.05 mol of NH3. Which of the following is closest to the Keq of the reaction?

A. 0.0001

B. 0.001

C. 0.01

D. 0.1



11. The overall reaction depicted by this energy diagram is

A. endothermic, because point B is higher than point A.

B. endothermic, because point C is higher than point A.

C. exothermic, because point D is higher than point E.

D. exothermic, because point A is higher than point E.

12. Which process has the highest activation energy?

A. The first step of the forward reaction

B. The first step of the reverse reaction

C. The second step of the forward reaction

D. The second step of the reverse reaction

13. Which of the following components of the reaction mechanism will NEVER be present in the reaction vessel when the reaction coordinate is at point B?

A. Reactants

B. Products

C. Intermediates

D. Catalysts

14. Consider the following two reactions:

3A + 2B 137 3C + 4D (Reaction 1)
4D + 3C 138 3A + 2B (Reaction 2)

If Keq for reaction 1 is equal to 0.1, what is Keq for reaction 2?

A. 0.1

B. 1

C. 10

D. 100

15. Which of the following statements would best describe the experimental result if the temperature of the following theoretical reaction were decreased?

A + B 139 C + D imageH = -1.12 kJ/mol

A. [C] + [D] would increase.

B. [A] + [B] would increase.

C. imageH would increase.

D. imageH would decrease.

16. Compound A has a Ka of approximately 10-4. Which of the following compounds is most likely to react with a solution of compound A?


B. NO2

C. NH3

D. N2O5

17. The following system obeys second-order kinetics.

2NO2image NO3 + NO (slow)

NO3 + CO image NO2 + CO2 (fast)

What is the rate law for this reaction?

A. Rate = k [NO2] [CO].

B. Rate = k [NO2]2 [CO].

C. Rate = k [NO2] [NO3].

D. Rate = k [NO2]2.

18. The potential energy diagram shown represents four different reactions.


Assuming identical conditions, which of the reactions displayed on the energy diagram proceeds the fastest?

A. A

B. B

C. C

D. D

Small Group Questions

1. Because catalysts affect reaction rate, why don’t we include their concentrations in reaction rate laws?

2. Do zero-order reactions contradict the collision theory of chemical kinetics?

3. How can Le Châtelier’s principle be used to manipulate reactions?

Explanations to Practice Questions

1. D

Based on the information given in the question, the rate is first-order with respect to the concentration of the first reactant; when the concentration of that reactant doubles, the rate also doubles. Because the reaction is third-order, the sum of the exponents in the rate law must be equal to 3. Therefore, the rate law is defined as follows:

Rate = k [reactant 1]1 [reactant 2]2

So the concentration of reactant 2 must be squared in order to write a rate law that represents a third-order reaction. When the concentration of reactant 2 is multiplied by ½, the rate will be multiplied by (½)2 = ¼.

2. A

Before you try to answer this question, you should draw a potential energy diagram for the system.


If the activation energy of the forward reaction is greater than the activation energy of the reverse reaction, then the products must have a higher enthalpy than the reactants. The overall energy of the system is higher at the end than it was in the beginning. The net enthalpy change is positive, indicating an endothermic reaction. Spontaneity is correlated with Gibbs free energy (imageG).

3. A

Carbon dioxide gas evolves and leaves the bottle, which decreases the total pressure of the reactants. Le Châtelier’s principle explains that a decrease in pressure shifts the equilibrium so as to increase the number of moles of gas present. This particular reaction will shift to the left, thereby decreasing the amount of carbonic acid and increasing the amount of carbon dioxide and water. Oxygen and nitrogen are not highly reactive and are unlikely to combine spontaneously with CO2 or H2CO3.

4. D

imageH is always positive for an endothermic reaction (meaning that option I is correct), and imageG is always negative for a spontaneous reaction (meaning that option II is correct). Based on these two facts, imageS can be determined by the free energy equation: imageG = imageH-TimageS, which can be rewritten as: image = imageH-imageG. If imageH is positive and imageG is negative, imageH-imageG must be positive. This means that image is positive. T (the temperature of the system in Kelvin) is always positive, so imageS must also be positive. Therefore, option III is also correct. (A), (B), and (C) all include true but incomplete statements.

