Chapter 15 Lecture- Chemical Equilibrium

Chapter 15 - Chemical Equilibrium

Sections 15.1 - 15.4 Equilibrium & the Equilibrium Constant (K)

The Concept of Equilibrium

Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate.

The Concept of Equilibrium• As a system

approaches equilibrium, both the forward and reverse reactions are occurring.

The Concept of Equilibrium• As a system

approaches equilibrium, both the forward and reverse reactions are occurring.

• At equilibrium, the forward and reverse reactions are proceeding at the same rate.

A System at Equilibrium

Once equilibrium is achieved, the amount of each reactant and product remains constant.

Depicting Equilibrium In a system at equilibrium, both

the forward and reverse reactions are being carried out; as a result, we write its equation

N2O4 (g) ⇋ 2 NO2 (g)

N2(g) + 3H2(g) ⇌ 2NH3(g)

1M N2 + 3M H2 2M NH3

The Equilibrium Constant


• Forward reaction:N2O4 (g) → 2 NO2 (g)

✤Rate law: Rate = kf [N2O4]


• Forward reaction:N2O4 (g) → 2 NO2 (g)

✤Rate law: Rate = kf [N2O4]

• Reverse reaction:2 NO2 (g) → N2O4 (g)

✤Rate law: Rate = kr [NO2]2

The Equilibrium Constant• Therefore, at equilibrium

Ratef = Rater

kf [N2O4] = kr [NO2]2

The Equilibrium Constant• Therefore, at equilibrium

Ratef = Rater

kf [N2O4] = kr [NO2]2

• Rewriting this, it becomeskfkr

[NO2]2

[N2O4]=


The ratio of the rate constants is a constant at that temperature, and the expression becomes

Keq =kfkr

[NO2]2

[N2O4]=

For the gas-phase reaction A ⇌ Bthe forward reaction rate is 3.0 × 10−4 s−1 and the reverse reaction rate is 1.5 × 10−2 s−1. What is the value of the equilibrium constant, Keq?

1. 0.02 2. 503. 0.0004 4. 2500

Correct Answer:

1. 0.02 2. 503. 0.0004 4. 2500

N2O4 (g) → 2 NO2 (g) Eq 15.1

Eq 15.5

1. moles of reactant always equals moles of product.

2. mass of reactant always equals mass of product.3. the rates of opposing reactions are equal.4. cannot be determined without temperature

information.

Keq =kfkr

[NO2]2[N2O4]

=

b) the constant will be______ than 1

Keq =kfkr

[NO2]2[N2O4]

=

N2O4 (g) → 2 NO2 (g) Eq 15.1

Eq 15.5

b) the constant will be greater than 1

Keq =kfkr

[NO2]2[N2O4]

=

N2O4 (g) → 2 NO2 (g) Eq 15.1

Eq 15.1

the concentrations of reactants and products no longer change.

The Equilibrium Constant• To generalize this expression, consider

the reactionaA + bB cC + dD


the reaction

• The equilibrium expression for this reaction would be

Kc = [C]c[D]d

[A]a[B]b

aA + bB cC + dD


the reaction

• The equilibrium expression for this reaction would be

Kc = [C]c[D]d

[A]a[B]b

aA + bB cC + dD

a.k.a. the LAW of MASS ACTION

products

reactants = K

SAMPLE EXERCISE 15.1 Writing Equilibrium-Constant Expressions

Write the equilibrium expression for Kc for the following reactions:

Kc is independent of the starting concentrations of reactants and products.

The Equilibrium Constant Because pressure is proportional to

concentration for gases in a closed system, the equilibrium expression can also be written in terms of partial pressures

Kp = (PC)c (PD)d

(PA)a (PB)b

Relationship between Kc and Kp

• From the ideal gas law we know that

• Rearranging it, we get

PV = nRT

P = RTnV

Relationship between Kc and Kp

Plugging this into the expression for Kp for each substance, the relationship between Kc and Kp becomes

Where Δn = (moles of gaseous product) − (moles of gaseous reactant)

Kp = Kc (RT)Δn

Kc represents the equil. constant when equilibrium concentrations are expressed in molarity and Kp represents the equilibrium constant when equilibrium concentrations are expressed in atm.

