Chemistry Chapter 14 Notes



Chemistry Chapter 14 Notes

Section 1 Reversible Reactions and Equilibrium

Objectives

• Contrast reactions that go to completion with reversible ones.

• Describe chemical equilibrium.

• Give examples of chemical equilibria that involve complex ions.

Completion Reactions and Reversible Reactions

• If enough oxygen gas is provided for the following reaction, almost all of the sulfur will react:

S8(s) + 8O2(g) → 8SO2(g)

• Reactions such as this one, in which almost all of the reactants react, are called completion reactions.

• In other reactions, called reversible reactions, the products can re-form reactants.

Reversible Reactions Reach Equilibrium

• One reversible reaction occurs when you mix solutions of calcium chloride and sodium sulfate.

CaCl2(aq) + Na2SO4(aq) → CaSO4(s) + 2NaCl(aq)

• The net ionic equation best describes what happens.

• Solid calcium sulfate, the product, can break down to make calcium ions and sulfate ions in a reaction that is the reverse of the previous one.

• Use arrows that point in opposite directions when writing a chemical equation for a reversible reaction.

• The reactions occur at the same rate after the initial mixing of CaCl2 and Na2SO4.

• The amounts of the products and reactants do not change.

• Chemical equilibrium is a state of balance in which the rate of a forward reaction equals the rate of the reverse reaction and the concentrations of products and reactants remain unchanged.

Opposing Reaction Rates Are Equal at Equilibrium

• The reaction of hydrogen, H2, and iodine, I2, to form hydrogen iodide, HI, reaches chemical equilibrium.

Only a very small fraction of the collisions between H2 and I2 result in the formation of HI.

H2(g) + I2(g) → 2HI(g)

• After some time, the concentration of HI goes up.

• As a result, fewer collisions occur between H2 and I2 molecules, and the rate of the forward reaction drops.

• Similarly, in the beginning, few HI molecules exist in the system, so they rarely collide with each other.

• As more HI molecules are made, they collide more often and form H2 and I2 by the reverse reaction.

2HI(g) → H2(g) + I2(g)

• The greater the number of HI molecules that form, the more often the reverse reaction occurs.

• Rate Comparison for H2(g) + I2(g) ( 2HI(g)

[pic]

• When the forward rate and the reverse rate are equal, the system is at chemical equilibrium.

• If you repeated this experiment at the same temperature, starting with a similar amount of pure HI instead of the H2 and I2, the reaction would reach chemical equilibrium again and produce the same concentrations of each substance.

Chemical Equilibria Are Dynamic

• If you drop a ball into a bowl, it will bounce.

• When the ball comes to a stop it has reached static equilibrium, a state in which nothing changes.

• Chemical equilibrium is different from static equilibrium because it is dynamic.

• In a dynamic equilibrium, there is no net change in the system.

• Two opposite changes occur at the same time.

• In equilibrium, an atom may change from being part of the products to part of the reactants many times.

• But the overall concentrations of products and reactants stay the same.

• For chemical equilibrium to be maintained, the rates of the forward and reverse reactions must be equal.

• Arrows of equal length also show equilibrium.

• In some cases, the equilibrium has a higher concentration of products than reactants.

• This type of equilibrium is also shown by using two arrows.

• The forward reaction has a longer arrow to show that the products are favored.

More Examples of Equilibria

• Even when systems are not in equilibrium, they are continuously changing in order to reach equilibrium.

• For example, combustion produces carbon dioxide, CO2, and poisonous carbon monoxide, CO. After combustion, a reversible reaction produces soot.

• This reaction of gases and a solid will reach chemical equilibrium.

• Equilibria can involve any state of matter, including aqueous solutions.

Equilibria Involving Complex Ions

• Complex ion, or coordination compound, is the name given to any metal atom or ion that is bonded to more than one atom or molecule.

• Some ions have a metal ion surrounded by ligands, molecules or anions that readily bond to metal ions.

• Complex ions may be positively charged cations or negatively charged anions.

• In this complex ion, [Cu(NH3)4]2+, ammonia molecules bond to the central copper(II) ion.

[pic]Complex ions formed from transition metals are often deeply colored. [pic]

• The charge on a complex ion is a sum of the charges on the species from which the complex ion forms.

• For example, when the cobalt ion, Co2+, bonds with four Cl− ligands, the total charge is (+2) + 4(−1) = −2.

• Metal ions and ligands can form complexes that have no charge. These are not complex ions.

• Complex ions often form in systems that reach equilibrium.

• Consider zinc nitrate dissolving in water:

• In the absence of other ligands, water molecules bond with zinc ions. So, this reaction can be written:

• If another ligand, such as CN−, is added, the new system will again reach chemical equilibrium.

• Both water molecules and cyanide ions “compete” to bond with zinc ions, as shown in the equation below.

• All of these ions are colorless, so you cannot see which complex ion has the greater concentration.

• In the chemical equilibrium of nickel ions, ammonia, and water, the complex ions have different colors.

• You can see which ion has the greater concentration.

green blue-violet

• The starting concentration of NH3 will determine which one will have the greater concentration.

Section 2 Systems at Equilibrium

Objectives

• Write Keq expressions for reactions in equilibrium, and perform calculations with them.

• Write Ksp expressions for the solubility of slightly soluble salts, and perform calculations with them.

The Equilibrium Constant, Keq

• Limestone caverns form as rainwater, slightly acidified by H3O+, dissolves calcium carbonate.

• The reverse reaction also takes place, depositing calcium carbonate and forming stalactites and stalagmites.

• When the rates of the forward and reverse reactions become equal, the reaction reaches chemical equilibrium.

• There is a mathematical relationship between product and reactant concentrations at equilibrium.

• For limestone reacting with acidified water at 25°C:

• Keq is the equilibrium constant of the reaction.

• Keq for a reaction is unitless, applies only to systems in equilibrium, and depends on temperature and

must be found experimentally or from tables.

Determining Keq for Reactions at Chemical Equilibrium

1. Write a balanced chemical equation.

• Make sure that the reaction is at equilibrium before you write a chemical equation.

2. Write an equilibrium expression.

1. To write the expression, place the product concentrations in the numerator and the reactant concentrations in the denominator.

• The concentration of any solid or a pure liquid that takes part in the reaction is left out.

• For a reaction occurring in aqueous solution, water is omitted.

3. Complete the equilibrium expression.

• Finally, raise each substance’s concentration to the power equal to the substance’s coefficient in the balanced chemical equation.

• Sample Problem A ( page 504)

• An aqueous solution of carbonic acid reacts to reach equilibrium as described below.

• The solution contains the following solution concentrations: carbonic acid, 3.3 × 10−2 mol/L; bicarbonate ion, 1.19 × 10−4 mol/L; and hydronium ion, 1.19 × 10−4 mol/L. Determine the Keq.

For this reaction, the equilibrium constant expression is

Substitute the concentrations into the expression.

Keq Shows If the Reaction Is Favorable

• When Keq is large, the numerator of the equilibrium constant expression is larger than the denominator.

• Thus, the concentrations of the products will usually be greater than those of the reactants.

• In other words, when a reaction that has a large Keq reaches equilibrium, there will be mostly products.

• Reactions in which more products form than reactants form are said to be “favorable.”

• The synthesis of ammonia is very favorable at 25°C and has a large Keq value.

• However, the reaction of oxygen and nitrogen to give nitrogen monoxide is not favorable at 25°C.

• When Keq is small, the denominator of the equilibrium constant expression is larger than the numerator.

• The larger denominator shows that the concentrations of reactants at chemical equilibrium may be greater than those of products.

• A reaction that has larger concentrations of reactants than concentrations of products is an “unfavorable” reaction.

These pie charts show the relative amounts of reactants and products for three Keq values of a reaction.

[pic]

Calculating Concentrations of Products from Keq and Concentrations of Reactants

Sample Problem B (page 506)

The Solubility Product Constant, Ksp

• The maximum concentration of a salt in an aqueous solution is called the solubility of the salt in water.

• Solubilities can be expressed in moles of solute per liter of solution (mol/L or M).

• For example, the solubility of calcium fluoride in water is 3.4 × 10−4 mol/L.

• So, 0.00034 mol of CaF2 will dissolve in 1 L of water to give a saturated solution.

• If you try to dissolve 0.00100 mol of CaF2 in 1 L of water, 0.00066 mol of CaF2 will remain undissolved.

• Calcium fluoride is one of a large class of salts that are said to be slightly soluble in water.

• The ions in solution and any solid salt are at equilibrium.

Solids are not a part of equilibrium constant expressions, so Keq for this reaction is the product

of [Ca2+] and [F−]2, which is equal to a constant.

• Equilibrium constants for the dissolution of slightly soluble salts are called solubility product constants, Ksp, and have no units.

• The Ksp for calcium fluoride at 25°C is 1.6 ( 10−10.

Ksp = [Ca2+][F−]2 = 1.6 ( 10−10

• This relationship is true whenever calcium ions and fluoride ions are in equilibrium with calcium fluoride, not just when the salt dissolves.

