Chapter 18 Reaction Rates and Equilibrium 18.1 Rates of Reaction

Chapter 18 Reaction Rates and Equilibrium 18.1 Rates of Reaction
18.2 The Progress of Chemical Reactions 18.3 Reversible Reactions and Equilibrium 18.4 Solubility Equilibrium 18.5 Free Energy and Entropy Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

How can rusting be used to cook a meal?
CHEMISTRY & YOU How can rusting be used to cook a meal? There are products that use the rusting of an iron-magnesium alloy to heat packaged food. These products are known as Meals Ready to Eat, or MREs for short. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Describing Reaction Rates
How is the rate of a chemical reaction expressed? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

How is the rate of a chemical reaction expressed? When you strike a match, it erupts into flame almost instantly and burns quickly. Millions of years were required for plants buried beneath Earth’s surface to be converted to coal. The speed of chemical reactions can vary from very fast to extremely slow. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

In chemistry, the rate of a chemical reaction, or the reaction rate, is usually expressed as the change in the amount of reactant or product per unit time. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

The figure below illustrates the progress of a typical reaction. Over time, the amount of reactant decreases and the amount of product increases. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Collision Theory A model called collision theory is used to relate the properties of particles to the rates of chemical reactions. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Collision Theory A model called collision theory is used to relate the properties of particles to the rates of chemical reactions. According to collision theory, atoms, ions, and molecules can react to form products when they collide if the particles have enough kinetic energy. Particles that do not have enough energy to react bounce apart unchanged when they collide. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

The minimum energy that colliding particles must have in order to react is called the activation energy. You can think of the activation energy for a reaction as a barrier that reactants must cross before products can form. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

When two reactant particles collide, they may form an activated complex. An activated complex is an unstable arrangement of atoms that forms for a moment at the peak of the activation-energy barrier. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

When two reactant particles collide, they may form an activated complex. An activated complex is an unstable arrangement of atoms that forms for a moment at the peak of the activation-energy barrier. The activated complex forms only if the colliding particles have enough energy and if the atoms are oriented properly. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

When two reactant particles collide, they may form an activated complex. The lifetime of an activated complex is typically about seconds. Its brief existence ends with the reformation of the reactants or with the formation of products. Thus, the activated complex is sometimes called the transition state. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Collision theory explains why some reactions are extremely slow at room temperature. Carbon and oxygen react when charcoal burns, but the reaction has a high activation energy. The O—O and C—C bonds must be broken to form the activated complex. At room temperature, the collisions of oxygen and carbon molecules are not energetic enough to break the bonds. Thus, the reaction rate of carbon with oxygen at room temperature is essentially zero. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

What factor determines whether a molecular collision results in a reaction?
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What factor determines whether a molecular collision results in a reaction?
The molecules must collide with enough energy in order to react. The minimum amount of energy needed is called the activation energy. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Factors Affecting Reaction Rates
What four factors influence the rate of a chemical reaction? By varying the conditions, you can modify the rate of almost any reaction. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Factors that can affect the rate of a chemical reaction are temperature, concentration, particle size, and the use of a catalyst. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Temperature At higher temperatures, particles move faster. The frequency of collisions increases along with the percentage of particles that have enough kinetic energy to slip over the activation-energy barrier. Thus, an increase in temperature causes products to form faster. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Concentration The number of particles in a given volume affects the rate at which reactions occur. Cramming more particles into a fixed volume increases the concentration of reactants, and, thus, the frequency of collision. Increased collision frequency leads to a higher reaction rate. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

The lighted splint glows in air and soon dies out because air is only 20% oxygen. When the glowing splint is plunged into pure oxygen, it immediately bursts into flame. The increased concentration of oxygen greatly speeds up the combustion reaction. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Particle Size The total surface area of a solid or liquid reactant affects the rate of a reaction. The smaller the particle size, the greater the surface area is for a given mass of particles. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Particle Size The total surface area of a solid or liquid reactant affects the rate of a reaction. The smaller the particle size, the greater the surface area is for a given mass of particles. The result of an increase in surface area is an increase in the frequency of collisions and the reaction rate. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

When a piece of magnesium is placed in dilute acid, hydrogen ions can collide with magnesium atoms. Mg(s) + 2H+(aq)  Mg2+(aq) + H2(g) Only atoms at the surface of the metal are available for reaction. Dividing the metal into smaller pieces increases the surface area and the number of collisions. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Particle Size Another way to increase the surface area of solids is to dissolve them. In a solution, particles are separated and more accessible to other reactants. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Particle Size You can also increase the surface area of a solid by grinding it into a fine powder. Small dustlike particles, however, can be dangerous when suspended in air. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Particle Size An explosion destroyed this sugar refinery. The tiny size of the reactant particles (sugar dust) caused the reaction of the sugar with oxygen in the air to be explosive. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Catalysts Increasing the temperature is not always the best way to increase the rate of a reaction. A catalyst is often better. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Catalysts Increasing the temperature is not always the best way to increase the rate of a reaction. A catalyst is often better. Recall that a catalyst is a substance that increases the rate of a reaction without being used up during the reaction. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Catalysts Increasing the temperature is not always the best way to increase the rate of a reaction. A catalyst is often better. Recall that a catalyst is a substance that increases the rate of a reaction without being used up during the reaction. Catalysts permit reactions to proceed along a lower energy path. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Interpret Graphs The activation-energy barrier for the catalyzed reaction is lower than that of the uncatalyzed reaction. When the barrier is lower, a greater fraction of reactants have the energy to form products within a given time. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

CHEMISTRY & YOU When salt water is added to the metal alloy in an MRE, the rate of the rusting reaction increases, and heat is produced rapidly. Which factor that can affect reaction rates is being applied in this situation? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

CHEMISTRY & YOU When salt water is added to the metal alloy in an MRE, the rate of the rusting reaction increases, and heat is produced rapidly. Which factor that can affect reaction rates is being applied in this situation? Salt acts as a catalyst for the reaction between the metal and water, speeding up the reaction without being consumed. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Catalysts The rate of reaction of hydrogen and oxygen at room temperature is negligible. But with a small amount of platinum (Pt) as a catalyst, the reaction is rapid. 2H2(g) + O2(g) H2O(l) Pt Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Catalysts 2H2(g) + O2(g) H2O(l) Pt A catalyst is not consumed during a reaction. Therefore, it does not appear as a reactant in the chemical equation. Instead, the catalyst is often written above the yield arrow, as in the equation above. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Catalysts At normal body temperature (37C), reactions in the body would be too slow without catalysts. The catalysts that increase the rates of biological reactions are called enzymes. When you eat a meal containing protein, enzymes in your digestive tract help break down the protein molecules in a few hours. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Catalysts An inhibitor is a substance that interferes with the action of a catalyst. Some inhibitors work by reacting with, or “poisoning,” the catalyst itself. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Catalysts An inhibitor is a substance that interferes with the action of a catalyst. Some inhibitors work by reaction with, or “poisoning,” the catalyst itself. Thus, the inhibitor reduces the amount of catalyst available for a reaction. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Catalysts An inhibitor is a substance that interferes with the action of a catalyst. Some inhibitors work by reaction with, or “poisoning,” the catalyst itself. Thus, the inhibitor reduces the amount of catalyst available for a reaction. Reactions slow or even stop when a catalyst is poisoned. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

