chapter 16 principles of chemical reactivity: equilibria

Jeffrey MackCalifornia State University,

Sacramento

Chapter 16

Principles of Chemical Reactivity: Equilibria

• All chemical reactions are reversible, at least in principle.

• The concept of equilibrium is fundamental to chemistry.

• The general concept of equilibrium was introduced in Chapter 3 to explain the limited dissociation of weak acids.

• The goals of this and the following chapter will be to consider chemical equilibria in quantitative terms.

• The extent to which that equilibrium lies (product favored verses reactant favored) will be discussed.

Chemical Equilibrium: A Review

• At some point in time during the progress of a reaction, if the concentration of the reactants and products remains constant, equilibrium is said to be achieved.

• The concentrations are NOT equal.

Equilibrium

Moving towards

equilibrium

Equilibrium established

Equilibrium

Equilibrium systems are said to be:

• Dynamic (in constant motion)• Reversible • Equilibrium can be approached

from either direction.

Pink to blueCo(H2O)6Cl2 Co(H2O)4Cl2 + 2 H2O

Blue to pinkCo(H2O)4Cl2 + 2 H2O Co(H2O)6Cl2

Properties of Chemical Equilibria

• After a period of time, the concentrations of reactants and products are constant.

• The forward and reverse reactions continue after equilibrium is attained.

• They are equal and opposite in rate.

Chemical EquilibriumFe3+(aq) + SCN (aq) Fe(SCN)2+ (aq)

Phase changes such as H2O(s) vs. H2O(liq)

Examples of Chemical Equilibria

When a general chemical reaction

is at equilibrium, the equilibrium constant is given by:

If K > 0 then the reaction is said to be product favored.If K < 0 then the reaction is said to be reactant favored.

aA bB cC dD

The Equilibrium Constant

Product-favoredK > 1

Reactant-favoredK < 1


For the formation of HI(g) the equilibrium constant is given by:


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N2O4 (g) 2NO2 (g)

Start with NO2 Start with N2O4 Start with NO2 & N2O4

equilibrium

equilibrium equilibrium

The equilibrium between reactants and products may be disturbed in three ways:(1) by changing the temperature(2) by changing the concentration of a reactant(3) by changing the volume (for systems involving gases) A change in any of these factors will cause a system at equilibrium to shift back towards a state of equilibrium. This statement is often referred to as Le Chatelier’s principle.

Disturbing a Chemical Equilibrium

Effect of the Addition or Removal of a Reactant or Product• If the concentration of a reactant or product is

changed from its equilibrium value at a given temperature, equilibrium will be reestablished eventually.

• The new equilibrium concentrations of reactants and products will be different, but the value of the equilibrium constant expression will still equal K


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If an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset as the system reaches a new equilibrium position.

Le Châtelier’s Principle

• Changes in Concentration

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

AddNH3

Equilibrium shifts left to offset stress

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• Changes in Concentration continued

Change Shifts the EquilibriumIncrease concentration of product(s) leftDecrease concentration of product(s) right

Decrease concentration of reactant(s)Increase concentration of reactant(s) right

left

aA + bB cC + dD

AddAddRemove Remove

Effect of Volume Changes on Gas-Phase Equilibria• For a reaction that involves gases, what happens to

equilibrium concentrations or pressures if the size of the container is changed? (Such a change occurs, for example, when fuel and air are compressed in an automobile engine.)

• To answer this question, recall that concentrations are in moles per liter. If the volume of a gas changes, its concentration therefore must also change, and the equilibrium composition can change.


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• Changes in Volume and Pressure

A (g) + B (g) C (g)

Change Shifts the EquilibriumIncrease pressure Side with fewest moles of gasDecrease pressure Side with most moles of gas

Decrease volumeIncrease volume Side with most moles of gas

Side with fewest moles of gas

• When the number of gas moles on either side is the same, there is no effect.

Effect of Temperatue Changes on Gas-Phase EquilibriaConsider the reaction of nitrogen and oxygen to form nitric oxide:

As the temperature of the reaction is increased, the equilibrium constant increases.

