Chapter 13 Chemical Equilibrium AP*. AP Learning Objectives LO 6.1 The student is able to, given a...

Chapter 13

Chemical Equilibrium

AP*

AP Learning Objectives

LO 6.1 The student is able to, given a set of experimental observations regarding physical, chemical biological, or environmental processes that are reversible, construct an explanation that connects the observations to the reversibility of the underlying chemical reactions or processes. (Sec 13.1-13.7)

LO 6.2 The student can, given a manipulation of a chemical reaction or set of reactions (e.g. reversal of reaction or addition of two reactions), determine the effects of that manipulation on Q or K. (Sec 13.4-13.6)

LO 6.3 The student can connect kinetics to equilibrium by using reasoning about equilibrium, such as Le Châtelier’s Principle, to infer the relative rates of the forward and reverse reactions. (Sec 13.7)

LO 6.4 The student can, given a set of initial conditions (concentrations or partial pressures) and the equilibrium constant, K, use the tendency of Q to approach K to predict and justify the prediction as to whether the reaction will proceed toward products or reactants as equilibrium is approached. (Sec 13.5)

LO 6.5 The student can, given data (tabular, graphical, etc.) from which the state of a system at equilibrium can be obtained, calculate the equilibrium constant, K. (Sec 13.3, 13.5-13.6)

AP Learning Objectives

LO 6.6 The student can, given a set of initial conditions (concentrations or partial pressures) and the equilibrium constant, K, use stoichiometric relationships and the law of mass action (Q equals K at equilibrium) to determine qualitatively and/or quantitatively the conditions at equilibrium for a system involving a single reversible reaction. (Sec 13.5-13.6)

LO 6.7 The student is able, for a reversible reaction that has a large or small K to determine which chemical species will have very large versus very small concentrations at equilibrium. (Sec 13.2)

LO 6.8 The student is able to use Le Châtelier’s Principle to predict the direction of the shift resulting from various possible stresses on a system at chemical equilibrium. (Sec 13.7)

LO 6.9 The student is able to use Le Châtelier’s Principle to design a set of conditions that will optimize a desired outcome, such as product yield. (Sec 13.7)

LO 6.10 The student is able to connect Le Châtelier’s Principle to the comparison of Q to K by explaining the effects of the stress on Q and K. (Sec 13.7)

Section 13.1The Equilibrium Condition

AP Learning Objectives, Margin Notes and References Learning Objectives LO 6.1 The student is able to, given a set of experimental observations regarding physical, chemical biological, or

environmental processes that are reversible, construct an explanation that connects the observations to the reversibility of the underlying chemical reactions or processes.


© 2015 Pearson Education, Inc.

Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate. In the figure above, equilibrium is finally reached in the third picture.


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.

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


Copyright © Cengage Learning. All rights reserved 7

Equilibrium Is: On the molecular level, there is frantic activity. Equilibrium is not static, but is a highly dynamic situation.

Macroscopically static Microscopically dynamic


Writing the Equation for an Equilibrium Reaction

Since, in a system at equilibrium, both the forward and reverse reactions are being carried out, we write its equation with a double arrow:

N2O4(g) 2 NO⇌ 2(g)


Comparing Rates

For the forward reactionN2O4(g) → 2 NO2(g)

The rate law isRate = kf [N2O4]

For the reverse reaction2 NO2(g) → N2O4(g)

The rate law isRate = kr [NO2]2


The Meaning of Equilibrium

Therefore, at equilibriumRatef = Rater

kf[N2O4] = kr[NO2]2

Rewriting this, it becomes the expression for the equilibrium constant, Keq.

Section 13.1The Equilibrium ConditionAnother Equilibrium—The Haber Process Consider the Haber Process,

which is the industrial preparation of ammonia:N2(g) + 3 H2(g) 2 NH⇌ 3(g)

The equilibrium constant depends on stoichiometry:



Consider an equilibrium mixture in a closed vessel reacting according to the equation:

H2O(g) + CO(g) H2(g) + CO2(g)

You add more H2O(g) to the flask. How does the concentration of each chemical compare to its original concentration after equilibrium is reestablished? Justify your answer.

CONCEPT CHECK!CONCEPT CHECK!



Consider an equilibrium mixture in a closed vessel reacting according to the equation:

H2O(g) + CO(g) H2(g) + CO2(g)

You add more H2 to the flask. How does the concentration of each chemical compare to its original concentration after equilibrium is reestablished? Justify your answer.