A. Catalyst concentration B. Concentration C. Temperature
Which of the following factors could be increased in order to decrease a reaction rate? A. Catalyst concentration B. Concentration C. Temperature D. Particle size Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Key Concepts In chemistry, the rate of a chemical reaction, or the reaction rate, is usually expressed as the change in the amount of reactant or product per unit time. Factors that can affect the rate of a chemical reaction are temperature, concentration, particle size, and the use of a catalyst. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

rate: describes the speed of change over an interval of time
Glossary Terms rate: describes the speed of change over an interval of time collision theory: atoms, ions, and molecules can react to form products when they collide, provided that the particles have enough kinetic energy activation energy: the minimum energy colliding particles must have in order to react Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

inhibitor: a substance that interferes with the action of a catalyst
Glossary Terms activated complex: an unstable arrangement of atoms that exists momentarily at the peak of the activation-energy barrier; an intermediate or transitional structure formed during the course of a reaction inhibitor: a substance that interferes with the action of a catalyst Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Chemical Reactions, Matter, and Energy
BIG IDEA Chemical Reactions, Matter, and Energy The rate of a chemical reaction can be controlled by adjusting temperature, concentration, or particle size. Adding a catalyst speeds up a reaction by lowering the activation energy. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Chapter 18 Reaction Rates and Equilibrium
18.1 Rates of Reaction 18.2 The Progress of Chemical Reactions 18.3 Reversible Reactions and Equilibrium 18.4 Solubility Equilibrium 18.5 Free Energy and Entropy Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

How is a bicycle race like a chemical reaction?
CHEMISTRY & YOU How is a bicycle race like a chemical reaction? Riders in the Tour de France bicycle race must cross steep mountains with heights of 1900 meters or more. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Rate Laws Rate Laws What is the relationship between the value of the specific rate constant, k, and the speed of a chemical reaction? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Rate Laws The rate of a reaction depends in part on the concentrations of the reactants. Suppose there were a reaction with only one reactant and one product. A  B The rate at which A forms B can be expressed as the change in A (ΔA) with time. rate = ΔA Δt Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Rate Laws The rate of disappearance of A is proportional to the concentration of A. ΔA Δt [A] The proportionality can be expressed as the concentration of A, [A], multiplied by a constant, k. rate = = k × [A] ΔA Δt Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Rate Laws rate = = k × [A] ΔA Δt This equation is a rate law, an expression for the rate of a reaction in terms of the concentration of the reactants. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Rate Laws rate = = k × [A] ΔA Δt This equation is a rate law, an expression for the rate of a reaction in terms of the concentration of the reactants. The specific rate constant (k) for a reaction is a proportionality constant relating the concentrations of reactants to the rate of the reaction. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Rate Laws The value of the specific rate constant, k, in a rate law is large if the products form quickly; the value is small if the products form slowly. rate = = k × [A] ΔA Δt Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

First-Order Reactions
Rate Laws First-Order Reactions The order of a reaction is the power to which the concentration of a reactant must be raised to match the experimental data on concentration and rate. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

First-Order Reactions
Rate Laws First-Order Reactions The order of a reaction is the power to which the concentration of a reactant must be raised to match the experimental data on concentration and rate. In a first-order reaction, the rate is directly proportional to the concentration of only one reactant. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Interpret Graphs Over time, the rate of reaction decreases because the concentration of the reactant is decreasing. The reaction A  B is a first-order reaction. If [A] is reduced by one half, the reaction rate is reduced by one half. The rate (ΔA/Δt) at any point on the graph equals the slope of the tangent to the curve at that point. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Higher-Order Reactions
Rate Laws Higher-Order Reactions In some reactions, two substances react to produce products. In the general equation for a double-replacement reaction below, the coefficients are represented by lowercase letters. aA + bB  cC + dD For the reaction of A with B, the rate of reaction is dependent on the concentrations of both A and B. rate = k[A]x[B]y Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Rate Laws Higher-Order Reactions rate = k[A]x[B]y When each exponent in the rate law equals 1 (that is, x = y = 1) the reaction is said to be first order in A and first order in B. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Rate Laws Higher-Order Reactions The overall order of a reaction is the sum of the exponents for the individual reactants. A reaction that is first order in A and first order in B is thus second order overall. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Rate Laws Higher-Order Reactions The overall order of a reaction is the sum of the exponents for the individual reactants. A reaction that is first order in A and first order in B is thus second order overall. The actual order of a reaction must be determined by experiment. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Finding the Order of a Reaction from Experimental Data
Sample Problem 18.1 Finding the Order of a Reaction from Experimental Data Consider the reaction aA  B. The rate law for this reaction is Rate = k[A]x. From the data in the table, find the order of the reaction with respect to A and the overall order of the reaction. Trial Initial concentration of A (mol/L) Initial rate (mol/(L·s)) 1 0.050 3.0  10–4 2 0.10 12  10–4 3 0.20 48  10–4 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Analyze List the knowns and the unknowns.
Sample Problem 18.1 Analyze List the knowns and the unknowns. 1 Use the first two trials to calculate the order and the third to evaluate your answer. KNOWNS [A]1 = mol/L [A]2 = 0.10 mol/L Rate1 = 3.0  10–4 mol/(L·s) Rate2 = 12  10–4 mol/(L·s) UNKNOWN Order of reaction with respect to A = ? Overall order of the reaction = ? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Calculate Solve for the unknowns.
Sample Problem 18.1 Calculate Solve for the unknowns. 2 Start with the rate law for each initial concentration of A. The rate law of the reaction and the specific rate constant, k, is the same for any initial concentration of A. Rate1 = k [A1]x Rate2 = k [A2]x Divide the second expression by the first expression. = = ( ) Rate k [A1]x Rate k [A2]x [A2] [A1] x Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 18.1 Calculate Solve for the unknowns. 2 Substitute the known quantities into the equation. = ( ) 3.0  10–4 mol/(L·s) 12  10–4 mol/(L·s) 0.10 mol/L 0.050 mol/L x 4.0 = 2.0x Determine the value of x. x = 2 The reaction is second order in A. Since A is the only reactant, the reaction must be second order overall. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Evaluate Does this result make sense?
Sample Problem 18.1 Evaluate Does this result make sense? 3 If the reaction was first order in A, doubling the concentration would double the rate. However, Rate2 is four times Rate1. So, the reaction is second order for A and second order overall because A is the only reactant. As a further test, look at what happens to the rate when the concentration doubles again from 0.10 mol/L to 0.20 mol/L. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

aA  bB + cC The following reaction is a second-order reaction.
If the initial concentration of A is 2.0 mol/L and the initial rate is 9.6  10–7 mol/(L·s), what is the concentration when the rate is 1.2  10–7 mol/(L·s)? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

aA  bB + cC The following reaction is a second-order reaction.
If the initial concentration of A is 2.0 mol/L and the initial rate is 9.6  10–7 mol/(L·s), what is the concentration when the rate is 1.2  10–7 mol/(L·s)? 1.2  10–7 mol/(L·s) 9.6  10–7 mol/(L·s) = ( ) [A2] 2.0 mol/L 2 [A2] = 0.71 mol/L Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

