In a coordinate covalent bond, the shared electron pair comes from the lone pair of one of the atoms in the molecule, while the other atom involved in the bond contributes nothing to the relationship. (This works great for atoms but might not be the best way to form a lasting marriage!) Once such a bond forms, however, it is indistinguishable from any other covalent bond. The distinction is only helpful for keeping track of the valence electrons and formal charges (see Figure 3.2). Coordinate covalent bonds are typically found in Lewis acid-base compounds (see Chapter 10, Acids and Bases). A Lewis acid is any compound that will accept a lone pair of electrons, while a Lewis base is any compound that will donate a pair of electrons to form a covalent bond; for example, as in the reaction between borontrifluoride (BF₃) and ammonia (NH₃) shown in Figure 3.2.

The electrons that are involved in a covalent bond are in the valence shell and are called bonding electrons, while those electrons in the valence shell that are not involved in the covalent bond are called nonbonding electrons. The unshared electron pairs can also be called lone electron pairs, because they are associated only with one atomic nucleus. Because atoms can bond with other atoms in many different possible combinations, the Lewis structure system of notation has been developed to keep track of the bonded and nonbonded electron pairs. You may think of Lewis structures as a kind of bookkeeping. The number of valence electrons attributed to a particular atom in the Lewis structure of a molecule is not necessarily the same as the number of valence electrons the atom would have as an isolated atom, and the difference accounts for what is referred to as the formal charge of that atom in a particular Lewis structure. Often, more than one Lewis structure can be drawn for a molecule. If the possible Lewis structures differ in the way the atoms are connected (that is to say, they differ in their bond connectivity or arrangement), then the Lewis structures represent different possible compounds. If the Lewis structures show the same bond connectivity and differ only in the arrangement of the electron pairs, then these structures represent different resonance forms for a single compound. Lewis structures do not represent the actual or even theoretical geometry of a real compound. Their usefulness lies in showing you the different possible ways in which atoms may be combined to form different compounds or resonances of a single compound. When more than one arrangement can be made, you can assess the likelihood of each arrangement by checking the formal charges on the atoms in each arrangement. The arrangement that minimizes the number of formal charges, or minimizes the value of formal charges, is the arrangement that most likely represents the actual compound.

Key Concept

When dealing with Lewis dot structures, we only deal with the eight valence electrons (s- and p-orbitals of the outer shell) on each atom. Remember that some atoms can expand their octets by utilizing the d-orbitals in this outer shell, but this will only take place with atoms in period 3 or greater.

Lewis Structures

A Lewis structure, or Lewis dot symbol, is the chemical symbol of an element surrounded by dots, each representing one of the s and/or p valence electrons of the atom. The Lewis symbols of the elements in the second period of the periodic table are shown in Table 3.1.

Table 3.1

Because Lewis structures do not in any way suggest or reflect the actual geometric arrangement of atoms in a compound, we need another system that provides that information. One such system is known as valence shell electron pair repulsion theory (VSEPR theory). This theory is actually dependent upon Lewis structure, so they go hand in hand.

To finish our discussion of covalent bonds, we need to address the issue of atomic and molecular orbitals. If you remember back to the first chapter, we described the modern understanding of the atom as a dense, positively charged nucleus surrounded by clouds of electrons organized into orbitals (regions in space surrounding the nucleus within which there are certain probabilities of finding an electron). The four quantum numbers completely describe the energy and position of any electron of an atom. While the principal quantum number indicates the average energy level of the orbitals, the azimuthal quantum number, l, describes the subshells within each principal energy level, n. When l = 0, this indicates the s-subshell, which has one orbital that is spherical in shape. The 1s-orbital (n = 1, l = 0) is plotted in Figure 3.8.

Bridge

Quantum Numbers (Chapter 1) revisited:

• For any value of n, there are n values of l(0n- 1).

