18.3 Reversible Reactions and Equilibrium > 18.3 Reversible Reactions and Equilibrium > 1 Copyright...

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18.3 Reversible Reactions and 18.3 Reversible Reactions and Equilibrium > Equilibrium > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 18 Reaction Rates and Equilibr 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

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Chapter 18Reaction Rates and Equilibrium

18.1 Rates of Reaction18.2 The Progress of Chemical Reactions

18.3 Reversible Reactions and Equilibrium

18.4 Solubility Equilibrium18.5 Free Energy and Entropy

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How did chemists help farmers produce more food?

CHEMISTRY & YOUCHEMISTRY & YOU

Fertilizers can increase the amount of a crop per unit of land. Most fertilizers contain ammonia or nitrogen compounds made from ammonia.

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Reversible Reversible ReactionsReactions

Reversible Reactions

What happens at the molecular level in a chemical system at equilibrium?

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Reversible Reversible ReactionsReactions

• 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.

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Reversible Reversible ReactionsReactions

Here is an example of a reversible reaction.

• The first reaction is called the forward reaction.

• The second reaction is called the reverse reaction.

2SO2(g) + O2(g) 2SO3(g)

2SO2(g) + O2(g) 2SO3(g)

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Reversible Reversible ReactionsReactions

The two equations can be combined into one using a double arrow.

The double arrow tells you that the reaction is reversible.

2SO2(g) + O2(g) 2SO3(g)

2SO2(g) + O2(g) 2SO3(g)

2SO2(g) + O2(g) 2SO3(g)

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Reversible Reversible ReactionsReactions

Molecules of SO2 and O2 react to give SO3. Molecules of SO3 decompose to give SO2 and O2.

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Reversible Reversible ReactionsReactions

Establishing Equilibrium

When the rates of the forward and reverse reactions are equal, the reaction has reached a state of balance called chemical equilibrium.

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Notice that after a certain time, the concentrations remain constant.

Interpret GraphsInterpret 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.

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Reversible Reversible ReactionsReactions

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.

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Reversible Reversible ReactionsReactions

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.

Conditions at Equilibrium

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Reversible Reversible ReactionsReactions

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.

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Reversible Reversible ReactionsReactions

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% 99%

Concentrations at Equilibrium

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Reversible Reversible ReactionsReactions

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.

Concentrations at Equilibrium

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Why is equilibrium considered to be a

dynamic state?

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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.

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Factors Affecting Factors Affecting Equilibrium: Equilibrium: Le Châtelier’s PrincipleLe Châtelier’s Principle

Factors Affecting Equilibrium: Le Châtelier’s Principle

What three stresses can cause a change in the equilibrium position of a chemical system?

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Factors Affecting Factors Affecting Equilibrium: Equilibrium: Le Châtelier’s PrincipleLe Châtelier’s Principle

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.

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Factors Affecting Factors Affecting Equilibrium: Equilibrium: Le Châtelier’s PrincipleLe Châtelier’s Principle

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.

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Factors Affecting Factors Affecting Equilibrium: Equilibrium: Le Châtelier’s PrincipleLe Châtelier’s Principle

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.

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Factors Affecting Factors Affecting Equilibrium: Equilibrium: Le Châtelier’s PrincipleLe Châtelier’s Principle

Concentration

Consider the decomposition of carbonic acid (H2CO3) in aqueous solution.

• The system has reached equilibrium.

• The amount of carbonic acid is less than 1%.

H2CO3(aq) CO2(aq) + H2O(l)< 1% > 99%

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Factors Affecting Factors Affecting Equilibrium: Equilibrium: Le Châtelier’s PrincipleLe Châtelier’s Principle

Suppose carbon dioxide is added to the system.

• 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.

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

Add CO2

Direction of shift

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Factors Affecting Factors Affecting Equilibrium: Equilibrium: Le Châtelier’s PrincipleLe Châtelier’s Principle

Suppose carbon dioxide is removed.

• 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.

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

Add CO2

Direction of shift

Remove CO2

Direction of shift

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• 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.

Factors Affecting Factors Affecting Equilibrium: Equilibrium: Le Châtelier’s PrincipleLe Châtelier’s Principle

An equilibrium between carbonic acid, carbon dioxide, and water exists in your blood.

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Factors Affecting Factors Affecting Equilibrium: Equilibrium: Le Châtelier’s PrincipleLe Châtelier’s Principle

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.

Temperature

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Factors Affecting Factors Affecting Equilibrium: Equilibrium: Le Châtelier’s PrincipleLe Châtelier’s Principle

Temperature

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.

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

Add heatDirection of shift

Remove heat (cool)Direction of shift

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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.

Pressure

Factors Affecting Factors Affecting Equilibrium: Equilibrium: Le Châtelier’s PrincipleLe Châtelier’s Principle

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Factors Affecting Factors Affecting Equilibrium: Equilibrium: Le Châtelier’s PrincipleLe Châtelier’s Principle

Pressure

Initial equilibrium Equilibrium is disturbed by an increase in pressure.