How do most reactions progress from start to finish?
Reaction Mechanisms Reaction Mechanisms How do most reactions progress from start to finish? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

A balanced equation does not tell you how a reaction occurred.
Reaction Mechanisms A balanced equation does not tell you how a reaction occurred. Plants use photosynthesis to capture and store light energy. The process can be summarized by stating that carbon dioxide and water yield simple sugars and oxygen. However, the process of photosynthesis is not as simple as this summary implies. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

One-Step and Multistep Reactions
Reaction Mechanisms One-Step and Multistep Reactions An elementary reaction is a reaction in which reactants are converted to products in a single step. This type of reaction has only one activation-energy peak and one activated complex. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Reaction Mechanisms One-Step and Multistep Reactions Most chemical reactions consist of two or more elementary reactions. The series of elementary reactions or steps that take place during the course of a complex reaction is called a reaction mechanism. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Reaction Mechanisms One-Step and Multistep Reactions An intermediate is a product of one step in a reaction mechanism and a reactant in the next step. An intermediate has a more stable structure and longer lifetime than an activated complex. Intermediates do not appear in the overall chemical equation for a reaction. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Interpret Graphs The figure below shows a reaction progress curve for a complex chemical reaction. A reaction progress curve shows all the energy changes that occur as reactants are converted to products. The graph has a peak for each activated complex and a valley for each intermediate. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

CHEMISTRY & YOU In the mountain stage of the Tour de France, a rider encounters a series of peaks and valleys. In terms of energy, how does the trip through the mountains compare to what happens during a multistep reaction? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

CHEMISTRY & YOU In the mountain stage of the Tour de France, a rider encounters a series of peaks and valleys. In terms of energy, how does the trip through the mountains compare to what happens during a multistep reaction? Riders need extra energy each time they must climb a peak. This extra energy compares to the activation energy needed in a chemical reaction. Each time they ride down into a valley, they are in an intermediate state. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Rate-Determining Steps
Reaction Mechanisms Rate-Determining Steps In a multistep chemical reaction, the steps do not all progress at the same rate. One step will be slower than the others. The slowest step will determine, or limit, the rate of the overall reaction. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Reaction Mechanisms Rate-Determining Steps Consider the reaction mechanism for the decomposition of nitrous oxide (N2O). Experiments have shown that the mechanism consists of the two steps shown below. N2O(g) N2(g) +O(g) N2O(g) + O(g) N2(g) + O2(g) 2N2O (g) 2N2(g) + O2(g) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Reaction Mechanisms Rate-Determining Steps N2O(g) N2(g) +O(g) N2O(g) + O(g) N2(g) + O2(g) 2N2O (g) 2N2(g) + O2(g) In the first step, nitrous oxide decomposes into nitrogen gas and oxygen atoms. The oxygen atoms are an intermediate. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Reaction Mechanisms Rate-Determining Steps N2O(g) N2(g) +O(g) N2O(g) + O(g) N2(g) + O2(g) 2N2O (g) 2N2(g) + O2(g) For the decomposition of nitrous oxide, the first step is the rate-determining step. To increase the rate of the overall reaction, you would need to increase the rate of the first step. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

In the following reaction mechanism, which is an intermediate?
A2 2A 2A + B2 2AB A2 + B2 2AB A. A2 C. A B. B2 D. AB Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Key Concepts and Key Equations
The value of the specific rate constant, k, in a rate law is large if the products form quickly; the value is small if the products form slowly. Most chemical reactions consist of two or more elementary reactions. Rate = = k  [A] DA Dt Rate = k[A]x[B]y Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Glossary Terms rate law: an expression relating the rate of a reaction to the concentration of the reactants specific rate constant: a proportionality constant relating the concentrations of reactants to the rate of the reaction first-order reaction: a reaction in which the reaction rate is proportional to the concentration of only one reactant Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Glossary Terms elementary reaction: a reaction in which reactants are converted to products in a single step reaction mechanism: a series of elementary reactions that take place during the course of a complex reaction intermediate: a product of one of the steps in a reaction mechanism; it becomes a reactant in the next step Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Chapter 18 Reaction Rates and Equilibrium 18.3 Reversible Reactions

How did chemists help farmers produce more food?
CHEMISTRY & YOU How did chemists help farmers produce more food? Fertilizers can increase the amount of a crop per unit of land. Most fertilizers contain ammonia or nitrogen compounds made from ammonia. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

This inference is not true. Some reactions are reversible.
Reversible Reactions You may have inferred that chemical reactions always progress in one direction. This inference is not true. Some reactions are reversible. A reversible reaction is one in which the conversion of reactants to products and the conversion of products to reactants occur at the same time. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Here is an example of a reversible reaction.
Reversible Reactions Here is an example of a reversible reaction. 2SO2(g) + O2(g)  2SO3(g) 2SO2(g) + O2(g)  2SO3(g) The first reaction is called the forward reaction. The second reaction is called the reverse reaction. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

The two equations can be combined into one using a double arrow.
Reversible Reactions 2SO2(g) + O2(g)  2SO3(g) 2SO2(g) + O2(g)  2SO3(g) The two equations can be combined into one using a double arrow. 2SO2(g) + O2(g) 2SO3(g) The double arrow tells you that the reaction is reversible. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Establishing Equilibrium
Reversible Reactions Establishing Equilibrium When the rates of the forward and reverse reactions are equal, the reaction has reached a state of balance called chemical equilibrium. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Notice that after a certain time, the concentrations remain constant.
Interpret Graphs This graph shows the progress of a reaction that starts with concentrations of SO2 and O2, but with zero SO3. This graph shows the progress of the reaction that begins with an initial concentration of SO3, and zero concentrations for SO2 and O2. Notice that after a certain time, the concentrations remain constant. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Conditions at Equilibrium
Reversible Reactions Conditions at Equilibrium Chemical equilibrium is a dynamic state. When the store opens, only the forward reaction occurs as shoppers head to the second floor. Equilibrium is reached when the rate at which shoppers move from the first floor to the second is equal to the rate at which shoppers move from the second floor to the first. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Conditions at Equilibrium
Reversible Reactions Conditions at Equilibrium At chemical equilibrium, both the forward and reverse reactions continue, but because their rates are equal, no net change occurs in the concentrations of the reaction components. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Concentrations at Equilibrium
Reversible Reactions Concentrations at Equilibrium Although the rates of the forward and reverse reactions are equal at equilibrium, the concentrations of the components usually are not. The relative concentrations of the reactants and products at equilibrium mark the equilibrium position of a reaction. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Reversible Reactions Concentrations at Equilibrium The equilibrium position tells you whether the forward or reverse reaction is more likely to happen. Suppose a single reactant, A, forms a single product, B. If the equilibrium mixture contains 1% A and 99% B, then the formation of B is said to be favored. A B 1% % Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Reversible Reactions Concentrations at Equilibrium In principle, almost all reactions are reversible to some extent under the right conditions. In practice, one set of components is often so favored at equilibrium that the other set cannot be detected. When no reactants can be detected, you can say that the reaction has gone to completion, or is irreversible. When no products can be detected, you can say that no reaction has taken place. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Why is equilibrium considered to be a dynamic state?