•

l= 0s

The bonding electrons in nonpolar covalent bonds may appear, on paper, to be shared equally between two atoms, but at any point in time, they will be located randomly throughout the orbital. For these instantaneous moments, then, the electron density may be unequally distributed between the two atoms. This results in rapid polarization and counterpolarization of the electron cloud and the formation of short-lived dipole moments. These dipoles interact with the electron clouds of neighboring compounds, inducing the formation of more dipoles. The momentarily negative end of one molecule will cause the closest region in any neighboring molecule to become temporarily positive itself, thus causing the other end of this neighboring molecule to become temporarily negative, which in turn induces other molecules to become temporarily polarized, and the entire process begins all over again. The attractive interactions of these short-lived and rapidly shifting dipoles are called dispersion forces or London forces.

Real World

While London forces (a type of van der Waals force) are the weakest of the intermolecular attractions, when there are millions of these interactions, as there are on the bottom of a gecko’s foot, there is an amazing power of adhesion, which is demonstrated by the animal’s ability to climb smooth vertical, even inverted, surfaces.

Dispersion forces are the weakest of all the intermolecular interactions because they are the result of induced dipoles that change and shift moment by moment. They do not extend over long distances and are, therefore, significant only when molecules are close together. The strength of the London force also depends on the degree to which and the ease by which the molecules can be polarized (that is to say, how easily the electrons can be shifted around). Large molecules in which the electrons are far from the nucleus are relatively easy to polarize and, thus, possess greater dispersion forces. Think about two plates—a big dinner plate and a small teacup saucer—with nearly equal numbers of marbles on each. The marbles will roll around more, and may be distributed more unequally, on the bigger plate than on the smaller saucer. Nevertheless, don’t underestimate the importance of the dispersion forces. If it weren’t for them, the noble gases would not liquefy at any temperature because no other intermolecular forces exist between the noble gas atoms. The low temperatures at which the noble gases liquefy is to some extent indicative of the magnitude of the dispersion forces between the atoms.

Polar molecules tend to orient themselves in such a way that the opposite ends of the respective molecular dipoles are closest to each other: The positive region of one molecule is close to the negative region of another molecule. This arrangement is energetically favorable because an attractive electrostatic force is formed between the two molecules.

Hydrogen bonds are a favorite of the MCAT! So understand them well, and you will be able to answer every question pertaining to their nature, behavior, and effects. A hydrogen bond is actually only a specific, unusually strong form of dipole–dipole interaction, which may be intra- or intermolecular. Please understand that hydrogen bonds are not actually bonds—there is no sharing or transfer of electrons between two atoms. When hydrogen is bound to a highly electronegative atom, such as fluorine, oxygen, or nitrogen, the hydrogen atom carries little of the electron density of the covalent bond. The hydrogen atom acts essentially as a naked proton (it is a little scandalous!). The positively charged hydrogen atom interacts with the partial negative charge of oxygen, nitrogen, or fluorine on nearby molecules. Substances that display hydrogen bonding tend to have unusually high boiling points compared with compounds of similar molecular formula that do not hydrogen bond. The difference derives from the energy required to break the hydrogen bonds. A very good example of this difference, which you absolutely need to understand for the MCAT, is the difference between a carboxylic acid, which can form hydrogen bonds between molecules of itself, and the acyl halide, a derivative of the carboxylic acid, which has substituted a halogen for the hydroxyl group and which cannot form hydrogen bonds with molecules of itself. The boiling point of propanoic acid (MW = 74 g/mol) is 141°C, but the boiling point of propanoyl chloride (MW = 92.5 g/mol) is 80°C. Hydrogen bonding is particularly important in the behavior of water, alcohols, amines, and carboxylic acids. It is no overstatement to say that if it weren’t for water’s ability to form hydrogen bonds and exist in the liquid state at room temperature, we would not exist (at least not in the form we recognize as “human”).

