Topic 22

Topic

22

Table of Contents

Topic

22

Topic 22: Chemical Equilibrium

Basic Concepts

Additional Concepts

Chemical Equilibrium: Basic Concepts

Topic

22 What is equilibrium?

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


Topic


• 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.

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


Topic


• But spontaneous reactions are not always fast.

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

• 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.


Topic


• 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?

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


Topic


• The concentrations of the reactants (H

2

N

2 of the product (NH

3

) increases. and

) decrease at first while the concentration

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


Topic


• The reactants, H

2 and N

2

, are consumed in the reaction, so their concentrations gradually decrease.

• After a period of time, however, the concentrations of H longer change.

2

, N

2

, and NH

3 no


Topic


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

• The concentrations of H

2 and N

2 are not zero, so not all of the reactants were converted to product even though

∆

G

 for this reaction is negative.


Topic

22 Reversible reactions

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

• But most reactions, including the ammoniaforming reaction, do not go to completion.

They appear to stop.

• The reason is that these reactions are reversible.


Topic


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


Topic


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


Topic


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

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


Topic


• 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.


Topic


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


Topic


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

• 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.


Topic


• 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.


Topic


• 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.


Topic


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


Topic


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

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


Topic


•

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

Rate forward reaction

= Rate reverse reaction


Topic


• 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.

• 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.


Topic

22 Equilibrium Expressions and Constants

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

Click box to view movie clip.


Topic


• 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.


Topic


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

• 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.


Topic


• 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.


Topic


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

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


Topic


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

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


Topic


• 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.


Topic


• The equilibrium constant , K eq

, 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

K eq is constant only at a specified temperature.


Topic


•

K eq

> 1: More products than reactants at equilibrium.

•

K eq

< 1: More reactants than products at equilibrium.


Topic

22 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.


Topic


• 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 K eq when you add the coefficients from the balanced chemical equation as exponents.


Topic


•

K eq 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.


Topic

22 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


Topic



Topic


• 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 K eq

.


Topic

22 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.


Topic


• 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 .


Topic


• 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.


Topic


• At any given temperature, density does not change.

• No matter how much or how little C

2

H

5

OH is present, its concentration remains constant.

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


Topic


• The equilibrium constant expression for this phase change is


Topic


• 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.


Topic

22 Equilibrium Constant Expressions for Heterogeneous Equilibria

• Write the equilibrium constant expression for the decomposition of baking soda

(sodium hydrogen carbonate).


Topic


• 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.


Topic


• Known

• Unknown

• equilibrium constant expression = ?


Topic


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

• Leave out [NaHCO

3

[Na

2

CO solids.

3

] and

] because they are


Topic


• Because the coefficients of [CO

2 are 1, the expression is complete.

] and [H

2

O]


Topic

22 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 K eq at a particular temperature, it has an unlimited number of equilibrium positions.

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


Topic

22 Calculating the Value of Equilibrium Constants

• Calculate the value of

K eq constant expression for the equilibrium given concentration data at one equilibrium position: [NH

3 mol/L, [H

2

] = 0.933 mol/L, [N

] = 1.600 mol/L.

2

] = 0.533


Topic


• 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, H

2

, has the largest concentration and is raised to the third power in the denominator, than 1.

K eq is likely to be less


Topic


• Known

• [NH

3

] = 0.933 mol/L

• [N

2

] = 0.533 mol/L

• [H

2

] = 1.600 mol/L

• Unknown

•

K eq

= ?


Topic


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


Topic

22 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.

• For example, consider the equilibrium system

• If an additional amount of reactant (NO or

Br

2

) is added to the system, the equilibrium will shift to the right, that is, more product

(NOBr) will be formed.


Topic

22 Factors Affecting Chemical Equilibrium

• Conversely, adding more NOBr to the system will result in a shift to the left, forming more

NO and Br

2

.

• 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.


Topic

22 Changes in volume

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

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

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


Topic


• 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.


Topic


• 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


Topic

22 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 K eq

.


Topic


• 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.


Topic


• 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

K eq 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 K eq

.

Topic

22

Basic Assessment Questions

Question 1

Write equilibrium constant expressions for the following homogeneous equilibria.

Topic

22


Question 1a

Answer 1a

Topic

22


Question 1b

Answer 1b

Topic

22


Question 2

Write equilibrium constant expressions for the following heterogeneous equilibria.

Topic

22


Question 2a

Answer 2a

Topic

22


Question 2b

Answer 2b


Topic

22

Question 3

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

Calculate K eq if the equilibrium concentrations are [H

2

] = 0.205 mol/L, [CS

2

] = 0.0664 mol/L,

[CH

4

] = 0.0196 mol/L, and [H

2

S] = 0.0392 mol/L.

Topic

22


Answer

0.257

Topic

22


Question 4

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

Topic

22


Question 4a

Answer 4a

shift to the right

Topic

22


Question 4b

Answer 4b

shift to the left

Topic

22

Chemical Equilibrium: Additional Concepts

Additional Concepts

Topic

22


Using Equilibrium Constants

• When

K eq 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.

Topic

22


Calculating Equilibrium Concentrations

• At 350 

C, K eq

= 66.9 for the formation of hydrogen iodide from its elements.

• What is the concentration of HI if [H

2

0.0295 mol/L and [I

2

] =

] = 0.0174 mol/L?


Topic

22 Calculating Equilibrium Concentrations

• Write the equilibrium constant expression.

• Multiply both sides of the equation by

[H

2

] [I

2

].


Topic

22 Calculating Equilibrium Concentrations

• 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.

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


Topic

22 Solubility equilibria

• The solubility product constant ( K sp

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


Topic


• 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.


Topic


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

• The coefficient of Cu 2+ is 1, and the coefficient of OH

– is 2, so the following is the solubility product constant expression.


Topic


• Tabulated

K sp 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.

Topic

22


Calculating Molar Solubility and Ion Concentration from K sp

• The

K sp

3.3 x 10 for lead(II) fluoride (PbF

–8 at 25



C. Use this K s calculate the following.

2

) is value to a. The solubility in mol/L of PbF

2 b. The fluoride ion concentration in a saturated solution of PbF

2

.


Topic

22 Calculating Molar Solubility and Ion Concentration from K sp a. Write the balanced equation for the solubility equilibrium, and write the K sp expression.

• The moles of Pb 2+ ions in solution equal the moles of PbF

2 that dissolved.

Topic

22



• Therefore, let [Pb 2+ ] equal s , where s represents the solubility of PbF

2

.

• Because there are two F – ions for every Pb 2+ ion, [F

–

] = 2 s .


Topic

22 Calculating Molar Solubility and Ion Concentration from K sp

• Substitute these terms into the

K sp and solve for s . expression

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

Topic

22



• The molar solubility of PbF

2

25°C is 2.0 x 10

–3 mol/L.

in water at


Topic

22 Calculating Molar Solubility and Ion Concentration from K sp b. As shown above, the fluoride ion concentration in a saturated solution of

PbF

2 at 25



C is as follows.

Topic

22


Predicting precipitates

• 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,

Q sp

.

Topic

22


Predicting precipitates

• If

Q sp

< K sp

, a precipitate will form, reducing the ion concentrations until the system reaches equilibrium and the solution is saturated.

• If

Q sp

< K sp

, no precipitate forms.


Topic

22 Predicting precipitates

• The following example problem demonstrates how to use Q sp and K to sp determine whether a precipitate will form.

Predicting a Precipitate

• Predict whether a precipitate will form if 200 mL of 0.030

M CaCl

0.080

M NaOH.

2 is added to 200 mL of


Topic

22 Predicting a Precipitate

• 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

K sp

= 5.0 x 10

–6 is sparingly soluble with

, so it might precipitate if the concentrations of its ions are high enough.


Topic


• Write the equation for the dissolving of

Ca(OH)

2

.

• The ion product expression is as follows.

•

Q sp is a trial value that will be compared to K sp

.


Topic


• Next, find the concentrations of the Ca 2+ and OH

– ions.

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


Topic


• Calculate

Q sp

.

• Calculate

Q sp with K sp

.

• The concentrations of Ca 2+ and OH

– high enough to cause a precipitate of are

Ca(OH)

2 to form.


Topic

22 Common ion effect

• 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, PbI

2 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 Pb 2+ and thus reduces the solubility of PbI

2

.

Topic

22

Additional Assessment Questions

Question 1

At a certain temperature, K eq following reaction.

= 0.118 for the

Calculate the concentration of [H

2

] in an equilibrium mixture with [CH

4

] = 0.0492 mol/L and [C

2

H

2

] = 0.0755 mol/L.

Topic

22


Answer

0.156 mol/L


Topic

22

Question 2

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

298 K.


Topic

22

Question 2a

MgCO

3

Answer 2a

2.6 x 10

–3 mol/L


Topic

22

Question 2b

AlPO

4

Answer 2b

9.9 x 10

–11 mol/L

Topic

22


Question 3

Will a precipitate form when 125 mL of

0.010

M K

2

SO

4 is mixed with 250 mol of

0.015

M CaBr

2

? (Hint: Note that the volumes of the two solutions are not equal, and the volume after mixing is 375 mL.)


Topic

22

Answer

For CaSO

4

: Q sp

(3.3 x 10

–5

) < K sp

(4.9 x 10

–5

); no precipitate forms

Help

To advance to the next item or next page click on any of the following keys: mouse, space bar, enter, down or forward arrow.

Click on this icon to return to the table of contents

Click on this icon to return to the previous slide

Click on this icon to move to the next slide

Click on this icon to open the resources file.

Click on this icon to go to the end of the presentation.

End of Topic Summary File

Topic 22

Topic 22: Chemical Equilibrium

Question 1

Question 1a

Answer 1a

Question 1b

Answer 1b

Question 2

Question 2a

Answer 2a

Question 2b

Answer 2b

Question 3

Answer

Question 4

Question 4a

Answer 4a

Question 4b

Answer 4b

Additional Concepts

Question 1

Answer

Question 2

Question 2a

Answer 2a

Question 2b

Answer 2b

Question 3

Answer

End of Topic Summary File

Related documents

Products

Support

Topic 22

Topic 22: Chemical Equilibrium

Question 1

Question 1a

Answer 1a

Question 1b

Answer 1b

Question 2

Question 2a

Answer 2a

Question 2b

Answer 2b

Question 3

Answer

Question 4

Question 4a

Answer 4a

Question 4b

Answer 4b

Additional Concepts

Question 1

Answer

Question 2

Question 2a

Answer 2a

Question 2b

Answer 2b

Question 3

Answer

End of Topic Summary File

Related documents

Add this document to collection(s)

Add this document to saved

Suggest us how to improve StudyLib