Why is equilibrium considered to be a dynamic state?
Both the forward and reverse reactions are constantly taking place, but their rates are equal, so no net change occurs in the concentrations of the products or reactants. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Factors Affecting Equilibrium: Le Châtelier’s Principle
What three stresses can cause a change in the equilibrium position of a chemical system? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

The French chemist Henri Le Châtelier (1850–1936) proposed what has come to be called Le Châtelier’s principle: If a stress is applied to a system in dynamic equilibrium, the system changes in a way that relieves the stress. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Stresses that upset the equilibrium of a chemical system include changes in the concentration of reactants or products, changes in temperature, and changes in pressure. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Concentration Changing the amount, or concentration, of any reactant or product in a system at equilibrium disturbs the equilibrium. The system will adjust to minimize the effects of the change. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Concentration Consider the decomposition of carbonic acid (H2CO3) in aqueous solution. H2CO3(aq) CO2(aq) + H2O(l) < 1% > 99% The system has reached equilibrium. The amount of carbonic acid is less than 1%. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Suppose carbon dioxide is added to the system. H2CO3(aq) CO2(aq) + H2O(l) Add CO2 Direction of shift This increase in the concentration of CO2 causes the rate of the reverse reaction to increase. Adding a product to a reaction at equilibrium pushes a reversible reaction in the direction of the reactants. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Suppose carbon dioxide is removed. H2CO3(aq) CO2(aq) + H2O(l) Add CO2 Direction of shift Remove CO2 This decrease in the concentration of CO2 causes the rate of the reverse reaction to decrease. Removing a product always pulls a reversible reaction in the direction of the products. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

An equilibrium between carbonic acid, carbon dioxide, and water exists in your blood. During exercise, the concentration of CO2 in the blood increases. This shifts the equilibrium in the direction of carbonic acid. The increase in the level of CO2 also triggers an increase in the rate of breathing. With more breaths per minute, more CO2 is removed through the lungs. The removal of CO2 causes the equilibrium to shift toward the products, which reduces the amount of H2CO3. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Temperature Increasing the temperature causes the equilibrium position of a reaction to shift in the direction that absorbs heat. In other words, it will shift in the direction that reduces the stress. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Temperature N2(g) + 3H2(g) NH3(g) + heat Add heat Direction of shift Remove heat (cool) Heat can be considered to be a product, just like NH3. Heating the reaction mixture at equilibrium pushes the equilibrium position to the left, which favors the reactants. Cooling, or removing heat, pulls the equilibrium position to the right, and the product yield increases. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Pressure Equilibrium systems in which some reactants and products are gases can be affected by a change in pressure. A shift will occur only if there are an unequal number of moles of gas on each side of the equation. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Pressure When the plunger is pushed down, the volume decreases and the pressure increases. Initial equilibrium Equilibrium is disturbed by an increase in pressure. A new equilibrium position is established with fewer molecules. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Pressure You can predict which way the equilibrium position will shift by comparing the number of molecules of reactants and products. N2(g) + 3H2(g) NH3(g) Add pressure Direction of shift Reduce pressure When two molecules of ammonia form, four molecules of reactants are used up. A shift toward ammonia (the product) will reduce the number of molecules. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

CHEMISTRY & YOU Fritz Haber and Karl Bosch figured out how to increase the yield of ammonia when nitrogen and hydrogen react. Their success came from controlling the temperature and pressure. In which direction did they adjust each factor and why? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

CHEMISTRY & YOU Fritz Haber and Karl Bosch figured out how to increase the yield of ammonia when nitrogen and hydrogen react. Their success came from controlling the temperature and pressure. In which direction did they adjust each factor and why? An increase in pressure and a decrease in temperature would increase the yield of ammonia by shifting the equilibrium toward the production of ammonia. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Catalysts and Equilibrium Catalysts decrease the time it takes to establish equilibrium. However, they do not affect the amounts of reactants and products present at equilibrium. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

PCl5(g) + heat PCl3(g) + Cl2(g)
Sample Problem 18.2 Applying Le Châtelier’s Principle What effect will each of the following changes have on the equilibrium position for this reversible reaction? PCl5(g) + heat PCl3(g) + Cl2(g) a. Cl2 is added. b. Pressure is increased. c. Heat is removed. d. PCl3 is removed as it forms. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Analyze Identify the relevant concepts.
Sample Problem 18.2 Analyze Identify the relevant concepts. 1 According to Le Châtelier’s principle, the equilibrium position will shift in a direction that minimizes the imposed stress. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 18.2 Solve Apply the concepts to this problem. 2 Start with the addition of Cl2. Cl2 is a product. Increasing the concentration of a product shifts the equilibrium to the left. PCl5(g) + heat PCl3(g) + Cl2(g) Add Cl2 Direction of shift Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 18.2 Solve Apply the concepts to this problem. 2 Analyze the effect of an increase in pressure. Reducing the number of molecules that are gases decreases the pressure. The equilibrium shifts to the left. For a change in pressure, compare the number of molecules of gas molecules on both sides of the equation. Increase pressure Direction of shift PCl5(g) + heat PCl3(g) + Cl2(g) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 18.2 Solve Apply the concepts to this problem. 2 Analyze the effect of removing heat. The reverse reaction produces heat. The removal of heat causes the equilibrium to shift to the left. Remove heat Direction of shift PCl5(g) + heat PCl3(g) + Cl2(g) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 18.2 Solve Apply the concepts to this problem. 2 Analyze the effect of removing PCl3. PCl3 is a product. Removal of a product as it forms causes the equilibrium to shift to the right. Remove PCl3 Direction of shift PCl5(g) + heat PCl3(g) + Cl2(g) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

4HCl(g) + O2(g) 2Cl2(g) +2H2O(g)
In the following equilibrium reaction, in which direction would the equilibrium position shift with an increase in pressure? 4HCl(g) + O2(g) Cl2(g) +2H2O(g) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

4HCl(g) + O2(g) 2Cl2(g) +2H2O(g)
In the following equilibrium reaction, in which direction would the equilibrium position shift with an increase in pressure? 4HCl(g) + O2(g) Cl2(g) +2H2O(g) Reducing the number of molecules that are gases decreases the pressure. The equilibrium will shift to the right. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Equilibrium Constants
What does the size of an equilibrium constant indicate about a system at equilibrium? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Chemists express the equilibrium position as a numerical value. This value relates the amounts of reactants to products at equilibrium. In this general reaction, the coefficients a, b, c, and d represent the number of moles. aA + bB cC + dD Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

The equilibrium constant (Keq) is the ratio of product concentrations to reactant concentrations at equilibrium. aA + bB cC + dD From the general equation, each concentration is raised to a power equal to the number of moles of that substance in the balanced chemical equation. Keq = [C]c x [D]d [A]a x [B]b Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

The value of Keq depends on the temperature of the reaction. The flask on the left is in a dish of hot water. The flask on the right is in ice. Dinitrogen tetroxide is a colorless gas. Nitrogen dioxide is a brown gas. N2O4(g) NO2(g) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

The size of the equilibrium constant indicates whether reactants or products are more common at equilibrium. When Keq has a large value, such as 3.1 x 1011, the reaction mixture at equilibrium will consist mainly of product. When Keq has a small value, such as 3.1 x 10–11, the mixture at equilibrium will consist mainly of reactant. When Keq has an intermediate value, such as 0.15 or 50, the mixture will have significant amounts of both reactant and product. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Expressing and Calculating Keq
Sample Problem 18.3 Expressing and Calculating Keq The colorless gas dinitrogen tetroxide (N2O4) and the brown gas nitrogen dioxide (NO2) exist in equilibrium with each other. N2O4(g) NO2(g) A liter of the gas mixture at equilibrium contains mol of N2O4 and mol of NO2 at 10oC. Write the expression for the equilibrium constant (Keq) and calculate the value of the constant for the reaction. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 18.3 Analyze List the knowns and the unknowns. 1 Modify the general expression for the equilibrium constant and substitute the known concentrations to calculate Keq. KNOWNS UNKNOWN [N2O4] = mol/L [NO2] = mol/L Keq (algebraic expression) = ? Keq (numerical value) = ? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 18.3 Calculate Solve for the unknowns. 2 Start with the general expression for the equilibrium constant. Place the concentration of the product in the numerator and the concentration of the reactant in the denominator. Raise each concentration to the power equal to its coefficient in the chemical equation. Keq = [C]c x [D]d [A]a x [B]b Write the equilibrium constant expression for this reaction. Keq = [NO2]2 [N2O2] Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 18.3 Calculate Solve for the unknowns. 2 Substitute the concentrations that are known and calculate Keq. Keq = (0.030 mol/L)2 ( mol/L) = (0.030 mol/L x mol/L) Keq = 0.20 mol/L = 0.20 You can ignore the unit mol/L; chemists report equilibrium constants without a stated unit. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Evaluate Does the result make sense?
Sample Problem 18.3 Evaluate Does the result make sense? 3 Each concentration is raised to the correct power. The numerical value of the constant is correctly expressed to two significant figures. The value for Keq is appropriate for an equilibrium mixture that contains significant amounts of both gases. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Finding the Equilibrium Constant
Sample Problem 18.4 Finding the Equilibrium Constant One mole of colorless hydrogen gas and one mole of violet iodine vapor are sealed in a 1-L flask and allowed to react at 450oC. At equilibrium, 1.56 mol of colorless hydrogen iodide is present, together with some of the reactant gases. Calculate Keq for the reaction. H2(g) + I2(g) HI(g) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Analyze List the knowns and the unknown.
Sample Problem 18.4 Analyze List the knowns and the unknown. 1 Find the concentrations of the reactants at equilibrium. Then substitute the equilibrium concentrations in the expression for the equilibrium constant for this reaction. KNOWNS UNKNOWN [H2] (initial) = 1.00 mol/L [I2] (initial) = 1.00 mol/L [HI] (equilibrium) = 1.56 mol/L Keq = ? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Calculate Solve for the unknown.
Sample Problem 18.4 Calculate Solve for the unknown. 2 First find out how much H2 and I2 are consumed in the reaction. x + x = 1.56 mol 2x = 1.56 mol x = mol Let mol H2 used = mol I2 used = x. The number of mol H2 and mol I2 used must equal the number of mol HI formed (1.56 mol). Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 18.4 Calculate Solve for the unknown. 2 Calculate how much H2 and I2 remain in the flask at equilibrium. mol H2 = mol I2 = (1.00 mol – mol) = 0.22 mol Use the general expression for Keq as a guide: Keq = [C]c x [D]d [A]a x [B]b Write the expression for Keq. Keq = [HI]2 [H2] x [I2] Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 18.4 Calculate Solve for the unknown. 2 Substitute the equilibrium concentrations of the reactants and products into the equation and solve for Keq. Keq = (1.56 mol/L)2 0.22 mol/L x 0.22 mol/L 1.56 mol/L x 1.56 mol/L Keq = 5.0 x 101 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 18.4 Evaluate Does the result make sense? 3 Each concentration is raised to the correct power. The value of the constant reflects the presence of significant amounts of the reactions and product in the equilibrium mixture. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Finding Concentrations at Equilibrium
Sample Problem 18.5 Finding Concentrations at Equilibrium Bromine chloride (BrCl) decomposes to form bromine and chlorine. 2BrCl(g) Br2(g) + Cl2(g) At a certain temperature, the equilibrium constant for the reaction is A sample of pure BrCl is placed in a 1-L container and allowed to decompose. At equilibrium, the reaction mixture contains 4.00 mol Cl2. What are the equilibrium concentrations of Br2 and BrCl? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 18.5 Analyze List the knowns and the unknowns. 1 Use the balanced equation, the equilibrium constant, and the equilibrium constant expression to find the unknown concentrations. According to the balanced equation, when BrCl decomposes, equal numbers of moles of Br2 and Cl2 are formed. UNKNOWN KNOWNS [Br2] (equilibrium) = ? mol/L [BrCl] (equilibrium) = ? mol/L [Cl2] (equilibrium) = 4.00 mol/L Keq = 11.1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 18.5 Calculate Solve for the unknowns. 2 The volume of the container is 1 L, so calculate [Br2] at equilibrium. [Br2] = = 4.00 mol/L 4.00 mol 1 L Write the equilibrium expression for the reaction. Keq = [BrCl]2 [Br2] x [Cl2] Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 18.5 Calculate Solve for the unknowns. 2 Rearrange the equation to solve for [BrCl]2. [BrCl]2 = Keq [Br2] x [Cl2] Substitute the known values for Keq, [Br2], and [Cl2]. [BrCl]2 = = 1.44 mol2/L2 11.1 4.00 mol/L x 4.00 mol/L Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 18.5 Calculate Solve for the unknowns. 2 Calculate the square root. [BrCl] = = 1.20 mol/L 1.44 mol2/L2 Use your calculator to find the square root. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 18.5 Evaluate Does the result make sense? 3 It makes sense that the equilibrium concentration of the reactant and the products are both present in significant amounts because Keq has an intermediate value. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

HCl is formed when H2 and Cl2 react at high temperatures.
H2(g) + Cl2(g) HCl(g) At equilibrium, [HCl] = 1.76 x 10–2 mol/L, and [H2] = [Cl2] = 1.60 x 10–3 mol/L. What is the value of the equilibrium constant? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

HCl is formed when H2 and Cl2 react at high temperatures.
H2(g) + Cl2(g) HCl(g) At equilibrium, [HCl] = 1.76 x 10–2 mol/L, and [H2] = [Cl2] = 1.60 x 10–3 mol/L. What is the value of the equilibrium constant? [HCl]2 (1.76 x 10–2 mol/L)2 Keq = = [H2] x [Cl2] (1.60 x 10–3 mol/L) x (1.60 x 10–3 mol/L) Keq = 121 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Key Concepts At chemical equilibrium, both the forward and reverse reactions continue, but because their rates are equal, no net change occurs in the concentrations of the reactant components. Stresses that upset the equilibrium of a chemical system include changes in concentration of reactants or products, changes in temperature, and changes in pressure. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Key Concept and Key Equation
The size of the equilibrium constant indicates whether reactants or products are more common at equilibrium. Keq = [C]c x [D]d [A]a x [B]b Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Glossary Terms reversible reaction: a reaction in which the conversion of reactants into products and the conversion of products into reactants occur simultaneously chemical equilibrium: a state of balance in which the rates of the forward and reverse reactions are equal; no net change in the amount of reactants and products occurs in the chemical system Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Glossary Terms equilibrium position: the relative concentrations of reactants and products of a reaction that has reached equilibrium; indicates whether the reactants or products are favored in the reversible reaction Le Châtelier’s principle: when a stress is applied to a system in dynamic equilibrium, the system changes in a way that relieves the stress Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Glossary Terms equilibrium constant: the ratio of product concentrations to reactant concentrations at equilibrium, with each concentration raised to a power equal to the number of moles of that substance in the balanced chemical equation Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Chapter 18 Reaction Rates and Equilibrium 18.4 Solubility Equilibrium