This chapter built upon our knowledge of the atom and the trends demonstrated by the elements in the periodic table to understand the different ways by which atoms partner together to form compounds, either (a) exchanging electrons to form ions, which are then held together by the electrostatic force of attraction that exists between opposite charges, or (b) sharing electrons to form covalent bonds. We discussed the nature and characteristics of covalent bonds, noting their relative lengths and energies as well as polarities. The review of Lewis structures and VSEPR theory will prepare you for predicting likely bond arrangements, resonance structures, and molecular geometry. Finally, we compared the relative strengths of the most important intermolecular electrostatic interactions, noting that even the strongest of these, the hydrogen bond, is still much weaker than an actual covalent bond. The next time you’re searing meat in a skillet or browning toast in a toaster, take a moment to consider what’s actually going on at the atomic and molecular level. It’s not just cooking; it’s science!

CONCEPTS TO REMEMBER

1. Why does bond length decrease with increasing bond strength?

2. Predict two elements whose hydrides (H^-) would contain incomplete octets.

Explanations to Practice Questions

We’ve mentioned already that the term atomic weight is a misnomer, because it is a measurement of mass, not weight (another distinction not worth any sacrifice of sleep), and the same applies here to our discussion of molecular weight: It’s really a measurement of mass. Molecular weight, then, is simply the sum of the atomic weights of all the atoms in a molecule, and its units are atomic mass units (amu). Similarly, the formula weight of an ionic compound is found by adding up the atomic weights of the constituent ions according to its empirical formula (see the following example), and its units are grams.

Example: What is the molecular weight of SOCl₂?

Solution: To find the molecular weight of SOCl₂, add together the atomic weights of each of the atoms.

We defined the term mole in Chapter 1, Atomic Structure, but let’s briefly review it. A mole is a quantity of any thing (molecules, atoms, dollar bills, chairs, etc.) equal to the number of particles that are found in 12 grams of carbon-12. That seems like an awfully strange point of comparison, so all you really need to remember is that “mole” is a quick way of indicating that we have an amount of particles equal to Avogadro’s number, 6.022 × 10²³. One mole of a compound has a mass in grams equal to the molecular weight of the compound expressed in amu and contains 6.022 × 10²³ molecules of that compound. For example, 62 grams of H₂CO₃ (carbonic acid) represents one mole of the acid compound and contains 6.022 × 10²³ molecules of H₂CO₃. The mass of one mole of a compound, called its molar weight or molar mass, is usually expressed as g/mol. Therefore, the molar mass of H₂CO₃ is 62 g/mol. You may be accustomed to using the term molecular weight to imply molar mass. Technically this is not correct, but nobody is going to rap your knuckles with a ruler for this minor infraction.

Equivalent weight and the related concept of equivalents are a source of confusion for many students. Part of the problem may be the context in which equivalents and equivalent weights are discussed: acid-base reactions, redox reactions, and precipitation reactions, all three of which themselves can be sources of much student confusion and anxiety. So let’s start with some very basic discussion, and then, in later chapters, we will see how these concepts and calculations apply to these three types of reactions.

Bridge

The idea of equivalents is similar to the concept of normality, which we will see when we study Acids and Bases in Chapter 10.

Generally, some compounds or elements have different capacities to act in certain ways, not in terms of their characteristics or behaviors, like electronegativity or ionization energy, but rather the ability of certain elements or compounds to act more potently than others in performing certain reactions. For example, one mole of HCl has the ability to donate one mole of hydrogen ions (H⁺) in solution, but one mole of H₂SO₄ has the ability to donate two moles of hydrogen ions, and one mole of H₃PO₄ has the ability to donate three moles of hydrogen ions. Another example to consider is the difference between Na and Mg: One mole of sodium has the ability to donate one mole of electrons, while one mole of magnesium has the ability to donate two moles of electrons. To find one mole of hydrogen ions for a particular acid-base reaction, we could “source” those protons from one mole of HCl, or we could instead use a half-mole of H₂SO₄. If we’re using H₃PO₄, we’d only need one-third of a mole. This is what we mean by “different capacities to act in certain ways.” If we need only one mole of hydrogen ions, it would be a waste to use an entire mole of H₃PO₄, which could donate three times what we need. We are defining the concept of equivalents: How many moles of the “thing we are interested in” (i.e., protons, hydroxide ions, electrons, or ions) will the number of moles of the compound present produce? One mole of hydrogen ions (one equivalent) will be donated by one mole of HCl, but two moles of hydrogen ions (two equivalents) will be donated by one mole of H₂SO₄, and three moles of hydrogen ions (three equivalents) will be donated by one mole of H₃PO₄. Simply put, an equivalent is a mole of charge.