• For example, if you mix solutions of calcium nitrate and sodium fluoride, calcium fluoride precipitates.

• The net ionic equation for this precipitation is the reverse of the dissolution.

• This equation is the same equilibrium. So, the Ksp for the dissolution of CaF2 in this system is the same and is 1.6 × 10−10.

[pic]

Determining Ksp for Reactions at Chemical Equilibrium

1. Write a balanced chemical equation.

• Solubility product is only for salts that have low solubility. Soluble salts do not have Ksp values.

• Make sure that the reaction is at equilibrium.

• Equations are always written so that the solid salt is the reactant and the ions are products.

2. Write a solubility product expression.

• Write the product of the ion concentrations.

• Concentrations of solids or liquids are omitted.

3. Complete the solubility product expression.

1. Raise each concentration to a power equal to the substance’s coefficient in the balanced chemical equation.

Sample Problem C ( page 509)

Sample Problem D (page 510)

Using Ksp to Make Magnesium

• Though slightly soluble hydroxides are not salts, they have solubility product constants.

• Magnesium hydroxide is an example.

[Mg2+][OH−]2 = Ksp = 1.8 × 10−11

• This equilibrium is the basis for obtaining

magnesium.

• The table below lists the most abundant ions in ocean water and their concentrations.

• Mg2+ is the third most abundant ion in the ocean.

[pic]

• To get magnesium, calcium hydroxide is added to sea water, which raises the hydroxide ion concentration to a large value so that [Mg2+][OH−]2 would be greater than 1.8 × 10−11.

• Magnesium hydroxide precipitates.

• Magnesium hydroxide is treated with hydrochloric acid to make magnesium chloride, MgCl2.

• Finally, magnesium is obtained by the electrolysis of MgCl2 in the molten state.

• One cubic meter of sea water yields 1 kg of magnesium metal.

• Because of magnesium’s low density and rigidity, alloys of magnesium are used when light weight and strength are needed.

Section 3 Equilibrium Systems and Stress

Objectives

• State Le Châtelier’s principle.

• Apply Le Châtelier’s principle to determine whether the forward or reverse reaction is favored when a stress such as concentration, temperature, or pressure is applied to an equilibrium system.

• Discuss the common-ion effect in the context of Le Châtelier’s principle.

• Discuss practical uses of Le Châtelier’s principle.

• Stress is another word for something that causes a change in a system at equilibrium.

• Chemical equilibrium can be disturbed by a stress, but the system soon reaches a new equilibrium.

• Le Châtelier’s principle states that when a system at equilibrium is disturbed, the system adjusts in a way to reduce the change.

• Chemical equilibria respond to three kinds of stress:

• changes in the concentrations of reactants or products

• changes in temperature

• changes in pressure

• When a stress is first applied to a system, equilibrium is disturbed and the rates of the forward and backward reactions are no longer equal.

• The system responds to the stress by forming more products or by forming more reactants.

• A new chemical equilibrium is reached when enough reactants or products form.

• At this point, the rates of the forward and backward reactions are equal again.

Changes in Concentration Alter Equilibrium Composition

• If you increase a reactant’s concentration, the system will respond to decrease the concentration of the reactant by changing some of it into product.

• Therefore, the rate of the forward reaction must be greater than the rate of the reverse reaction.

• The equilibrium is said to shift right, and the reactant concentration drops until the reaction reaches equilibrium.

• In a reaction of two colored complex ions:

pale blue blue-purple

• When the reaction mixture in a beaker is pale blue, we know that chemical equilibrium favors the formation of reactants.

If ammonia is added, the system responds by forming more product and the solution becomes blue-purple.

[pic]

• The equilibrium below occurs in a bottle of soda.

• After you uncap the bottle, the dissolved carbon dioxide leaves the solution and enters the air.

• The forward reaction rate of this system will increase to produce more CO2.

• This increase in the rate of the forward reaction decreases the concentration of H3O+.

• As a result, the drink gets “flat.”

• If you could increase the concentration of CO2 in the bottle, the reverse reaction rate would increase, and [H3O+]and [HCO3-] would increase.

• Temperature Affects Equilibrium Systems

• The effect of temperature on the gas-phase equilibrium of nitrogen dioxide, NO2, and dinitrogen tetroxide, N2O4, can be seen because of the difference in color of NO2 and N2O4.

• The intense brown NO2 gas is the pollution that is responsible for the colored haze that you sometimes see on smoggy days.

•

•

• Recall that endothermic reactions absorb energy and have positive ∆H values.

• Exothermic reactions release energy and have negative ∆H values.

• The forward reaction is an exothermic process, as the equation below shows.