Section 13.2The Equilibrium Constant



LO 6.7 The student is able, for a reversible reaction that has a large or small K to determine which chemical species will have very large versus very small concentrations at equilibrium.


Consider the following reaction at equilibrium:

jA + kB lC + mD

A, B, C, and D = chemical species. Square brackets = concentrations of species at equilibrium. j, k, l, and m = coefficients in the balanced equation. K = equilibrium constant (given without units).


j

l

k

m

[B][A]

[D] [C]K =



Forward and reverse reaction rates, not reactant and product concentrations, that are equal.


The Direction of the Chemical Equation and K

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]

Kc = = 4.72 at 100 C[N2O4][NO2]2

N2O4(g)2 NO2(g)

N2O4(g) 2 NO2(g)⇌

⇌


Stoichiometry and Equilibrium Constants

To find the new equilibrium constant of a reaction when the equation has been multiplied by a number, simply raise the original equilibrium constant to that power. Here, the stoichiometry is doubled; the constant is the squared!

Kc = = 0.212 at 100 C[NO2]2

[N2O4]

Kc = = (0.212)2 at 100 C[NO2]4

[N2O4]24 NO2(g)2 N2O4(g)

N2O4(g) 2 NO2(g)⇌

⇌


Consecutive Equilibria When two consecutive equilibria occur, the

equations can be added to give a single equilibrium.

The equilibrium constant of the new reaction is the product of the two constants:K3 = K1 × K2

Example 2 NOBr 2 NO + Br⇌ 2 K1 = 0.014

Br2 + Cl2 2 BrCl⇌ K2 = 7.2

2 NOBr + Cl2 2 NO + 2 BrCl⇌

K3 = K1 × K2 = 0.014 × 7.2 = 0.10


Conclusions About the Equilibrium Expression

Equilibrium expression for a reaction is the reciprocal of that for the reaction written in reverse.

When the balanced equation for a reaction is multiplied by a factor of n, the equilibrium expression for the new reaction is the original expression raised to the nth power; thus Knew = (Koriginal)n

.

K values are usually written without units.



K always has the same value at a given temperature regardless of the amounts of reactants or products that are present initially.

For a reaction, at a given temperature, there are many equilibrium positions but only one equilibrium constant, K. Equilibrium position is a set of equilibrium

concentrations.



Equilibrium Can Be Reached from Either Direction

N2O4(g) 2 NO⇌ 2(g)

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.

Section 13.3Equilibrium Expressions Involving Pressures



LO 6.5 The student can, given data (tabular, graphical, etc.) from which the state of a system at equilibrium can be obtained, calculate the equilibrium constant, K.


Kc involves concentrations.

Kp involves pressures.

For a given reaction Kc is different from Kp .



Consider the generalized reactiona A + b B ⇌ d D + e E

The equilibrium expression for this reaction would be

• Also, since pressure is proportional to concentration for gases in a closed system, the equilibrium expression can also be written


Example

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


3

2 2

2

NH

p 3

N H

P =

P PK

2

33

2 2

NH =

N HK


Example

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

Equilibrium pressures at a certain temperature:


3

2

2

2NH

1N

3H

= 2.9 10 atm

= 8.9 10 atm

= 2.9 10 atm

P

P

P


Example

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


3

2 2

2

NH

p 3

N H

P =

P PK

22

p 31 3

2.9 10 =

8.9 10 2.9 10

K

4p = 3.9 10K


More with Gases and Equilibrium We can compare the equilibrium constant

based on concentration to the one based on pressure.

For gases, PV = nRT (the Ideal Gas Law). Rearranging, P = (n/V)RT; (n/V) is [ ].


The Relationship Between K and Kp

The result is

where

R = 0.08206 L·atm/mol·K T = temperature (in Kelvin)



Example

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

Using the value of Kp = 3.9 × 104, calculate the value of K at 35°C.


p

2 44

7

=

3.9 10 = 0.08206 L atm/mol K 308K

= 2.5 10

nK K RT

K

K

Section 13.4Heterogeneous Equilibria



LO 6.2 The student can, given a manipulation of a chemical reaction or set of reactions (e.g. reversal of reaction or addition of two reactions), determine the effects of that manipulation on Q or K.


Homogeneous Equilibria

Homogeneous equilibria – involve the same phase:N2(g) + 3H2(g) 2NH3(g)

HCN(aq) H+(aq) + CN-(aq)



Heterogeneous Equilibria

Heterogeneous equilibria – involve more than one phase:

2KClO3(s) 2KCl(s) + 3O2(g)

2H2O(l) 2H2(g) + O2(g)



The position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present. The concentrations of pure liquids and solids are

constant. (Therefore solids and liquids are not included in the equilibrium expression!)