SAMPLE EXERCISE 15.2 Converting Between Kc and Kp

In the synthesis of ammonia from nitrogen and hydrogen, Kc = 9.60 at 300°C. Calculate Kp for this reaction at this temperature.

SAMPLE EXERCISE 15.2 Converting Between Kc and Kp

In the synthesis of ammonia from nitrogen and hydrogen, Kc = 9.60 at 300°C. Calculate Kp for this reaction at this temperature.

Solve:

Δn = 2 – 4 = –2 (Remember that Δ functions are always products minus reactants)

T = 273 + 300 = 573 KR = 0.0821 L-atm/mol-K Kc = 9.60

Equilibrium Can Be Reached from Either Direction

As you can see, the ratio of [NO2]2 to [N2O4] remains constant at this temperature no matter what the initial concentrations of NO2 and N2O4 are.

N2O4 (g) → 2 NO2 (g)


This is the data from the last two trials from the table on the previous slide.

N2O4 (g) → 2 NO2 (g)


It does not matter whether we start with N2 and H2 or whether we start with NH3. We will have the same proportions of all three substances at equilibrium.

What Does the Value of K Mean?

• If K >> 1, the reaction is product-favored; product predominates at equilibrium.

What Does the Value of K Mean?

• If K >> 1, the reaction is product-favored; product predominates at equilibrium.

• If K << 1, the reaction is reactant-favored; reactant predominates at equilibrium.

SAMPLE EXERCISE 15.3 Interpreting the Magnitude of an Equilibrium Constant

The reaction of N2 with O2 to form NO might be considered a means of “fixing” nitrogen:The value for the equilibrium constant for this reaction at 25°C is Kc = 1 × 10–30. Describe the feasibility of fixing nitrogen by forming NO at 25°C.

Solve: Because Kc is so small, very little NO will form at 25°C. The equilibrium lies to the left, favoring the reactants. Consequently, this reaction is an extremely poor choice for nitrogen fixation, at least at 25°C.

SAMPLE EXERCISE 15.3 Interpreting the Magnitude of an Equilibrium Constant

The reaction of N2 with O2 to form NO might be considered a means of “fixing” nitrogen:The value for the equilibrium constant for this reaction at 25°C is Kc = 1 × 10–30. Describe the feasibility of fixing nitrogen by forming NO at 25°C.

Manipulating Equilibrium Constants The equilibrium constant of a

reaction in the reverse reaction is the reciprocal of the equilibrium constant of the forward reaction.

Kc = = 0.212 at 100°C[NO2]2[N2O4]

N2O4 (g) 2 NO2(g)

Kc = = 4.72 at 100°C

[N2O4][NO2]2

Manipulating Equilibrium Constants The equilibrium constant of a

reaction in the reverse reaction is the reciprocal of the equilibrium constant of the forward reaction.

10.212

=

Kc = = 0.212 at 100°C[NO2]2[N2O4]

N2O4 (g) 2 NO2(g)

N2O4(g)2 NO2 (g)

Manipulating Equilibrium Constants The equilibrium constant of a reaction that has

been multiplied by a number is the equilibrium constant raised to a power that is equal to that number.

Kc = = 0.212 at 100°C[NO2]2[N2O4]

N2O4(g) 2 NO2(g)

Manipulating Equilibrium Constants The equilibrium constant of a reaction that has

been multiplied by a number is the equilibrium constant raised to a power that is equal to that number.

Kc = = 0.212 at 100°C[NO2]2[N2O4]

N2O4(g) 2 NO2(g)

Kc = = (0.212)2 at 100°C[NO2]4[N2O4]2

2 N2O4(g) 4 NO2(g)

Manipulating Equilibrium Constants

The equilibrium constant for a net reaction made up of two or more steps is the product of the equilibrium constants for the individual steps.

the new equilibrium constant is equal to (Kp)3.