The law of constant composition states that any pure sample of a given compound will contain the same elements in the identical mass ratio. For example, every sample of H₂O will contain two hydrogen atoms for every one oxygen atom, or in terms of mass, for every one gram of hydrogen, there will be eight grams of oxygen. We’re sorry to say that it doesn’t matter whether the water you drink is sourced from deep springs in the Fiji Islands or flows out of a tap connected to a shallow well. The law of constant composition says that water is water.

There are two ways to express a formula—the elemental composition—for a compound. The empirical formula gives the simplest whole number ratio of the elements in the compound. The molecular formula gives the exact number of atoms of each element in the compound and is usually a multiple of the empirical formula. For example, the empirical formula for benzene is CH, while the molecular formula is C₆H₆. For some compounds, the empirical and molecular formulas are the same, as in the case of H₂O. For the reasons previously discussed, ionic compounds, such as NaCl or CaCO₃, will only have empirical formulas.

The percent composition by mass of an element is the weight percent of a given element in a specific compound. To determine the percent composition of an element X in a compound, the following formula is used:

% Composition = (Mass of X in formula/Formula weight of compound) × 100%

You can calculate the percent composition of an element by using either the empirical or the molecular formula; just be sure to use to the appropriate mass measurement:

MCAT Expertise

Percent composition is a common way for stoichiometry to be tested on the MCAT. Practice these problems to build up speed and efficiency for Test Day.

formula weight for empirical formula or molar mass for molecular formula. Formula weight is simply the mass of the atoms in the empirical formula of a compound.

A combination reaction has two or more reactants forming one product. The formation of sulfur dioxide by burning sulfur in air is an example of a combination reaction.

S (s) + O₂ (g) SO₂ (g)

A decomposition reaction is the opposite of a combination reaction: A single compound reactant breaks down into two or more products, usually as a result of heating or electrolysis. An example of decomposition is the breakdown of mercury (II) oxide. (The [delta] sign over a reaction arrow represents the addition of heat.)

Sometimes one atom in a molecule gets tired of the other (“I love you, but I’m not in love with you…”), and divorce becomes inevitable when something better comes along. This is a single-displacement reaction: an atom (or ion) of one compound is replaced by an atom of another element. For example, zinc metal will displace copper ions in a copper sulfate solution to form zinc sulfate.

Zn (s) + CuSO₄ (aq) Cu (s) + ZnSO₄

Single-displacement reactions are often further classified as redox reactions, which will be discussed in great detail in Chapter 11, Redox Reactions and Electrochemistry. Not to carry the analogy to the point of ridiculousness, but just to make the point: Cu in CuSO₄ has an oxidation state of +2, but when it leaves the compound, it gains two electrons (the Cu⁺² is reduced to Cu)—you may think of this as Cu⁺² getting an alimony settlement or gaining back its dignity for leaving that no-good cheater, SO₄^2-. On the other side, Zn loses its dignity (in the form of two electrons) when it desperately throws itself into the arms of SO₄^2-.

Well, if you think atoms are acting scandalously in single-displacement reactions, just wait until you get a load of the atomic depravity in double-displacement reactions. In double-displacement reactions, also called metathesis, elements from two different compounds swap places with each other (hence, the name double-displacement) to form two new compounds. This type of reaction occurs when one of the products is removed from the solution as a precipitate or gas or when two of the original species combine to form a weak electrolyte that remains undissociated in solution. For example, when solutions of calcium chloride and silver nitrate are combined, insoluble silver chloride forms in a solution of calcium nitrate.