K Temperature (K)4.5 1031 2986.7 1010 9001.7 103 2300

o2 2 rN (g) O (g) 2NO(g) H 180.6 kJ

Why?


Effect of Temperature Changes on Gas-Phase EquilibriaLets write the reaction in this manner:

Notice that energy is included as a reactant!

2 2N (g) O (g) 180.6 kJ 2NO(g)


Effect of Temperature Changes on Gas-Phase EquilibriaLets write the reaction in this manner:

As temperature (Energy) is increased, equilibrium shifts to the right, favoring products.

2 2N (g) O (g) 180.6 kJ 2NO(g)


Effect of Temperatue Changes on Gas-Phase EquilibriaLets write the reaction in this manner:

As the concentration of products increases so does the value of the equilibrium constant.

2 2N (g) O (g) 180.6 kJ 2NO(g)


Effect of Temperatue Changes on Gas-Phase EquilibriaThis explains the increase in the equilibrium constant with increasing temperature.

K Temperature (K)4.5 1031 2986.7 1010 9001.7 103 2300


Effect of Temperature Changes on Gas-Phase Equilibria

Conclusion:• Increasing the temperature of an endothermic

reaction favors the products, equilibrium shifts to the right.

• Increasing the temperature of an exothermic reaction favors the reactants, equilibrium shifts to the left.

• Lowering temperature results in the reverse effects.


Kc (273 K) = 0.00077Kc (298 K) = 0.0059

2 4 2

or

N O (colorless) + heat 2 NO (brown)

H = + 57.2 kJ

Temperature Effects on Equilibrium

25

Catalyst lowers Ea for both forward and reverse reactions.Catalyst does not change equilibrium constant or shift equilibrium.

• Adding a Catalyst• does not change K• does not shift the position of an equilibrium system• system will reach equilibrium sooner


2 2 3

8 orxn

N (g) + 3H (g) 2NH (g)

K = 3.5 10 H 91.8 kJ

Change Reaction ShiftAdding more N2(g) RightRemoving H2(g) LeftDecreasing the container volume RightIncreasing the container temperature LeftIncreasing the container volume LeftDecreasing the container temperature Rightadding a catalyst no effectadding argon to the container no effect

Le Chatelier’s Principle Practice

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Le Châtelier’s Principle - Summary

Change Shift EquilibriumChange Equilibrium

ConstantConcentration yes no

Pressure yes* no

Volume yes* no

Temperature yes yes

Catalyst no no

*Dependent on relative moles of gaseous reactants and products

• In an equilibrium constant expression, all concentrations are reported as equilibrium values.

• Product concentrations appear in the numerator, and reactant concentrations appear in the denominator.

• Each concentration is raised to the power of its stoichiometric balancing coefficient.

• Values of K are dimensionless. • The value of the constant K is particular to the

given reaction at a specific temperature.

Writing Equilibrium Constant Expressions

Reactions Involving Solids• So long as a solid is present in the course of

a reaction, its concentration is not included in the equilibrium constant expression.

• Equilibrium constant:

2 2S(s) O (g) SO (g)


Reactions in Solution• If water is a participant in the chemical

reaction, its concentration based on magnitude is considered to remain constant throughout.

• Equilibrium constant:

3 2 4NH (aq) H O(l) NH (aq) OH (aq)


Reactions Involving Gases: Kc and Kp

• Concentration data can be used to calculate equilibrium constants for both aqueous and gaseous systems.

• In these cases, the symbol K is sometimes given the subscript “c” for “concentration,” as in Kc.

• For gases, however, equilibrium constant expressions can be written in another way: in terms of partial pressures of reactants and products.


Reactions Involving Gases: Kp

• Notice that the basic form of the equilibrium constant expression is the same as for Kc. In some cases, the numerical values of Kc and Kp are the same. They are different when the numbers of moles of gaseous reactants and products are different.

2 2H (g) I (g) 2HI(g)


Reactions Involving Gases: Kp & Kc

• The general relationship between Kp and Kc is derived in chapter 26, pa726.