Solutions and gases are always included in the expression.

2KClO3(s) 2KCl(s) + 3O2(g)


3

2 = OK


The Decomposition of CaCO3— The equation for the reaction is

CaCO3(s) CaO(⇌ s) + CO2(g) This results in

Kc = [CO2]

andKp = PCO2


CaCO3(s) CaO(⇌ s) + CO2(g)

Section 13.5Applications of the Equilibrium Constant




LO 6.4 The student can, given a set of initial conditions (concentrations or partial pressures) and the equilibrium constant, K, use the tendency of Q to approach K to predict and justify the prediction as to whether the reaction will proceed toward products or reactants as equilibrium is approached.


LO 6.6 The student can, given a set of initial conditions (concentrations or partial pressures) and the equilibrium constant, K, use stoichiometric relationships and the law of mass action (Q equals K at equilibrium) to determine qualitatively and/or quantitatively the conditions at equilibrium for a system involving a single reversible reaction.


If K>>1, the reaction favors products; products predominate at equilibrium.

If K<<1, the reaction favors reactants; reactants predominate at equilibrium.

K indicates the tendency for a reaction to occur, but NOT the speed of the reaction!!!


It is important to note that the size of K and the time required to reach equilibrium are NOT directly related.

The time to reach equilibrium depends on the reaction rate, which is determined by Ea.

The size of K is determined by thermodynamic factors such as E.


Deducing Equilibrium Concentrations1) Tabulate all known initial and equilibrium

concentrations.2) For anything for which initial and equilibrium

concentrations are known, calculate the change.3) Use the balanced equation to find change for all other

reactants and products.4) Use initial concentrations and changes to find

equilibrium concentration of all species.5) Calculate the equilibrium constant using the

equilibrium concentrations.


An Example

A closed system initially containing 1.000 × 10–3 M H2 and 2.000×10–3 M I2 at 448 °C is allowed to reach equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is 1.87 × 10–3 M. Calculate Kc at 448 °C for the reaction taking place, which is

H2(g) + I2(g) 2 HI(⇌ g)



What Do We Know?

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

Initially 1.000 × 10–3 2.000 × 10–3 0

Change

At equilibrium 1.87 × 10–3



[HI] Increases by 1.87 × 10−3 M

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

Initially 1.000 × 10–3 2.000 × 10–3 0

Change +1.87 × 10−3




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

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

Initially 1.000 × 10–3 2.000 × 10–3 0

Change −9.35 × 10−4 −9.35 × 10−4 +1.87 × 10–3




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

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

Initially 1.000 × 10–3 2.000 × 10–3 0

Change –9.35 × 10–4 –9.35 × 10–4 +1.87 × 10–3

At equilibrium 6.5 × 10−5 1.065 × 10−3 1.87 × 10–3



And, therefore, the equilibrium constant…

Kc =[HI]2

[H2] [I2]

= 51

=(1.87 × 10–3)2

(6.5 × 10–5)(1.065 × 10–3)


Is a Mixture in Equilibrium? Which Way Does the Reaction Go?

To answer these questions, we calculate the reaction quotient, Q.

Q looks like the equilibrium constant, K, but the values used to calculate it are the current conditions, not necessarily those for equilibrium.

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


Comparing Q and K Nature wants Q = K. If Q = K, the reaction is in

equilibrium.


If Q < K, nature will make the reaction proceed to products. Consuming reactants and forming products, to attain equilibrium

If Q > K, nature will make the reaction proceed to reactants. Consuming products and forming reactants, until equilibrium is achieved

Section 13.6Solving Equilibrium Problems





LO 6.6 The student can, given a set of initial conditions (concentrations or partial pressures) and the equilibrium constant, K, use stoichiometric relationships and the law of mass action (Q equals K at equilibrium) to determine qualitatively and/or quantitatively the conditions at equilibrium for a system involving a single reversible reaction.


Solving Equilibrium Problems

1) Write the balanced equation for the reaction.2) Write the equilibrium expression using the law of mass

action.3) List the initial concentrations.4) Calculate Q, and determine the direction of the shift to

equilibrium.



Solving Equilibrium Problems

5) Define the change needed to reach equilibrium, and define the equilibrium concentrations by applying the change to the initial concentrations.

6) Substitute the equilibrium concentrations into the equilibrium expression, and solve for the unknown.

7) Check your calculated equilibrium concentrations by making sure they give the correct value of K.