How is it possible to ingest a poison without being harmed?
CHEMISTRY & YOU How is it possible to ingest a poison without being harmed? Barium sulfate, a poison, can absorb X-rays, so tissues coated with the liquid will appear as light areas in the X-ray images. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

The Solubility Product Constant
What is the relationship between the solubility product constant and the solubility of a compound? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Most ionic compounds containing alkali metals are soluble in water. For example, more than 35 g of sodium chloride will dissolve in only 100 g of water. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Most ionic compounds containing alkali metals are soluble in water. For example, more than 35 g of sodium chloride will dissolve in only 100 g of water. By contrast, some ionic compounds are insoluble in water. For example, compounds that contain phosphate, sulfite, or carbonate ions tend not to dissolve in water. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Solubility of Ionic Compounds in Water
Interpret Data This table provides some general rules for the solubility of ionic compounds in water. Solubility of Ionic Compounds in Water Compounds Solubility Exceptions Salts of Group 1A metals and ammonia Soluble Some lithium compounds Ethanoates, nitrates, chlorates, and perchlorates Few exceptions Sulfates Compounds of Pb, Ag, Hg, Ba, Sr, and Ca Chlorides, bromides, and iodides Compounds of Ag and some compounds of Hg and Pb Sulfides and hydroxides Most are insoluble Alkali metal sulfides and hydroxides are soluble. Compounds of Ba, Sr, and Ca are slightly soluble. Carbonates, phosphates, and sulfites Insoluble Compounds of the alkali metals and of ammonium ions Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Most insoluble ionic compounds will actually dissolve to some extent in water. These compounds are said to be slightly soluble in water. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

When the “insoluble” compound silver chloride is mixed with water, a very small amount of silver chloride dissolves in the water. AgCl(s) Ag+(aq) + Cl–(aq) An equilibrium is established between the solid and the dissolved ions in the saturated solution. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

When the “insoluble” compound silver chloride is mixed with water, a very small amount of silver chloride dissolves in the water. AgCl(s) Ag+(aq) + Cl–(aq) You can write an equilibrium-constant expression for this process. Keq = [Ag+]  [Cl–] [AgCl] Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

To compare the solubility of salts, it is useful to have a constant that reflects only the concentrations of the dissolved ions. This constant is called the solubility product constant (Ksp). It is equal to the product of the concentrations of the ions each raised to a power equal to the coefficient of the ion in the dissociation equation. Ksp = [A]a  [B]b Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

The table below lists the Ksp values for some ionic compounds that are slightly soluble in water. Solubility Product Constants (Ksp) at 25oC Ionic compound Ksp Halides AgCl AgBr AgI PbCl2 PbBr2 PbI2 PbF2 CaF2 1.8  10–10 5.0  10–13 8.3  10–17 1.7  10–5 2.1  10–6 7.9  10–9 3.6  10–8 3.9  10–11 Sulfates PbSO4 BaSO4 CaSO4 6.3  10–7 1.1  10–10 2.4  10–5 Hydroxides Al(OH)3 Zn(OH)2 Ca(OH)2 Mg(OH)2 Fe(OH)2 3.0  10–34 3.0  10–16 6.5  10–6 7.1  10–12 7.9  10–16 Sulfides NiS CuS Ag2S ZnS FeS CdS PbS 4.0  10–20 8.0  10–37 8.0  10–51 3.0  10–23 8.0  10–19 1.0  10–27 3.0  10–28 Carbonates CaCO3 SrCO3 ZnCO3 Ag2CO3 BaCO3 4.5  10–9 9.3  10–10 1.0  10–10 8.15  10–12 5.0  10–9 Chromates PbCrO4 Ag2CrO4 1.8  10–14 1.2  10–12 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Finding the Ion Concentrations in a Saturated Solution
Sample Problem 18.6 Finding the Ion Concentrations in a Saturated Solution What is the concentration of lead ions and chromate ions in a saturated solution of lead(II) chromate at 25oC? (Ksp = 1.8  10–14) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 18.6 Analyze List the knowns and the unknowns. 1 Write the expression for Ksp. Then modify it so that there is a single unknown. KNOWNS Ksp = 1.8  10–14 PbCrO4(s) Pb2+(aq) + CrO42–(aq) UNKNOWN [Pb2+] = ? M [CrO42–] = ? M Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 18.6 Calculate Solve for the unknowns. 2 Start with the general expression for the solubility product constant. The exponent for each ion is 1. Ksp = [A]a  [B]b Use the chemical equation to write the correct expression for Ksp for the reaction. Ksp = [Pb2+]  [CrO42–] = 1.8  10–14 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 18.6 Calculate Solve for the unknowns. 2 Substitute [Pb2+] for [CrO42–] in the expression to get an equation with one unknown. At equilibrium, [Pb2+] = [CrO42–]. Ksp = [Pb2+]  [Pb2+] = [Pb2+]2 = 1.8  10–14 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 18.6 Calculate Solve for the unknowns. 2 Solve for [Pb2+]. [Pb2+] = 1.8  10–14 [Pb2+] = [CrO42–] = 1.3  10–7 M Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 18.6 Evaluate Does the result make sense? 3 Calculate [Pb2+]  [CrO42–] to evaluate the answers. The result is 1.7  10–14, which is close to the value of Ksp. The result varies slightly from the actual value because the answers were rounded to two significant figures. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Which of the following compounds is least soluble at 25oC?
A. PbF2 B. ZnS C. SrCO3 D. CaSO4 Ksp at 25oC Ionic compound Ksp CaSO4 2.4  10–5 PbF2 3.6  10–8 SrCO3 9.3  10–10 ZnS 3.0  10–23 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

The Common Ion Effect The Common Ion Effect How can you predict whether precipitation will occur when two solutions are mixed? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

PbCrO4(s) Pb2+(aq) + CrO42–(aq)
The Common Ion Effect In a saturated solution of lead(II) chromate, an equilibrium is established between the solid lead(II) chromate and its ions in solution. PbCrO4(s) Pb2+(aq) + CrO42–(aq) What would happen if you added some lead nitrate to this solution? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

PbCrO4(s) Pb2+(aq) + CrO42–(aq)
The Common Ion Effect PbCrO4(s) Pb2+(aq) + CrO42–(aq) Lead(II) nitrite is soluble in water, so adding Pb(NO3)2 causes the concentration of lead ion to increase. The addition of lead ions is a stress on the equilibrium. Applying Le Châtelier’s principle, the stress can be relieved if the reaction shifts to the left. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

The Common Ion Effect The yellow solid in the test tube, which is PbCrO4, cannot dissolve because the solution is saturated with Pb2+ and CrO42– ions. When some Pb(NO3)2 is added, the excess lead ions combine with the chromate ions in solution to form additional solid PbCrO4. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

