Topic 22 Topic 22 Topic 22: Chemical Equilibrium Basic Concepts Additional Concepts Table of...

Topic 22Topic 22

Topic 22: Chemical Equilibrium

Basic Concepts

Additional Concepts

Table of ContentsTable of ContentsTopic 22Topic 22

• Consider the reaction for the formation of ammonia from nitrogen and hydrogen.

What is equilibrium?

Chemical Equilibrium: Basic ConceptsChemical Equilibrium: Basic ConceptsChemical Equilibrium: Basic ConceptsChemical Equilibrium: Basic ConceptsTopic 22Topic 22


Chemical Equilibrium: Basic ConceptsChemical Equilibrium: Basic ConceptsChemical Equilibrium: Basic ConceptsChemical Equilibrium: Basic Concepts

• Note that the equation for the production of ammonia has a negative standard free energy, ∆G°.

• Recall that a negative sign for ∆G° indicates that the reaction is spontaneous under standard conditions.

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• Standard conditions are defined as 298 K and one atmosphere pressure.



• But spontaneous reactions are not always fast.

• When carried out under standard conditions, this ammonia-forming reaction is much too slow.

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• To produce ammonia at a rate that is practical, the reaction must be carried out at a much higher temperature than 298 K and a higher pressure than one atmosphere.



• What happens when one mole of nitrogen and three moles of hydrogen, the amounts shown in the equation, are placed in a closed reaction vessel at 723 K?

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• Because the reaction is spontaneous, nitrogen and hydrogen begin to react.



• The concentrations of the reactants (H2 and N2) decrease at first while the concentration of the product (NH3) increases.

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• Then, before the reactants are used up, all concentrations become constant.



• The reactants, H2 and N2, are consumed in the reaction, so their concentrations gradually decrease.

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• After a period of time, however, the concentrations of H2, N2, and NH3 no longer change.



• All concentrations become constant, as shown by the horizontal lines on the right side of the diagram.

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• The concentrations of H2 and N2 are not zero, so not all of the reactants were converted to product even though ∆G for this reaction is negative.

Reversible reactions


• When a reaction results in almost complete conversion of reactants to products, chemists say that the reaction goes to completion.

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• But most reactions, including the ammonia-forming reaction, do not go to completion. They appear to stop.

• The reason is that these reactions are reversible.



• A reversible reaction is one that can occur in both the forward and the reverse directions.

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• Chemists combine these two equations into a single equation that uses a double arrow to show that both reactions occur.

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• When you read the equation, the reactants in the forward reaction are on the left.

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• In the reverse reaction, the reactants are on the right.



• In the forward reaction, hydrogen and nitrogen combine to form the product ammonia.

• In the reverse reaction, ammonia decomposes into the products hydrogen and nitrogen.



• Just as the reaction begins at a definite, initial rate; no ammonia is present so only the forward reaction can occur.

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• As hydrogen and nitrogen combine to form ammonia, their concentrations decrease.

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• The rate of a reaction depends upon the concentration of the reactants.

• The decrease in the concentration of the reactants causes the rate of the forward reaction to decrease.



• As soon as ammonia is present, the reverse reaction can occur, slowly at first, but at an increasing rate as the concentration of ammonia increases.

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• As the reaction proceeds, the rate of the forward reaction continues to decrease and the rate of the reverse reaction continues to increase until the two rates are equal.

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• At that point, ammonia is being produced as fast as it is being decomposed, so the concentrations of nitrogen, hydrogen, and ammonia remain constant.



• The system has reached a state of balance or equilibrium.

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• The word equilibrium means that opposing processes are in balance.



• Chemical equilibrium is a state in which the forward and reverse reactions balance each other because they take place at equal rates.

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Rateforward reaction = Ratereverse reaction



• You can recognize that the ammonia-forming reaction reaches a state of chemical equilibrium because its chemical equation is written with a double arrow like this.

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• At equilibrium, the concentrations of reactants and products are constant.

• However, that does not mean that the amounts or concentrations of reactants and products are equal.

Equilibrium Expressions and Constants


• You have learned that some chemical systems have little tendency to react and others go readily to completion.

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• In between these two extremes are the majority of reactions that reach a state of equilibrium with varying amounts of reactants unconsumed.

• If the reactants are not consumed, then not all the product predicted by the balanced chemical equation will be produced.



• According to the equation for the ammonia-producing reaction, two moles of ammonia should be produced when one mole of nitrogen and three moles of hydrogen react.

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• Because the reaction reaches a state of equilibrium, however, fewer than two moles of ammonia will actually be obtained.

• Chemists need to be able to predict the yield of a reaction.