Calculating Equilibrium Concentrations

If you know the equilibrium constant, you can find equilibrium concentrations from initial concentrations and changes (based on stoichiometry).

You will set up a table similar to the ones used to find the equilibrium concentration, but the “change in concentration” row will simple be a factor of “x” based on the stoichiometry.

Section 13.6Solving Equilibrium ProblemsAn Example

A 1.000 L flask is filled with 1.000 mol of H2(g) and 2.000 mol of I2(g) at 448 °C. Given a Kc of 50.5 at 448 °C, what are the equilibrium concentrations of H2, I2, and HI?

H2(g) + I2(g) 2 HI(⇌ g)initial concentration (M)

1.000 2.000 0

change in concentration (M)

–x –x +2x

equilibrium concentration (M)

1.000 – x 2.000 – x 2x


Example (continued)

Set up the equilibrium constant expression, filling in equilibrium concentrations from the table.

• Solving for x is done using the quadratic formula, resulting in x = 2.323 or 0.935.

See appendix A1.4


Example (completed)

Since x must be subtracted from 1.000 M, 2.323 makes no physical sense. (It results in a negative concentration!) The value must be 0.935.

So [H2]eq = 1.000 – 0.935 = 0.065 M

[I2]eq = 2.000 – 0.935 = 1.065 M

[HI]eq = 2(0.935) = 1.87 M

This problem could have been given with pressures instead of concentrations.


If K <<1, then can simplify the problem by assuming the reaction does not proceed significantly to the left. You may delete the –X term in the denominator (from the reactant).

Once you solve for X, you should confirm that the value of X is <5% of the initial amount of reactant to confirm the Mount of reactant lost compared to what you started with is insignificant.

See example page 631


A 1.00 mol sample of N2O4(g) is placed in a 10.0 L vessel and allowed to reach equilibrium according to the equation:

N2O4(g) 2NO2(g)

K = 4.00 × 10-4

Calculate the equilibrium concentrations of: N2O4(g) and NO2(g).

Concentration of N2O4 = 0.097 M

Concentration of NO2 = 6.32 × 10-3 M



Section 13.7Le Châtelier’s Principle



LO 6.3 The student can connect kinetics to equilibrium by using reasoning about equilibrium, such as Le Châtelier’s Principle, to infer the relative rates of the forward and reverse reactions.

LO 6.8 The student is able to use Le Châtelier’s Principle to predict the direction of the shift resulting from various possible stresses on a system at chemical equilibrium.

LO 6.9 The student is able to use Le Châtelier’s Principle to design a set of conditions that will optimize a desired outcome, such as product yield.

LO 6.10 The student is able to connect Le Châtelier’s Principle to the comparison of Q to K by explaining the effects of the stress on Q and K.

Additional AP References LO 6.9 (see APEC #13, “Le Châtelier’s Principle”)


Studying the factors that control the position of equilibrium.

If a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that change.



Although max NH3 production is favored at low temperature and high pressure, this is not ideal. The reaction rate to obtain equilibrium is too slow. This stresses the need to study both the thermodynamics and kinetics of a chemical reaction.


Effects of Changes on the System1. Concentration

(reactant or product): The system will shift away from the added component. If a component is removed, the will shift to replace the removed component. (K not affected by change in concentration!!)




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Effects of Changes on the System

2. Temperature: K will change depending upon the temperature (endothermic – energy is a reactant; exothermic – energy is a product).



Change in Temperature Is the reaction endothermic or exothermic

as written? That matters! Endothermic: Heats acts like a reactant;

adding heat drives a reaction toward products.

Exothermic: Heat acts like a product; adding heat drives a reaction toward reactants.


An Endothermic Equilibrium


An Exothermic Equilibrium The Haber Process for producing ammonia

from the elements is exothermic. One would think that cooling down the

reactants would result in more product. However, the activation energy for this

reaction is high! This is the one instance where a system in

equilibrium can be affected by a catalyst!



Catalysts increase the rate of both the forward and reverse reactions. Equilibrium is achieved faster, but the equilibrium composition remains unaltered. Activation energy is lowered, allowing equilibrium to be established at lower

temperatures.


Effects of Changes on the System

3. Pressure: (K not affected by a change in pressure)a) The system will shift away from the added gaseous

component. If a component is removed, the opposite effect occurs.

b) Addition of inert gas does not affect the equilibrium position.

c) Decreasing the volume shifts the equilibrium toward the side with fewer moles of gas.


Chapter 13 Chemical Equilibrium AP*. AP Learning Objectives LO 6.1 The student is able to, given a...

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