In this example, the lead ion is a common ion.
The Common Ion Effect In this example, the lead ion is a common ion. A common ion is an ion that is found in both ionic compounds in a solution. The lowering of the solubility of an ionic compound as a result of the addition of a common ion is called the common ion effect. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

CHEMISTRY & YOU The Ksp of BaSO4 is 1.1  10–10. Why can a patient ingest the toxic BaSO4 without being harmed? Barium sulfate is not very soluble. Therefore, very little of it can dissolve in bodily fluids and be absorbed as a toxin. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Finding Equilibrium Ion Concentrations in the Presence of a Common Ion
Sample Problem 18.7 Finding Equilibrium Ion Concentrations in the Presence of a Common Ion Small amounts of silver bromide can be added to the lenses used for eyeglasses. The silver bromide causes the lenses to darken in the presence of large amounts of UV light. The Ksp of silver bromide is 5.0  10–13. What is the concentration of bromide ion in a 1.00-L saturated solution of AgBr to which mol of AgNO3 is added? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Analyze List the knowns and the unknown.
Sample Problem 18.7 Analyze List the knowns and the unknown. 1 Use one unknown to express both [Ag+] and [Br–]. Let x be the equilibrium concentration of bromide ion and x be the equilibrium concentration of silver ion. KNOWNS UNKNOWN [Br–] = ? M Ksp = 5.0  10–13 moles of AgNO3 added = mol AgBr(s) Ag+(aq) + Br–(aq) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 18.7 Calculate Solve for the unknown. 2 Write the expression for Ksp. Ksp = [Ag+]  [Br–] Substitute x for [Br–] in the solubility product expression. Ksp = [Ag+]  x Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 18.7 Calculate Solve for the unknown. 2 Rearrange the equation to solve for x. Ksp x = [Ag+] Substitute the values for Ksp and [Ag+] in the expression and solve. Based on the small value of Ksp, you can assume that x will be very small compared to Thus, [Ag+] ≈ M. (5.0  10–13) x = 0.020 [Br–] = 2.5  10–11 M Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 18.7 Evaluate Does the result make sense? 3 The concentration of Br– in a saturated solution of AgBr is 7.0  10–7 M (the square root of the Ksp). It makes sense that the addition of AgNO3 would lower the concentration of Br– because the presence of a common ion, Ag+, causes AgBr to precipitate from solution. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

The Common Ion Effect A precipitate will form if the product of the concentrations of two ions in the mixture is greater than the Ksp value for the compound formed from the ions. As solutions of barium nitrate and sodium sulfate are mixed, a precipitate of BaSO4 is formed. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Does a precipitate form when 1 L of 0
Does a precipitate form when 1 L of M Na2SO4 is mixed with 1 L of M Ba(NO3)2? The Ksp for BaSO4 is 1.1  10–10. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Does a precipitate form when 1 L of 0
Does a precipitate form when 1 L of M Na2SO4 is mixed with 1 L of M Ba(NO3)2? The Ksp for BaSO4 is 1.1  10–10. [Ba2+]  [SO42–] = ( M)  (0.017 M) = 1.0  10–5 The result is larger than the Ksp value for BaSO4, so a precipitate will form. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Key Concepts and Key Equation
The smaller the value of the solubility product constant, the lower the solubility of the compound. A precipitate will form if the product of the concentrations of two ions in the mixture is greater than the Ksp value of the compound formed from the ions. Ksp = [A]a × [B]b Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Glossary Terms solubility product constant: an equilibrium constant applied to the solubility of electrolytes; it is equal to the product of the concentrations of the ions each raised to a power equal to the coefficient of the ion in the dissociation equation common ion: an ion that is common to both salts in a solution; in a solution of silver nitrate and silver chloride, Ag+ would be a common ion common ion effect: a decrease in the solubility of an ionic compound caused by the addition of a common ion Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Chapter 18 Reaction Rates and Equilibrium 18.5 Free Energy and Entropy

How can a fire start on its own?
CHEMISTRY & YOU How can a fire start on its own? Sometimes a fire can occur without an external source of ignition, such as a match or an electrical spark. Spontaneous combustion is the term used to describe these fires. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Some of the energy released in a chemical reaction can be harnessed to do work, such as pushing the pistons in an internal-combustion engine. The energy that is available to do work is called free energy. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Spontaneous Versus Nonspontaneous Reactions You can write a balanced equation for a chemical reaction, but the reaction may not actually take place. CO2(g)  C(s) + O2(g) Experience tells you that this reaction does not tend to occur. Carbon and oxygen react to form carbon dioxide, not the reverse. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Spontaneous Versus Nonspontaneous Reactions The world of balanced chemical equations is really divided into two groups. One group contains equations representing reactions that actually occur. The other contains equations representing reactions that do not tend to occur, or at least not efficiently. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Spontaneous Versus Nonspontaneous Reactions A spontaneous reaction occurs naturally and favors the formation of products at the stated conditions. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Spontaneous Versus Nonspontaneous Reactions Spontaneous reactions produce large amounts of products and release free energy. Fireworks displays are the result of highly favored spontaneous reactions. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Spontaneous Versus Nonspontaneous Reactions A chemical reaction that does not favor the formation of products at the stated conditions is called a nonspontaneous reaction. Such reactions produce little, if any, product. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Reversible Reactions Consider the decomposition of carbonic acid in water. H2CO3(aq) CO2(g) + H2O(l) <1% >99% The forward reaction is spontaneous and releases free energy. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Reversible Reactions Consider the decomposition of carbonic acid in water. H2CO3(aq) CO2(g) + H2O(l) <1% >99% The forward reaction is spontaneous and releases free energy. The combination of carbon dioxide and water to form carbonic acid is a nonspontaneous reaction. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Reversible Reactions When solutions of cadmium nitrate and sodium sulfide are mixed, the products are aqueous sodium nitrate and solid yellow cadmium sulfide. Cd(NO3)2(aq) + Na2S(aq) CdS(s) + 2NaNO3(aq) A precipitate of cadmium sulfide forms spontaneously. The reverse reaction is nonspontaneous. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

The Rate of Spontaneous Reactions The terms spontaneous and nonspontaneous do not refer to the rate of a reaction. Some spontaneous reactions are so slow that they appear to be nonspontaneous. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Changing the conditions of a chemical reaction can affect whether a reaction will occur. A reaction that is nonspontaneous in one set of conditions may be spontaneous in other conditions. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Photosynthesis is a multistep reaction that takes place in plant leaves. Outside of plants, carbon dioxide and water do not normally combine to produce sugar and oxygen. This complex process could not happen without the energy supplied by sunlight and plant pigments such as chlorophyll. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Coupled Reactions Sometimes a nonspontaneous reaction can be made to occur if it is coupled to a spontaneous reaction. One reaction releases energy that is used by the other reaction. Coupled reactions are common in the complex biological processes that take place in living organisms. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

CHEMISTRY & YOU Decomposition reactions that occur inside a pile of oily rags or a damp stack of hay cause heat to build up. If the heat cannot escape, the temperature within the pile or stack will rise. How can a rise in temperature cause a fire to start on its own? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

CHEMISTRY & YOU Decomposition reactions that occur inside a pile of oily rags or a damp stack of hay cause heat to build up. If the heat cannot escape, the temperature within the pile or stack will rise. How can a rise in temperature cause a fire to start on its own? The combustion reaction is a nonspontaneous reaction that can be made to occur when it is coupled to the spontaneous decomposition reaction. The decomposition reaction releases energy that is used by the combustion reaction. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Which of the following is ALWAYS true of spontaneous reactions?
They produce heat and are not reversible at the stated conditions. They release free energy and favor the formation of products at the stated conditions. They are coupled with a nonspontaneous reaction and are easily reversible at the stated conditions. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

What part does entropy play in a reaction?

Entropy You might expect that only exothermic reactions are spontaneous. Some processes, however, are spontaneous even though they absorb heat. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Entropy You might expect that only exothermic reactions are spontaneous. Some processes, however, are spontaneous even though they absorb heat. Consider what happens as ice melts. As it changes from a solid to a liquid, 1 mol of ice at 25oC absorbs 6.0 kJ of heat from its surroundings. If you consider only enthalpy changes, it is difficult to explain why the ice melts. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Entropy Some factor other than the enthalpy change must help determine whether a physical or chemical process is spontaneous. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Entropy Some factor other than the enthalpy change must help determine whether a physical or chemical process is spontaneous. The other factor is related to order. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Entropy is a measure of the disorder of a system.
The law of disorder states that the natural tendency is for systems to move in the direction of increasing disorder or randomness. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

This situation represents disorder.
Entropy A dog walker with several dogs could represent relative order and disorder. This situation represents disorder. All of the dogs are on leashes and are strolling orderly along the path. The dogs are no longer wearing leashes and are running freely. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Entropy can affect the direction of a reaction.
Reactions in which entropy increases as reactants form products tend to be favored. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Entropy For a given substance, the entropy of the gas is greater than the entropy of the liquid or the solid. Thus, entropy increases in reactions in which solid reactants form liquid or gaseous products. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Entropy increases when a substance is divided into parts.
For instance, entropy increases when an ionic compound dissolves in water. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Entropy Entropy tends to increase in chemical reactions in which the total number of product molecules is greater than the total number of reactant molecules. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Entropy Entropy tends to increase when the temperature increases. As the temperature rises, the molecules move faster and faster, which increases the disorder. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

A. 2NH4NO3(s)  2N2(g) + 4H2O(l) +O2(g) B. 2H2(g) + O2(g)  2H2O(l)
Which of the following would have an increase in the entropy of the reaction system? A. 2NH4NO3(s)  2N2(g) + 4H2O(l) +O2(g) B. 2H2(g) + O2(g)  2H2O(l) C. C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l) D. 2Fe(s) + O2(g) + 2H2O(l)  2Fe(OH)2(s) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

What two factors determine whether a reaction is spontaneous?
Enthalpy and Entropy Enthalpy and Entropy What two factors determine whether a reaction is spontaneous? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Enthalpy and Entropy The size and direction of enthalpy changes and entropy changes together determine whether a reaction is spontaneous. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Consider an exothermic reaction in which entropy increases.
Enthalpy and Entropy Consider an exothermic reaction in which entropy increases. The reaction will be spontaneous because both factors are favorable. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

A reaction can be spontaneous if:
Enthalpy and Entropy A reaction can be spontaneous if: A decrease in entropy is offset by a large release of heat. An increase in enthalpy is offset by an increase in entropy. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

How Enthalpy Changes and Entropy Changes Affect Reaction Spontaneity
Enthalpy and Entropy The table below summarizes the effect of enthalpy and entropy changes on the spontaneity of reactions. How Enthalpy Changes and Entropy Changes Affect Reaction Spontaneity Enthalpy change Entropy change Is the reaction spontaneous? Decreases (exothermic) Increases (more disorder in products than in reactants) Yes Increases (endothermic) Increases Only if unfavorable enthalpy change is offset by favorable entropy change Decreases (less disorder in products than in reactants) Only if unfavorable entropy change is offset by favorable enthalpy change Decreases No Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Would the following exothermic reaction be spontaneous
Would the following exothermic reaction be spontaneous? Explain why or why not. 2KClO3(s) KCl(s) +3O2(g) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Would the following exothermic reaction be spontaneous
Would the following exothermic reaction be spontaneous? Explain why or why not. 2KClO3(s) KCl(s) +3O2(g) Two molecules of solid are transformed into 2 molecules of solid and 3 molecules of gas, so entropy is increased in the reaction. A reaction that is exothermic with an increase in entropy will be spontaneous. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

How is the value of ΔG related to the spontaneity of a reaction?
Free Energy Change Free Energy Change How is the value of ΔG related to the spontaneity of a reaction? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Free energy is often expressed as Gibbs free energy.
Free Energy Change Free energy is often expressed as Gibbs free energy. This term is named for Josiah Gibbs, the scientist who defined this thermodynamic property. The symbol for Gibbs free energy is G. Free energy can either be released or absorbed during a physical or chemical process. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Free Energy Change The equation below is used to calculate the change in Gibbs free energy (ΔG). ΔS is the change in entropy. ΔH is the change in enthalpy. T is the temperature in Kelvins. ΔG = ΔH – TΔS Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Free Energy Change When the value of ΔG is negative, the process is spontaneous. When the value is positive, the process is nonspontaneous. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

The entropy change for the following reaction at 298 K is 3
The entropy change for the following reaction at 298 K is 3.0 J/mol·K, and the enthalpy change is –394 kJ/mol. C(s) + O2(g) CO2(g) Calculate the Gibbs free energy change and determine whether the reaction will occur spontaneously. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

The entropy change for the following reaction at 298 K is 3
The entropy change for the following reaction at 298 K is 3.0 J/mol·K, and the enthalpy change is –394 kJ/mol. C(s) + O2(g) CO2(g) Calculate the Gibbs free energy change and determine whether the reaction will occur spontaneously. ΔG = –394 kJ/mol – (298 K  kJ/mol·K) ΔG = –395 kJ/mol The reaction is spontaneous. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Key Concepts Spontaneous reactions produce large amounts of products and release free energy. Reactions in which entropy increases as reactants form products tend to be favored. The size and direction of enthalpy changes and entropy changes together determine whether a reaction is spontaneous. When the value of ΔG is negative, a process is spontaneous. When the value is positive, a process is nonspontaneous. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

ΔG = ΔH – TΔS Key Equation

free energy: the energy available to do work
Glossary Terms free energy: the energy available to do work spontaneous reaction: a reaction that favors the formation of products at the specified conditions; spontaneity depends on enthalpy and entropy changes nonspontaneous reaction: a reaction that does not favor the formation of products at the specified conditions Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Glossary Terms entropy: a measure of the disorder of a system; systems tend to go from a state of order (low entropy) to a state of maximum disorder (high entropy) law of disorder: it is a natural tendency of systems to move in the direction of maximum chaos or disorder Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

BIG IDEA Chemical Reactions, Matter, and Energy Changes in enthalpy and entropy can be used to explain why some reactions occur naturally and others do not. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Chapter 18 Reaction Rates and Equilibrium 18.1 Rates of Reaction

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Chapter 18 Reaction Rates and Equilibrium 18.1 Rates of Reaction

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