• When the number of gas mole is equivalent on either side of the chemical equation, the two equilibrium constants are the same value.

R = the gas constant

T = the absolute temperature

n = (mols gas product) mols gas reactant)


In general, ALL reacting chemical systems are characterized by their REACTION QUOTIENT, Q.

If Q = K, then system is at equilibrium.

aA bB cC dD

The Reaction Quotient, Q

• If Q < K then the system is heading towards equilibrium: There are more reactants than products as expected at equilibrium. The reaction is said to be headed to the “right”.

• If Q > K the system has gone past equilibrium. There are more products than reactants as expected at equilibrium. The reaction is said to be headed to the left.

• If Q = K then the system is at equilibrium.

The Reaction Quotient, Q

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K >> 1

K << 1

Lie to the right Favor products

Lie to the left Favor reactants

Equilibrium Will

K = [C]c[D]d

[A]a[B]baA + bB cC + dD

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The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form COCl2 (g) at 740C are [CO] = 0.012 M, [Cl2] = 0.054 M, and [COCl2] = 0.14 M. Calculate the equilibrium constants Kc and Kp.

CO (g) + Cl2 (g) COCl2 (g)

Kc = [COCl2]

[CO][Cl2]=

0.140.012 x 0.054

= 220

Kp = Kc(RT)n

n = 1 – 2 = -1 R = 0.0821 T = 273 + 74 = 347 K

Kp = 220 x (0.0821 x 347)-1 = 7.7

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Heterogenous equilibrium applies to reactions in which reactants and products are in different phases.

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

[CaCO3] = constant[CaO] = constant

Kc = [CO2] = Kp = PCO2

The concentration of solids and pure liquids are not included in the expression for the equilibrium constant.

[CaO][CO2][CaCO3]

Kc =′

[CaCO3][CaO]Kc x′

• The value of a reaction’s equilibrium constant is determined by measuring the concentrations of the reactants and products when a system is at equilibrium.

• The equilibrium constant can also determined by looking at the changes in concentrations as a system achieves equilibrium.

• This is know as an “ICE” table.

Determining the Equilibrium Constant

• ICE tables:• I = Initial concentration• C = change in concentration• E = concentrations at equilibrium

Reactants Products

I

C

E

Determining the Equilibrium Constant

2.00 mol of NOCl is added to 1.00 L flask. At equilibrium the concentration of NO(g) is found to be 0.66 mol/L. What is the value of K?

Determining K

22NOCl(g) 2NO(g) + Cl (g)

[NOCl] [NO] [Cl2]InitialChange

Equilibrium

Determining K




[NOCl] [NO] [Cl2]Initial 2.00 0 0Change

Equilibrium

Determining K



[NOCl] [NO] [Cl2]Initial 2.00 0 0Change -0.66 +0.66 0.33

Equilibrium

note the reaction stoichiometry

Determining K



[NOCl] [NO] [Cl2]Initial 2.00 0 0Change -0.66 +0.66 +0.33

Equilibrium 1.34 0.66 0.33

Determining K


22NOCl(g) 2NO(g) + Cl (g)[NOCl] [NO] [Cl2]

Initial 2.00 0 0Change -0.66 +0.66 +0.33

Equilibrium 1.34 0.66 0.33

Determining K

1.00 mol each of H2 and I2 in a 1.00 L flask. What are the equilibrium concentrations of all species. Kc = 55.3?

2 2H (g) I (g) 2HI(g)

Equilibrium Concentrations from K


2 2H (g) I (g) 2HI(g)

[H2] [I2] [HI]Initial 1.00 1.00 0Change

Equilibrium



2 2H (g) I (g) 2HI(g)


[H2] [I2] [HI]Initial 1.00 1.00 0Change - x - x + 2x

Equilibrium


2 2H (g) I (g) 2HI(g)



Equilibrium 1.00 - x 1.00 - x + 2 x

Where x is defined as amount of H2 and I2 consumed on approaching equilibrium in moles.