For the following hypothetical reaction:Y(g) + Z(g) 2 A(g)

Keq = 4.0 × 10−2. Given that, what is the value of Keq for the reaction:

4 A(g) 2 Y(g) + 2 Z(g)

1. 252. 5.0

3. 0.0164. 6.3 × 102

Correct Answer:

The second equation is twice the first and reversed. Thus, Keq of the second equation is related to the first as shown below:

1. 252. 5.03. 0.0164. 6.3 × 102

K2 = 630

SAMPLE EXERCISE 15.5 Combining Equilibrium Expressions

Given the following information,

determine the value of Kc for the reaction

Solve: Multiplying the first equation by 2 and raising K to the power 2, we get








Reversing the second equation and taking the reciprocal of K gives





With two equations that sum to give the net equation, we multiply the individual Kc values to get the desired equilibrium constant.

Reversing the second equation and taking the reciprocal of K gives

Heterogeneous Equilibrium

The Concentrations of Solids and Liquids Are Essentially Constant

Both can be obtained by dividing the density of the substance by its molar mass—and both of these are constants at constant temperature.

The Concentrations of Solids and Liquids Are Essentially Constant

Therefore, the concentrations of solids and liquids do not appear in the equilibrium expression

Kc = [Pb2+] [Cl−]2

PbCl2 (s) Pb2+ (aq) + 2 Cl−(aq)

As long as some CaCO3 or CaO remain in the system, the amount of CO2 above the solid will remain the same.

CaCO3 (s) CO2 (g) + CaO(s)

Kp = PH2O

PRACTICE EXERCISE

Write the following equilibrium-constant expressions:

SAMPLE EXERCISE 15.7 Analyzing a Heterogeneous Equilibrium

Each of the following mixtures was placed in a closed container and allowed to stand. Which is capable of attaining the equilibrium

(a) pure CaCO3, (b) CaO and a CO2 pressure greater than the value of Kp, (c) some CaCO3 and a CO2 pressure greater than the value of Kp, (d) CaCO3 and CaO?

Solve: Equilibrium can be reached in all cases except (c) as long as sufficient quantities of solids are present. (a) CaCO3 simply decomposes, forming CaO(s) and CO2(g) until the equilibrium pressure of CO2 is attained. There must be enough CaCO3, however, to allow the CO2 pressure to reach equilibrium. (b) CO2 continues to combine with CaO until the partial pressure of the CO2 decreases to the equilibrium value. (c) There is no CaO present, so equilibrium can’t be attained because there is no way the CO2 pressure can decrease to its equilibrium value (which would require some of the CO2 to react with CaO). (d) The situation is essentially the same as in (a): CaCO3 decomposes until equilibrium is attained. The presence of CaO initially makes no difference.

SAMPLE EXERCISE 15.7 Analyzing a Heterogeneous Equilibrium

Each of the following mixtures was placed in a closed container and allowed to stand. Which is capable of attaining the equilibrium

(a) pure CaCO3, (b) CaO and a CO2 pressure greater than the value of Kp, (c) some CaCO3 and a CO2 pressure greater than the value of Kp, (d) CaCO3 and CaO?


Sections 15.5 & 15.6

Equilibrium calculations

BellworkWrite the equilibrium expression for the

following equilibrated systems.

2 NO(g) + O2(g) ⇌ 2NO2(g)

6 CO2(g) + 6 H2O(l) ⇌ C6H12O6(s) + 6 O2(g)

Equilibrium Calculations

For the following hypothetical reaction:2 Y(aq) + 3 Z(aq) A(aq)

Calculate Keq given the following equilibrium concentrations: [A] = 0.50 M, [Y] = 0.10 M, [Z] = 1.0 M

[ ] [ ]

[ ] [ ]ba

dc

eqKBA

DC=1. 2.5

2. 5.03. 0.404. 0.0255. 50.

[ ][ ] [ ] 32 1.00.10

0.50=eqK

Correct Answer:

1. 2.52. 5.03. 0.404. 0.0255. 50.

Equilibrium CalculationsA closed system initially containing1.000 x 10−3 M H2 and 2.000 x 10−3 M I2 at 448°C is allowed to reach equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is 1.87 x 10−3 M. Calculate Kc at 448°C for the reaction

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

What Do We Know?