5. C

Recall that the Ksp of a salt is equal to the product of the concentrations of each of the salt’s ions in saturated solution. Each molecule of this salt dissociates into three A+ ions and one B3- ion, so Ksp is equal to [A+] [A+] [A+] [B3-] = [A+]3[B3-]. The molar solubility of the salt is equal to the total number of moles of the salt in solution. The molar solubility of the salt is 10 M. Thus, the A+ ion is present in a concentration of 30 M (3 A+ ions per molecule), and the B3- ion is present in a concentration of 10 M (one B3- ion per molecule). These values can be substituted into the Ksp expression to yield Ksp = [30 M]3[10 M] = (27 × 103) (101) = 27 × 104 = 2.7 × 105.

6. B

Adding sodium acetate increases the number of acetate ions present. According to Le Châtelier’s principle, this change will push this reaction to the left, resulting in a decrease in the number of free H+ ions. Because pH is determined by the hydrogen ion concentration, a decrease in the number of free protons will increase the pH. This problem can also be solved with the Ka equation: Ka = [CH 3COO-] [H+]/[CH3COOH]. An acid’s Ka will remain constant under a given temperature and pressure, eliminating (C) and (D). For Ka to remain unchanged while [CH3COO-] increases, [H+] must decrease or [CH3COOH] must increase. A decrease in products would require an increase in reactants, and vice versa, so the final effect would be both an increase in [CH3COOH] and a decrease in [H+]. Again, removing hydrogen ions will increase the pH of the solution.

7. A

If the sum of the exponents (orders) on the concentration of each species in the rate law is equal to 2, then the reaction is second-order. That’s the case in this situation, so option I is correct. Option II is incorrect because the exponents in the rate law are unrelated to stoichiometric coefficients, so NO2 and Br2 could be present in any ratio in the original reaction and still be first-order. Option III is incorrect because the rate can be affected by a wide variety of compounds. A catalyst, for example, could increase the rate, but it wouldn’t be included in the rate law. Any compound that would preferentially react with NO2 or Br2 (including strong acids/bases and strong oxidizing/reducing agents) would decrease the concentration of reactants and decrease the rate. Only I is valid; therefore, (A) is the correct answer.

8. C

In the first two trials, the concentration of XH4 is held constant, while the concentration of O2 is multiplied by 4. Because the rate of the reaction is also increased by a factor of approximately 4, oxygen must be a first-order reactant. Analyze the other reactant in the last two trials, XH4. When you double its concentration, the rate of the reaction quadruples. That means XH4 is a second-order reactant.

Until it becomes intuitive, use math to get to the right answer.

Write the following equations:

Trial 1: rate1 = 12.4 = k[XH4]x [O2]y = k(1.00)x (0.6)y
Trial 2: rate2 = 49.9 = k[XH4]x [O2]y = k(1.00)x (2.4)y

Plug in the values given in the question stem. When you divide Trial 2 by Trial 1, it simplifies to 4.02 = 4y, so y is approximately equal to 1. Next, a similar procedure allows you to calculate the order of [XH4] using trials 2 and 3.

Trial 2: rate2 = 49.9 = k[XH4]x [O2]y = k(0.6)x (2.4)1
Trial 3: rate3 = 198.3 = k[XH4]x [O2]y = k(1.2)x (2.4)1

Inserting the values you know, you find that the x exponent must be 2. Based on this, conclude that the experimental rate law is = k [XF4]2 [O2].

9. B

The equilibrium of a reaction can be changed by several factors. If you recall Le Châtelier’s principle, you can rule out any answer choice that would shift the direction of an equilibrium. Adding or subtracting heat would shift the equilibrium based on the enthalpy change of the reaction. Adding reactant concentration would shift the equilibrium in the direction of the product, and the opposite would occur if concentrations were decreased. Changing the volume of a reactant would change its concentration, so that would have the same effect. Only (B), adding or removing a catalyst, would change the reaction rate without changing where the equilibrium lies.