CaCl₂ (aq) + 2AgNO₃ (aq) Ca(NO₃)₂ (aq) + 2AgCl (s)

Neutralization reactions are a specific type of double displacement in which an acid reacts with a base to produce a salt. For example, hydrochloric acid and sodium, hydroxide will react to form sodium chloride and water.

HCl (aq) + NaOH NaCl (aq) + H₂O

We began our consideration of compounds with a particularly odoriferous one: benzaldehyde. As a compound, it is made from constituent atoms of different elements in a set ratio defined by its empirical or molecular formula. Each molecule of a compound has a defined mass that is measured as its molecular weight. The mass of one mole of any compound is determined from its molar mass in the units of grams/mole. We reviewed the basic classifications of reactions commonly tested on the MCAT: combination, decomposition, single-displacement, and double-displacement reactions. Furthermore, we are now confident in our understanding of the steps necessary to balance any chemical reaction; we are ready to tackle more stoichiometric problems in preparation for Test Day.

1. Can the Law of Constant Composition be applied to solutions?

2. What is the purpose of calculating a reactant’s gram equivalent weight?

Explanations to Practice Questions

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 the mechanism 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 2AB

If we consider a generic reaction, 2A + B 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 -[A]/t, with respect to B is -[B]/t, and with respect to C is [C]/t. 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 _-[A] = 2 rate_-[B]. Furthermore, for every two moles of A consumed, only one mole of C is produced; thus, we can say that the rate_-[A] = 2 rate_[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:

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 cC + 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.

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 cC + 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,

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.

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 reactant^’s order before making this assumption.

Temperature

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 E_a (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!)

For a generic reversible reaction aA + bB cC + 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 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, Q_c. 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:

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.

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

• Generally, the larger the value of K_eq, 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 K_eq, the equilibrium constant for the reaction written in reverse is 1/K_eq.

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 Q_c ≠ K_eq. By adding reactant or removing product, you have created a situation in which Q_c < K_eq, and the reaction will spontaneously move in the forward direction, increasing the value of Q_c until Q_c = K_eq. By removing reactant or adding product, you have created a situation in which Q_c > K_eq, and the reaction will spontaneously react in the reverse direction, thereby decreasing the value of Q_c until once again Q_c = K_eq. 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.

Bridge

Remember this equation:

CO₂ + H₂O 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 Q_p ≠ K_eq. 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.

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, Q_c or Q_p, but a change in K_eq. 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 K_eq; now after the temperature change, K_eq is a different value (because it depends on temperature), so Q ≠ K_eq. 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 (+H) and as a product if the reaction is exothermic (-H). 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 B + heat (-H)

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) C (g) + heat

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 K_eq 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.

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

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.

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

2. Why is H 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. Gases are made up of particles whose volumes are negligible compared to the container volume.

2. Gas atoms or molecules exhibit no intermolecular attractions or repulsions.

3. Gas particles are in continuous, random motion, undergoing collisions with other particles and the container walls.

4. Collisions between any two gas particles are elastic, meaning that there is conservation of both momentum and kinetic energy.

5. The average kinetic energy of gas particles is proportional to the absolute temperature (in Kelvin) of the gas, and it is the same for all gases at a given temperature, irrespective of chemical identity or atomic mass.

It’s fairly straightforward to imagine, based on the assumptions just listed, gas particles as lots of little rubber balls bouncing into and off each other and off the walls of their container. Of course, rubber balls, like real gas particles, have measurable mass and volume, and not even the bounciest rubber balls will collide in a completely elastic manner.

Average Molecular Speeds

According to the kinetic molecular theory of gases, the average kinetic energy of a gas particle is proportional to the absolute temperature of the gas: