Chapter 18 Reaction Rates and Equilibrium 18.3 Reversible Reactions

Chapter 18 Reaction Rates and Equilibrium 18.3 Reversible Reactions
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 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?
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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. 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.

Analyze List the knowns and the unknowns.
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.

Calculate Solve for the unknowns.
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.

Analyze List the knowns and the unknowns.
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 18.3 Reversible Reactions

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Chapter 18 Reaction Rates and Equilibrium 18.3 Reversible Reactions

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