10. A

The first step to answering this question is to write out the balanced equation for the reaction of H2 and N2 to produce NH3 [N2 + 3H2image 2NH3]. This means that Keq is equal to [NH3]2/([H2]3[N2]). Because the volume is 1 L, the amount of each gas (in moles) is equal to the value of the concentration of each gas (in M). We can plug these concentrations back into the Keq expression to get Keq = (0.05)2/([3]3[1]), which is equal to (0.0025)/(27). This is approximately equal to 0.0001, and approximations are appropriate for the MCAT.

11. D

A system is exothermic if energy is released by the reaction. For exothermic reactions, the net energy change is negative, and the potential energy stored in the final products is lower than the potential energy stored in the initial reactants. Point E, which represents the energy of the final products, is lower on the energy diagram than point A, which represents the energy of initial reactants. Thus, energy must have been given off. While point A is useful for determining the energy of the overall reaction, point B represents the activation energy of the first transition state, and point C suggests an intermediate. (C) references the difference between points D and E, which only indicates the change in energy from the transition state of the second reaction step to the final products.

12. B

The activation energy of a reaction is equal to the distance on the y-axis from the energy of the reactants to the peak energy prior to formation of products. The activation energy of the first step of the forward reaction, for example, is equal to the distance along the y-axis from point A to point B. The largest energy increase on this graph occurs during the progress between point E and point D, which represents the first step of the reverse reaction.

13. B

This energy diagram presents a two-step system. The first reaction proceeds from point A to point C, and the second reaction proceeds from point C to point E. This means that the reactants predominate at point A, the intermediates predominate at point C, and the products predominate at point E. Point B, which is between points A and C, is the energy threshold at which most of the reactant starts to be converted into intermediate, so the reactant and the intermediate will both be present at this point. No product is produced until after point C, so it will not be present in the reaction mixture. Catalysts may be present in the mixture at any point, depending on the nature and the quantity of the catalyst.

14. C

Reaction 2 is simply the reverse of reaction 1. Because Keq of reaction 1 is equal to [products]/[reactants], Keq of reaction 2 must be equal to [reactants]/[products]. This means that Keq for reaction 2 is the inverse of Keq of reaction 1, so the answer is 1/0.1 = 10.

15. A

A negative imageH value always indicates an exothermic reaction, meaning that the forward reaction produces heat. Visualize that as follows:

A + B 142 C + D + heat

This means that removing heat by decreasing the temperature is similar to removing any other product of the reaction. To compensate for this loss, the reaction will shift to the right, causing an increase in the concentrations of C and D as well as a decrease in the concentrations of A and B.

16. C

Ka is equal to the ratio of products to reactants in a dissociated acid. A compound with a Ka greater than 10-7 contains more H+ cations than HA- anions, which makes it a weak acid (unless Ka is several orders of magnitude higher than 1, which would indicate a strong acid). This means that the compound in question is acidic and that it is likely to react with a compound that is basic. Of the four answer choices, NH3 is the only base. Also remember that a Ka of 10-4 leads to a pKa of 4, which is a good approximation for the pKa of several organic acids.

17. D

To answer this question, you will need to recall that the slow step of a reaction is the rate-determining step. The rate is always related to the concentrations of the reactants in the slow step, so NO2 is the only compound that should be included in the correct answer. The concentration of NO2 is squared in the rate law because, according to the question, the reaction obeys second-order kinetics.

18. D

The faster a reaction can reach its activation energy, the faster it will proceed to completion. Because this question states that all conditions are equal, the reaction with the lowest activation energy will have the fastest rate. (D) illustrates the smallest difference between the initial and peak potential energies, so that reaction can overcome its activation energy more easily than the other proposed scenarios on the energy diagram. (A), (B), and (C) have higher activation energies.