• In 1864, the Norwegian chemists Cato Maximilian Guldberg and Peter Waage proposed the law of chemical equilibrium, which states that at a given temperature, a chemical system may reach a state in which a particular ratio of reactant and product concentrations has a constant value.

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• For example, the general equation for a reaction at equilibrium can be written as follows.

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• A and B are the reactants; C and D the products.



• The coefficients in the balanced equation are a, b, c, and d.

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• If the law of chemical equilibrium is applied to this reaction, the following ratio is obtained.



• This ratio is called the equilibrium constant expression.

• The square brackets indicate the molar concentrations of the reactants and products at equilibrium in mol/L.



• The equilibrium constant, Keq, is the numerical value of the ratio of product concentrations to reactant concentrations, with each concentration raised to the power corresponding to its coefficient in the balanced equation.

• The value of Keq is constant only at a specified temperature.



• Keq > 1: More products than reactants at equilibrium.

• Keq < 1: More reactants than products at equilibrium.

Constants for homogeneous equilibria


• How would you write the equilibrium constant expression for this reaction in which hydrogen and iodine react to form hydrogen iodide?

• This reaction is a homogeneous equilibrium, which means that all the reactants and products are in the same physical state.



• All participants are gases. • To begin writing the equilibrium constant

expression, place the product concentration in the numerator and the reactant concentrations in the denominator.

• The expression becomes equal to Keq when you add the coefficients from the balanced chemical equation as exponents.



• Keq for this homogeneous equilibrium at 731 K is 49.7.

• Note that 49.7 has no units. In writing equilibrium constant expressions, it’s customary to omit units.

Equilibrium Constant Expressions for Homogeneous Equilibria


• Write the equilibrium constant expression for the reaction in which ammonia gas is produced from hydrogen and nitrogen.

• The form of the equilibrium constant expression is



• Place the product concentration in the numerator and the reactant concentrations in the denominator. • Raise the concentration of each reactant and product to a power equal to its coefficient in the balanced chemical equation and set the ratio equal to Keq.

Constants for heterogeneous equilibria


• When the reactants and products of a reaction are present in more than one physical state, the equilibrium is called a heterogeneous equilibrium.

• When ethanol is placed in a closed flask, a liquid-vapor equilibrium is established.



• To write the equilibrium constant expression for this process, you would form a ratio of the product to the reactant.

• At a given temperature, the ratio would have a constant value K.



• Note that the term in the denominator is the concentration of liquid ethanol.

• Because liquid ethanol is a pure substance, its concentration is constant at a given temperature.

• That’s because the concentration of a pure substance is its density in moles per liter.



• At any given temperature, density does not change.

• No matter how much or how little C2H5OH is present, its concentration remains constant.

• Therefore, the term in the denominator is a constant and can be combined with K.



• The equilibrium constant expression for this phase change is



• Solids also are pure substances with unchanging concentrations, so equilibria involving solids can be simplified in the same way.

• For example, notice the experiment involving the sublimation of iodine crystals.

• The equilibrium depends only on the concentration of gaseous iodine in the system.

Equilibrium Constant Expressions for Heterogeneous Equilibria


• Write the equilibrium constant expression for the decomposition of baking soda (sodium hydrogen carbonate).



• You are given a heterogeneous equilibrium involving gases and solids.

• The general form of the equilibrium constant expression for this reaction is

• Because the reactant and one of the products are solids with constant concentrations, they can be omitted from the equilibrium constant expression.



• Known

• Unknown • equilibrium constant expression = ?



• Write a ratio with the concentrations of the products in the numerator and the concentration of the reactant in the denominator.

• Leave out [NaHCO3] and [Na2CO3] because they are solids.



• Because the coefficients of [CO2] and [H2O] are 1, the expression is complete.

Determining the Value of Equilibrium Constants


• When equilibrium is established, the concentration of each substance is determined experimentally.

• Although an equilibrium system has only one value for Keq at a particular temperature, it has an unlimited number of equilibrium positions.

• Equilibrium positions depend upon the initial concentrations of the reactants and products.

Calculating the Value of Equilibrium Constants


• Calculate the value of Keq for the equilibrium constant expression

given concentration data at one equilibrium position: [NH3] = 0.933 mol/L, [N2] = 0.533 mol/L, [H2] = 1.600 mol/L.



• You have been given the equilibrium constant expression and the concentration of each reactant and product.

• You must calculate the equilibrium constant.

• Because the reactant, H2, has the largest concentration and is raised to the third power in the denominator, Keq is likely to be less than 1.



• Known

• [NH3] = 0.933 mol/L• [N2] = 0.533 mol/L• [H2] = 1.600 mol/L

• Unknown• Keq = ?



• Substitute the known values into the equilibrium constant expression and calculate its value.

Factors Affecting Chemical Equilibrium


• Le Châtelier’s principle states that if a stress is applied to a system at equilibrium, the system shifts in the direction that relieves the stress.

• If an additional amount of reactant (NO or Br2) is added to the system, the equilibrium will shift to the right, that is, more product (NOBr) will be formed.

• For example, consider the equilibrium system

Factors Affecting Chemical Equilibrium


• Conversely, adding more NOBr to the system will result in a shift to the left, forming more NO and Br2.

• The removal of a reactant or product also results in a shift in the equilibrium.

• Removing a reactant causes the equilibrium to shift to the left, forming more reactants.

• Removing the product causes a shift to the right, forming more product.

Changes in volume


• Le Châtelier’s principle also applies to changes in the volume of the reaction vessel containing the equilibrium system.

• The equilibrium will shift to relieve the stress of increased pressure.

• Suppose the volume of the reaction vessel for the system is decreased, resulting in an increase in pressure.

Changes in volume


• In this case, the shift will be to the right because three moles of reactant gas combine to form only two moles of product gas.

• Thus, a shift toward the product will reduce the pressure of the system.

• If the volume of the reaction vessel was increased, the equilibrium would shift to the left, and more of the reactants would be formed.

Changes in volume


• Note that changing the volume of the reaction vessel causes no shift in the equilibrium when the number of moles of product gas equals the number of moles of reactant gas; an example is the equilibrium

Changes in temperature


• Even though equilibrium may shift to the right or left in response to a change in concentration or volume, the value of the equilibrium constant remains the same.

• A change in temperature, however, alters both the equilibrium position and the value of Keq.



• For example, consider the thermochemical equation for the reversible formation of hydrogen chloride gas from its elements.

• The forward reaction releases heat, so you can consider heat as a product in the forward reaction and a reactant in the reverse reaction.



• Raising the temperature of this system requires the addition of heat, which shifts the equilibrium to the left and reduces the concentration of hydrogen chloride.

• Thus, the value of Keq decreases. • Lowering the temperature of the system

means that heat is removed, so the equilibrium relieves the stress by shifting to the right, increasing both the concentration of hydrogen chloride and Keq.

Basic Assessment QuestionsBasic Assessment Questions

Question 1

Write equilibrium constant expressions for the following homogeneous equilibria.

Topic 22Topic 22

Answer 1a

Question 1a Topic 22Topic 22


Answer 1b

Question 1b Topic 22Topic 22



Question 2

Write equilibrium constant expressions for the following heterogeneous equilibria.

Topic 22Topic 22

Answer 2a



Answer 2b




Question 3

Hydrogen and carbon disulfide react to form methane and hydrogen sulfide according to this equation.

Topic 22Topic 22

Calculate Keq if the equilibrium concentrations are [H2] = 0.205 mol/L, [CS2] = 0.0664 mol/L, [CH4] = 0.0196 mol/L, and [H2S] = 0.0392 mol/L.


Answer

0.257

Topic 22Topic 22


Question 4

How would decreasing the volume of the reaction vessel affect these equilibria?

Topic 22Topic 22

shift to the right

Answer 4a



shift to the left

Answer 4b



Chemical Equilibrium: Additional ConceptsChemical Equilibrium: Additional Concepts Chemical Equilibrium: Additional ConceptsChemical Equilibrium: Additional Concepts Topic 22Topic 22

Additional Concepts

Chemical Equilibrium: Additional ConceptsChemical Equilibrium: Additional Concepts Chemical Equilibrium: Additional ConceptsChemical Equilibrium: Additional Concepts

Using Equilibrium Constants

• When Keq is known, the equilibrium concentration of a substance can be calculated if you know the concentrations of all other reactants and products.

• The following example problem shows you how to determine an equilibrium concentration.

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Calculating Equilibrium Concentrations

• At 350C, Keq = 66.9 for the formation of hydrogen iodide from its elements.

• What is the concentration of HI if [H2] = 0.0295 mol/L and [I2] = 0.0174 mol/L?

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• Write the equilibrium constant expression.

• Multiply both sides of the equation by [H2] [I2].

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• Substitute the known quantities into the equation and solve for [HI].

• An extra digit is retained here for accuracy, but the final answer will be rounded to three digits.

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• The equilibrium concentration of HI is 0.185 mol/L.


Solubility equilibria• The solubility product constant (Ksp) is

an equilibrium constant for the dissolving of a sparingly soluble ionic compound in water.

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Solubility equilibriaTopic 22Topic 22

• The solubility product constant expression is the product of the concentrations of the ions with each concentration raised to a power equal to

the coefficient of the ion in the chemical equation.


Solubility equilibria

• For example, copper (II) hydroxide dissolves in water according to this equation.

• The coefficient of Cu2+ is 1, and the coefficient of OH– is 2, so the following is the solubility product constant expression.

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Solubility equilibria

• Tabulated Ksp values may be used to calculate the molar solubility of a sparingly soluble ionic compound and also to calculate ion concentrations in a saturated solution.

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Calculating Molar Solubility and Ion Concentration from Ksp

• The Ksp for lead(II) fluoride (PbF2) is 3.3 x 10–8 at 25C. Use this Ks value to calculate the following.

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a. The solubility in mol/L of PbF2

b.The fluoride ion concentration in a saturated solution of PbF2.



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a. Write the balanced equation for the solubility equilibrium, and write the Ksp expression.

• The moles of Pb2+ ions in solution equal the moles of PbF2 that dissolved.



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• Therefore, let [Pb2+] equal s, where s represents the solubility of PbF2.

• Because there are two F– ions for every Pb2+ ion, [F–] = 2s.



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• Substitute these terms into the Ksp expression and solve for s.

• Here three digits are retained for accuracy, but the final answer will be rounded to two digits.



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• The molar solubility of PbF2 in water at 25°C is 2.0 x 10–3 mol/L.



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b. As shown above, the fluoride ion concentration in a saturated solution of PbF2 at 25C is as follows.


Predicting precipitatesTopic 22Topic 22

• The solubility product constant expression can also be used to predict whether a precipitate will form when two solutions of ionic compounds are mixed.

• The molar concentrations of the ions in a solution are used to calculate the ion product, Qsp.



• If Qsp < Ksp, a precipitate will form, reducing the ion concentrations until the system reaches equilibrium and the solution is saturated.

• If Qsp < Ksp, no precipitate forms.



• The following example problem demonstrates how to use Qsp and Ksp to determine whether a precipitate will form.

• Predict whether a precipitate will form if 200 mL of 0.030M CaCl2 is added to 200 mL of 0.080M NaOH.

Predicting a Precipitate


Predicting a Precipitate Topic 22Topic 22

• A double-replacement reaction might occur according to this equation.

• You know that NaCl is a soluble compound and will not form a precipitate.

• However, Ca(OH)2 is sparingly soluble with Ksp = 5.0 x 10–6, so it might precipitate if the concentrations of its ions are high enough.



• Write the equation for the dissolving of Ca(OH)2.

• The ion product expression is as follows.

• Qsp is a trial value that will be compared to Ksp.



• Next, find the concentrations of the Ca2+ and OH– ions.

• Divide the initial concentrations in half because the volume doubles on mixing.



• Calculate Qsp.

• Calculate Qsp with Ksp.

• The concentrations of Ca2+ and OH– are high enough to cause a precipitate of Ca(OH)2 to form.


Common ion effectTopic 22Topic 22

• The solubility of a substance is reduced when the substance is dissolved in a solution containing a common ion. This is called the common ion effect.

• For example, PbI2 is less soluble in an aqueous solution of NaI than in pure water because the common ion I–, already present in the NaI solution, reduces the maximum possible concentration of Pb2+ and thus reduces the solubility of PbI2.

Additional Assessment QuestionsAdditional Assessment Questions

At a certain temperature, Keq = 0.118 for the

following reaction.

Question 1 Topic 22Topic 22

Calculate the concentration of [H2] in an equilibrium mixture with [CH4] = 0.0492 mol/L and [C2H2] = 0.0755 mol/L.


0.156 mol/L

AnswerTopic 22Topic 22


Use the data in this table to calculate the solubility in mol/L of these ionic compounds at 298 K.

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Question 2

MgCO3

2.6 x 10–3

mol/L

Answer 2a

Question 2a

Additional Assessment QuestionsAdditional Assessment QuestionsTopic 22Topic 22

Question 2b

AlPO4

9.9 x 10–11

mol/L

Answer 2b

Additional Assessment QuestionsAdditional Assessment QuestionsTopic 22Topic 22


Will a precipitate form when 125 mL of 0.010M K2SO4 is mixed with 250 mol of

0.015 M CaBr2? (Hint: Note that the volumes

of the two solutions are not equal, and the volume after mixing is 375 mL.)

Question 3 Topic 22Topic 22


For CaSO4 : Qsp (3.3 x 10–5) < Ksp (4.9 x 10–5);

no precipitate forms

AnswerTopic 22Topic 22

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