[H2], M [I2], M [HI], M

Initially 1.000 x 10-3 2.000 x 10-3 0

Change

At equilibrium

1.87 x 10-3

[HI] Increases by 1.87 x 10-3 M

[H2], M [I2], M [HI], M

Initially 1.000 x 10-3 2.000 x 10-3 0

Change +1.87 x 10-3

At equilibrium

1.87 x 10-3

Stoichiometry tells us [H2] and [I2]decrease by half as much

[H2], M [I2], M [HI], M

Initially 1.000 x 10-3 2.000 x 10-3 0

Change -9.35 x 10-4 -9.35 x 10-4 +1.87 x 10-3

At equilibrium

1.87 x 10-3

We can now calculate the equilibrium concentrations of all three compounds…

[H2], M [I2], M [HI], M

Initially 1.000 x 10-3 2.000 x 10-3 0

Change -9.35 x 10-4 -9.35 x 10-4 +1.87 x 10-3

At equilibrium

6.5 x 10-5 1.065 x 10-3 1.87 x 10-3

…and, therefore, the equilibrium constant

Kc =[HI]2

[H2] [I2]

= 51

= (1.87 x 10-3)2

(6.5 x 10-5)(1.065 x 10-3)

PRACTICE EXERCISESulfur trioxide decomposes at high temperature in a sealed container: Initially, the vessel is charged at 1000 K with SO3(g) at a partial pressure of 0.500 atm. At equilibrium the SO3 partial pressure is 0.200 atm. Calculate the value of Kp at 1000 K.

Answer: 0.338

PRACTICE EXERCISESulfur trioxide decomposes at high temperature in a sealed container: Initially, the vessel is charged at 1000 K with SO3(g) at a partial pressure of 0.500 atm. At equilibrium the SO3 partial pressure is 0.200 atm. Calculate the value of Kp at 1000 K.

The Reaction Quotient (Q)

• To calculate Q, one substitutes the initial concentrations on reactants and products into the equilibrium expression.

The Reaction Quotient (Q)

• To calculate Q, one substitutes the initial concentrations on reactants and products into the equilibrium expression.

• Q gives the same ratio the equilibrium expression gives, but for a system that is not at equilibrium.

If Q = K,the system is at equilibrium.

If Q > K,there is too much product and the

equilibrium shifts to the left.

If Q < K,there is too much reactant, and the

equilibrium shifts to the right.

For the following hypothetical reaction:2 X(g) + Y(g) A(g)

Keq = 250. At a point during the reaction, the concentrations are [A] = 1.0 M, [X] = 0.50 M, and [Y] = 0.10 M. How will the reaction proceed to achieve equilibrium?

1. From left to right (toward products)2. From right to left (toward reactants)3. Already at equilibrium

Correct Answer:

[ ][ ] [ ]0.100.50

1.0 2

=Qc

Qc = 40Because Qc < Keq, the reaction will shift from reactants toward products to achieve equilibrium.

1. From left to right (toward products)2. From right to left (toward reactants)3. Already at equilibrium

SAMPLE EXERCISE 15.11 Calculating Equilibrium Concentrations

For the Haber process, at 500°C. In an equilibrium mixture of the three gases at 500°C, the partial pressure of H2 is 0.928 atm and that of N2 is 0.432 atm. What is the partial pressure of NH3 in this equilibrium mixture?


For the Haber process, at 500°C. In an equilibrium mixture of the three gases at 500°C, the partial pressure of H2 is 0.928 atm and that of N2 is 0.432 atm. What is the partial pressure of NH3 in this equilibrium mixture?

Comment: We can always check our answer by using it to recalculate the value of the equilibrium constant:


A 1.000-L flask is filled with 1.000 mol of H2 and 2.000 mol of I2 at 448°C. The value of the equilibrium constant Kc for the reaction

at 448°C is 50.5. What are the equilibrium concentrations of H2 , I2 , and HI in moles per liter?

from Initial Concentrations

If you have an equation-solving calculator, you can solve this equation directly for x. If not, expand this expression to obtain a quadratic equation in x:

Solving the quadratic equation (Appendix A.3) leads to two solutions for x:

If you have an equation-solving calculator, you can solve this equation directly for x. If not, expand this expression to obtain a quadratic equation in x:

Solving the quadratic equation (Appendix A.3) leads to two solutions for x:

When we substitute x = 2.323 into the expressions for the equilibrium concentrations, we find negative concentrations of H2 and I2. Because a negative concentration is not chemically meaningful, we reject this solution. We then use x = 0.935 to find the equilibrium concentrations:

Check: We can check our solution by putting these numbers into the equilibrium-constant expression:


Section 15.7 Le Chatelier’s Principle

BELLWORKEquilibrium is established between gases in the

reaction 2A(g) + 3B(g) ⇌ 1C(g) + 2D(g)

a) Write an expression for the equilibrium constantb) What happens to the forward reaction rate if

more A is added? All reactants are involved in the forward reaction’s rate law.

c) Would the equilibrium concentrations be the same as they were before more A was added? Explain.

d) Would [D] increase, decrease, or stay the same?

a) [C][D]2

[A]2[B]3

b) Forward reaction rate increases if A is addedc) No, to keep K constant when more A is added,

new equilibrium concentrations will develop.If all added A stays as A the ratio is “bottom heavy”.If all added A reacts to form products the ratio is

“top heavy”So SOME A will react to form products but not all. All of the concentrations will change, the

amount of change will depend on K, stoichiometry, and the amount of A added.

USE ICEBOX!d) D will increase as more products are formed.

Le Châtelier’s Principle

Le Châtelier’s Principle

“If a system at equilibrium is disturbed by a change in temperature, pressure, or the concentration of one of the components, the system will shift its equilibrium position so as to counteract the effect of the disturbance.”

An equilibrium position is a set of concentrations which fulfill an equilibrium

expression.


expression.

For a given reaction and temperature, there are infinite equilibrium positions, yet the value of K never changes.


expression.


Some things can shift the equilibrium position, but not the value of K.


expression.


Some things can shift the equilibrium position, but not the value of K.

Only a change in temperature will affect the value of K.

Le Chatelier’s principle

If a change is applied to a system at equilibrium, the position of the equilibrium will shift in a direction which counteracts the change.

The system will shift to undo the change

Changing concentrationA(g) + B(g) ⇌ C(g) + D(g)

Adding reactant causes a shift to the right as the forward rate increases and reactants form products.



Removing reactant causes a shift to the left as the forward reaction rate slows, causing a shift towards reactants.




Adding products causes a shift towards reactants as the reverse reaction rate increases.




Adding products causes a shift towards reactants as the reverse reaction rate increases.

Removing products causes a shift to the right and more product formation.

Changing concentration

A(g) + B(g) ⇌ C(g) + D(g)


Increase or decrease in concentration

Equilibrium position shift to right or left

a. the equilibrium shifts to the right, using up some of the added O2.

b. the equilibrium shifts to the left, forming more NO.

The following reaction is at equilibrium:N2(g) + 3 H2(g) 2 NH3(g)

If we remove NH3(g), in what direction will the reaction move to reestablish equilibrium?1. From left to right (toward products)2. From right to left (toward reactants)

3. No change in equilibrium

Correct Answer:

If [NH3] is decreased, to return to equilibrium, more NH3 must be produced from the reactants. Thus, the reaction shifts from the reactants to the products.

1. From left to right (toward products)2. From right to left (toward reactants)


Changing volume N2(g) + 3H2(g) ⇌ 2NH3(g)


4moles of gas 2moles of gas



Increasing pressure (decreasing volume) will shift equilibrium towards fewer moles of gas. Ex. Shifts right to decrease pressure




Increasing pressure (decreasing volume) will shift equilibrium towards fewer moles of gas. Ex. Shifts right to decrease pressure

Decreasing pressure shifts equilibrium towards more moles of gas.

Ex. Shifts left to increase pressure




Changing volume/pressure only affects gaseous equilibria

Adding an inert gas does not change the equilibrium position

the equilibrium shifts to the left.

The following reaction is at equilibrium:N2(g) + 3 H2(g) 2 NH3(g)

If we increase the volume while holding the temperature constant, in what direction will the reaction move to reestablish equilibrium?



Correct Answer:

Increasing the volume causes a shift in the equilibrium in the direction that produces more gas molecules, in this case on the reactants side.



The following reaction is at equilibrium:N2(g) + O2(g) 2 NO(g)

If we increase the pressure, in what direction will the reaction move to reestablish equilibrium?



Correct Answer:

The number of gas molecules on both sides of the equation are equal; thus, changing the pressure will not change the equilibrium position.



The equilibrium constant (K) does not change with changing concentration or pressure.

ONLY A TEMPERATURE

CHANGE AFFECTS K

The equilibrium constant (K) does not change with changing concentration or pressure.

Changes that cause no shift

• A shift does not occur when a catalyst is added because both forward and reverse rates are increased equally, but equilibrium will be established faster.

Changes that cause no shift

• A shift does not occur when a catalyst is added because both forward and reverse rates are increased equally, but equilibrium will be established faster.

• Adding pure liquid, solid or an inert gas to an equilibrium system will not cause a shift.

A temperature change will change K

Rates for forward & reverse reactions are affected differently by temperature change because of activation energy differences.

A temperature change will change K

Rates for forward & reverse reactions are affected differently by temperature change because of activation energy differences.

Temperature change is the only thing that will affect the value of K for an ideal closed system.


•Rewrite the equation with energy as a reactant for endothermic processes or as a product for exothermic processes.

Changing temperatureN2(g) + 3H2(g) ⇌ 2NH3(g) ΔH=-92kJ

Changing temperatureN2(g) + 3H2(g) ⇌ 2NH3(g) ΔH=-92kJ


•Rewrite the equation with energy as a reactant for endothermic processes or as a product for exothermic processes.

•Use Le Chatelier’s principle.

N2(g) + 3H2(g) ⇌ 2NH3(g) + 92kJ

Adding heat will shift the equilibrium to the left, in the direction that consumes energy

N2O4(g) ⇌ 2NO2(g) ΔH = +57kJ

57kJ + N2O4(g) ⇌ 2NO2(g)Heating the system shifts the

equilibrium right, consuming energy.


Changing temperature

N2O4(g) ⇌ 2NO2(g) ΔH = +57kJ

57kJ + N2O4(g) ⇌ 2NO2(g)Heating the system shifts the

equilibrium right, consuming energy.Cooling the system shifts the

equilibrium towards the reactants


Changing temperature

evaporation is an endothermic process and increasing temperature shifts equilibrium to the right.

The following reaction is at equilibrium:N2(g) + O2(g) 2 NO(g)

ΔH° = +180.8 kJ. In what direction will the reaction move to reestablish equilibrium if the temperature is decreased?



Correct Answer:

As the temperature is lowered, equilibrium will shift to the side of the equation that produces heat. Thus, for an endothermic reaction, the equilibrium shifts from right to left toward the reactants.



Catalysts increase the rate of both the forward and reverse reactions.

Equilibrium is achieved faster, but the equilibrium composition remains

unaltered.

SAMPLE EXERCISE 15.13 Using Le Châtelier’s Principle to Predict Shifts in Equilibrium

Consider the equilibriumIn which direction will the equilibrium shift when (a) N2O4 is added, (b) NO2 is removed, (c) the total pressure is increased by addition of N2(g), (d) the volume is increased, (e) the temperature is decreased?

SAMPLE EXERCISE 15.13 Using Le Châtelier’s Principle to Predict Shifts in Equilibrium

Consider the equilibriumIn which direction will the equilibrium shift when (a) N2O4 is added, (b) NO2 is removed, (c) the total pressure is increased by addition of N2(g), (d) the volume is increased, (e) the temperature is decreased?

Solve: (a) The system will adjust to decrease the concentration of the added N2O4, so the equilibrium shifts to the right, in the direction of products.(b) The system will adjust to the removal of NO2 by shifting to the side that produces more NO2; thus, the equilibrium shifts to the right.(c) Adding N2 will increase the total pressure of the system, but N2 is not involved in the reaction. The partial pressures of NO2 and N2O4 are therefore unchanged, and there is no shift in the position of the equilibrium.(d) If the volume is increased, the system will shift in the direction that occupies a larger volume (more gas molecules); thus, the equilibrium shifts to the right

(e) The reaction is endothermic, so we can imagine heat as a reagent on the reactant side of the equation. Decreasing the temperature will shift the equilibrium in the direction that produces heat, so the equilibrium shifts to the left, toward the formation of more N2O4. Note that only this last change also affects the value of the equilibrium constant, K.

Chapter 15 Lecture- Chemical Equilibrium

Education

Transcript of Chapter 15 Lecture- Chemical Equilibrium