When the plunger is pushed down, the volume decreases and the pressure increases.

A new equilibrium position is established with fewer molecules.

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• 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.

Factors Affecting Factors Affecting Equilibrium: Equilibrium: Le Châtelier’s PrincipleLe Châtelier’s Principle

You can predict which way the equilibrium position will shift by comparing the number of molecules of reactants and products.

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

Add pressureDirection of shift

Reduce pressureDirection of shift

Pressure

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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?

CHEMISTRY & YOUCHEMISTRY & YOU

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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.

CHEMISTRY & YOUCHEMISTRY & YOU

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Factors Affecting Factors Affecting Equilibrium: Equilibrium: Le Châtelier’s PrincipleLe Châtelier’s Principle

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.

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What effect will each of the following changes have on the equilibrium position for this reversible reaction?

Sample Problem 18.2Sample Problem 18.2

Applying Le Châtelier’s Principle

a. Cl2 is added.

b. Pressure is increased.

c. Heat is removed.d. PCl3 is removed as it forms.

PCl5(g) + heat PCl3(g) + Cl2(g)

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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.

Sample Problem 18.2Sample Problem 18.2

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Start with the addition of Cl2.

Solve Apply the concepts to this problem.2

• Cl2 is a product.

• Increasing the concentration of a product shifts the equilibrium to the left.

PCl5(g) + heat PCl3(g) + Cl2(g)

Add Cl2Direction of shift

Sample Problem 18.2Sample Problem 18.2

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Analyze the effect of an increase in pressure.

Solve Apply the concepts to this problem.2

• 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 pressureDirection of shift

PCl5(g) + heat PCl3(g) + Cl2(g)

Sample Problem 18.2Sample Problem 18.2

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Analyze the effect of removing heat.

Solve Apply the concepts to this problem.2

• The reverse reaction produces heat.

• The removal of heat causes the equilibrium to shift to the left.

Remove heatDirection of shift

PCl5(g) + heat PCl3(g) + Cl2(g)

Sample Problem 18.2Sample Problem 18.2

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Analyze the effect of removing PCl3.

Solve Apply the concepts to this problem.2

• PCl3 is a product.

• Removal of a product as it forms causes the equilibrium to shift to the right.

Remove PCl3Direction of shift

PCl5(g) + heat PCl3(g) + Cl2(g)

Sample Problem 18.2Sample Problem 18.2

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In the following equilibrium reaction, in which

direction would the equilibrium position shift

with an increase in pressure?

4HCl(g) + O2(g) 2Cl2(g) +2H2O(g)

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In the following equilibrium reaction, in which

direction would the equilibrium position shift

with an increase in pressure?

Reducing the number of molecules that are gases decreases the pressure. The equilibrium will shift to the right.

4HCl(g) + O2(g) 2Cl2(g) +2H2O(g)

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Equilibrium Equilibrium ConstantsConstants

Equilibrium Constants

What does the size of an equilibrium constant indicate about a system at equilibrium?

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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

Equilibrium Equilibrium ConstantsConstants

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The equilibrium constant (Keq) is the ratio of product concentrations to reactant concentrations at equilibrium.

aA + bB cC + dD

Keq = [C]c x [D]d

[A]a x [B]b

• 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.

Equilibrium Equilibrium ConstantsConstants

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• The flask on the left is in a dish of hot water.

• The flask on the right is in ice.

The value of Keq depends on the temperature of the reaction.

Dinitrogen tetroxide is a colorless gas. Nitrogen dioxide is a brown gas.

Equilibrium Equilibrium ConstantsConstants

N2O4(g) 2NO2(g)

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The size of the equilibrium constant indicates whether reactants or products are more common at equilibrium.

Equilibrium Equilibrium ConstantsConstants

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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.

Equilibrium Equilibrium ConstantsConstants

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The colorless gas dinitrogen tetroxide (N2O4) and the brown gas nitrogen dioxide (NO2)

exist in equilibrium with each other.

Expressing and Calculating Keq

A liter of the gas mixture at equilibrium contains 0.0045 mol of N2O4 and 0.030 mol of NO2 at 10oC. Write the expression for the equilibrium constant (Keq) and calculate the

value of the constant for the reaction.

N2O4(g) 2NO2(g)

Sample Problem 18.3Sample Problem 18.3

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KNOWNS UNKNOWNKeq (algebraic expression) = ?

Keq (numerical value) = ?

Analyze List the knowns and the unknowns.1

Modify the general expression for the equilibrium constant and substitute the known concentrations to calculate Keq.

[N2O4] = 0.0045 mol/L

[NO2] = 0.030 mol/L

Sample Problem 18.3Sample Problem 18.3

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• Start with the general expression for the equilibrium constant.