2NO2(g) → N2O4(g) ∆H = −55.3 kJ

Temperature Changes Affect an Equilibrium System

[pic]

• Now suppose that you heat the flask to 100°C.

• The mixture becomes dark brown because the reverse reaction rate increased to remove some of the energy that you added to the system.

• The equilibrium shifts to the left, toward the formation of NO2.

• Because this reaction is endothermic, the temperature of the flask drops as energy is absorbed.

• This equilibrium shift is true for all exothermic forward reactions:

• Increasing the temperature of an equilibrium mixture usually leads to a shift in favor of the reactants.

• The opposite statement is true for endothermic forward reactions:

• Increasing the temperature of an equilibrium mixture usually leads to a shift in favor of the products.

• The following is an endothermic reaction involving two colored cobalt complex ions:

• The forward reaction is endothermic, so the forward reaction is favored at high temperatures.

• The reverse reaction is favored at low temperatures.

• Equilibrium changes with changes in temperature they affect the value of equilibrium constants.

• Consider Keq for the ammonia synthesis equilibrium:

• The forward reaction is exothermic (∆H = −91.8 kJ), so the equilibrium constant decreases a lot as temperature increases.

Pressure Changes May Alter Systems in Equilibrium

• Changes in pressure can affect gases in equilibrium.

• The NO2 and N2O4 equilibrium can show the effect of a pressure stress on a chemical equilibrium.

2NO2(g) → N2O4(g)

• When the gas mixture has a larger concentration of N2O4 than of NO2, it has a pale color.

Pressure Changes May Alter Systems in Equilibrium

[pic]

When the gas is suddenly compressed to about half its former volume, the pressure doubles.

• Before the system has time to adjust to the pressure stress, the concentration of each gas doubles, which results in a darker color.

• Le Châtelier’s principle predicts that the system will adjust in an attempt to reduce the pressure.

• According to the equation, 2 mol of NO2 produce 1 mol of N2O4.

• At constant volume and temperature, pressure is proportional to the number of moles.

• So, the pressure reduces when there are fewer moles of gas.

• Thus, the equilibrium shifts to the right, and more N2O4 is produced.

• In an equilibrium, a pressure increase favors the reaction that produces fewer gas molecules, which for the following equilibrium is the reverse reaction.

• When there is no change in the number of molecules, a change in pressure will not affect equilibrium.

• The solubility of CuCl in water is 1.1 ( 10−3 mol/L.

• The solubility of CuCl in sea water is 2.2 ( 10−6 mol/L

• CuCl is 500 times less soluble in sea water.

• The dramatic reduction in solubility of CuCl shows Le Châtelier’s principle and is described below.

• [Cu+][Cl−] = Ksp = 1.2 × 10−6

• If you add chloride-rich ocean water to a saturated solution of copper(I) chloride, [Cl−] increases.

• However, the Ksp, or [Cl−] ( [Cu+], remains constant.

• [Cu+] must decrease.

• This decrease can occur only by the precipitation of the CuCl salt.

• The ion Cl− is the common-ion in this case.

• The reduction of the solubility of a salt in the solution due to the addition of a common ion is called the common-ion effect.

• Doctors use solutions of barium sulfate, BaSO4, to diagnose problems in the digestive tract.

• BaSO4 is in equilibrium with a small concentration of Ba2+(aq), a poison.

• To reduce the Ba2+ concentration to a safe level, a common ion is added.

• Doctors add sodium sulfate, Na2SO4, into the BaSO4 solution that a patient must swallow.

• Doctors use solutions of barium sulfate, BaSO4, to diagnose problems in the digestive tract.

• BaSO4 is in equilibrium with a small concentration of Ba2+(aq), a poison.

• To reduce the Ba2+ concentration to a safe level, a common ion is added.

• Doctors add sodium sulfate, Na2SO4, into the BaSO4 solution that a patient must swallow.

• So, CaSO4 precipitates to establish equilibrium:

•

• The chemical industry makes use of Le Châtelier’s principle in the synthesis of ammonia by the Haber Process.

• High pressure is used to drive the following equilibrium to the right.

•

• The forward reaction converts 4 mol of gas into 2 mol of another gas, so it is favored at high pressures.

• The ammonia synthesis is an exothermic reaction, so the forward reaction is favored at low temperatures.

• 0°C Keq = 6.5 × 108

• 250°C Keq = 52

• 500°C Keq = 5.8 × 10−2

•

• The Haber Process is operated at temperatures of 500°C even though the Keq is small at that temperature.

• The reaction proceeds too slowly at lower temperatures.

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