Calculate Solve for the unknowns.2

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.

• Write the equilibrium constant expression for this reaction.

Keq = [C]c x [D]d

[A]a x [B]b

Keq = [NO2]2

[N2O2]

Sample Problem 18.3Sample Problem 18.3

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Substitute the concentrations that are known and calculate Keq.

Calculate Solve for the unknowns.2

You can ignore the unit mol/L; chemists report equilibrium constants without a stated unit.

Keq = (0.030 mol/L)2

(0.0045 mol/L) = (0.030 mol/L x 0.030 mol/L)

(0.0045 mol/L)

Keq = 0.20 mol/L = 0.20

Sample Problem 18.3Sample Problem 18.3

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• 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.

Evaluate Does the result make sense?3

Sample Problem 18.3Sample Problem 18.3

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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.

Finding the Equilibrium Constant

H2(g) + I2(g) 2HI(g)

Sample Problem 18.4Sample Problem 18.4

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KNOWNS UNKNOWNKeq = ?

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.

[H2] (initial) = 1.00 mol/L

[I2] (initial) = 1.00 mol/L

[HI] (equilibrium) = 1.56 mol/L

Sample Problem 18.4Sample Problem 18.4

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First find out how much H2 and I2 are consumed in the reaction.

Calculate Solve for the unknown.2

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).

x + x = 1.56 mol

2x = 1.56 mol

x = 0.780 mol

Sample Problem 18.4Sample Problem 18.4

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• Calculate how much H2 and I2 remain in the flask at equilibrium.

Calculate Solve for the unknown.2

mol H2 = mol I2 = (1.00 mol – 0.780 mol) = 0.22 mol

• Write the expression for Keq.

Keq = [HI]2

[H2] x [I2]

Use the general expression for Keq as a guide:

Keq = [C]c x [D]d

[A]a x [B]b

Sample Problem 18.4Sample Problem 18.4

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Substitute the equilibrium concentrations of the reactants and products into the equation and solve for Keq.

Calculate Solve for the unknown.2

Keq = (1.56 mol/L)2

0.22 mol/L x 0.22 mol/L

Keq = 1.56 mol/L x 1.56 mol/L

Keq = 5.0 x 101

0.22 mol/L x 0.22 mol/L

Sample Problem 18.4Sample Problem 18.4

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• 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.

Evaluate Does the result make sense?3

Sample Problem 18.4Sample Problem 18.4

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Bromine chloride (BrCl) decomposes to form bromine and chlorine.

Finding Concentrations at Equilibrium

At a certain temperature, the equilibrium constant for the reaction is 11.1. 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?

Sample Problem 18.5Sample Problem 18.5

2BrCl(g) Br2(g) + Cl2(g)

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KNOWNS UNKNOWN[Br2] (equilibrium) = ? mol/L

[BrCl] (equilibrium) = ? mol/L

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.

[Cl2] (equilibrium) = 4.00 mol/L

Keq = 11.1

Sample Problem 18.5Sample Problem 18.5

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• The volume of the container is 1 L, so calculate [Br2] at equilibrium.

Calculate Solve for the unknowns.2

• Write the equilibrium expression for the reaction.

Keq = [BrCl]2

[Br2] x [Cl2]

[Br2] = = 4.00 mol/L4.00 mol

1 L

Sample Problem 18.5Sample Problem 18.5

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• Rearrange the equation to solve for [BrCl]2.

• Substitute the known values for Keq, [Br2], and [Cl2].

[BrCl]2 = Keq [Br2] x [Cl2]

[BrCl]2 =

= 1.44 mol2/L2

11.14.00 mol/L x 4.00 mol/L

Sample Problem 18.5Sample Problem 18.5

Calculate Solve for the unknowns.2

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• Calculate the square root.

[BrCl] = = 1.20 mol/L1.44 mol2/L2

Use your calculator to find the square root.

Sample Problem 18.5Sample Problem 18.5

Calculate Solve for the unknowns.2

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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.

Evaluate Does the result make sense?3

Sample Problem 18.5Sample Problem 18.5

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HCl is formed when H2 and Cl2 react at

high temperatures.

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?

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

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HCl is formed when H2 and Cl2 react at

high temperatures.

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?

Keq = = [HCl]2

[H2] x [Cl2]

(1.76 x 10–2 mol/L)2

(1.60 x 10–3 mol/L) x (1.60 x 10–3 mol/L)

Keq = 121

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

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Key ConceptsKey Concepts

Stresses that upset the equilibrium of a chemical system include changes in concentration of reactants or products, changes in temperature, and changes in pressure.

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.

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Key Concept and Key Concept and Key EquationKey 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

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Glossary TermsGlossary 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

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• 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

Glossary TermsGlossary Terms

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• 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

Glossary TermsGlossary Terms

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END OF 18.3END OF 18.3