2 2H (g) I (g) 2HI(g)



Equilibrium 1.00 - x 1.00 - x + 2 x

Consider the following reaction:If initial concentration of N2O4 is 0.50 M, what are the equilibrium concentrations?

2 4 2N O (g) 2NO (g)



2 4 2N O (g) 2NO (g)


[N2O4] [NO2]Initial 0.50 0Change

Equilibrium


2 4 2N O (g) 2NO (g)


[N2O4] [NO2]Initial 0.50 0Change -x +2x

Equilibrium


2 4 2N O (g) 2NO (g)


[N2O4] [NO2]Initial 0.50 0Change -x +2x

Equilibrium 0.50 - x 2x


2 4 2N O (g) 2NO (g)



2 4 2N O (g) 2NO (g)


Solving this requires the Quadratic Equation:

20.0059 (0.0059) 4 (4) (-0.0030)x

2 (4)


2 4 2N O (g) 2NO (g)



2 4 2N O (g) 2NO (g)


x = 0.027 or 0.028

The negative value is not reasonable which gives the equilibrium values of concentration to be:


2 4 2N O (g) 2NO (g)


x = 0.027 or 0.028


x = 0.027 M

[N2O4] = 0.50 M x = 0.47 M

[NO2] = 2x = 0.054 M


2 4 2N O (g) 2NO (g)


x = 0.027 or 0.028


x = 0.027 M

[N2O4] = 0.50 M x = 0.47 M

[NO2] = 2x = 0.054 M

The results are in agreement with the magnitude of K: The reactant concentration is favored due to the small value of K


2 4 2N O (g) 2NO (g)


• What happens when the change in concentration for a reaction is much less than the initial concentration of the reactants?

• It turns out that when K 100 < [A]0, then the quadratic equation is not required.

since [A]0 x [A]0

Approximations in Equilibrium Concentrations

• Let’s look at what happens to the previous problem if the initial concentration of N2O4(g) is doubled to 1.00 M.



2 4 2N O (g) 2NO (g)



Consider the following reaction:If initial concentration of N2O4 is 1.00 M, what are the equilibrium concentrations?Solving for the equilibrium concentration using the quadratic equation yields:


Consider the following reaction:If initial concentration of N2O4 is 1.00 M, what are the equilibrium concentrations?Solving for the equilibrium concentration using the quadratic equation yields:

x = 0.038 M, [N2O4] = 0.096 & [N2O] = 0.076


Consider the following reaction:If initial concentration of N2O4 is 1.00 M, what are the equilibrium concentrations?Solving for the equilibrium concentration using the approximation, ([N2O4]0 x) = [N2O4]0 also yields:

x = 0.038 M, [N2O4] = 0.096 & [N2O] = 0.076

The two results are in agreement.

Multiplying the coefficients of a reaction:

The second reaction is 2 the first.

The general relationship is:

When “n” is the multiplication factor.

More About Balanced Equations & Equilibrium Constants

Reversing a reaction

The second reaction is the reverse the first.

The general relationship is:

The equilibrium constant is the inverse of the original.


Adding reactions at equilibrium:

Net.

The new equilibrium constant is the product of the individual.


butane

isobutane

Butane-Isobutane Equilibrium

At equilibrium [isobutane] = 1.25 M and [butane] = 0.50 M. K = 2.50If 1.50 M butane is added to the system, what is the mew concentration of each when the system returns to equilibrium?


Let’s first calculate the value of Q to see which direction the reaction will take to reestablish equilibrium.

At equilibrium [isobutane] = 1.25 M and [butane] = 0.50 M.

If 1.50 M butane is added to the system, what is the new concentration of each when the system returns to equilibrium?

Butane Isobutane K = 2.50


Let’s first calculate the value of Q to see which direction the reaction will take to reestablish equilibrium.

Q < K: There are more reactants than products as expected at equilibrium. The reaction shift to the “right”. Reactants must react to form products.





Let’s now calculate the changes in concentration by setting up an “ICE” table.

[butane] [isobutane]InitialChange

Equilibrium









[butane] [isobutane]Initial 0.5 +1.50 1.25Change

Equilibrium






[butane] [isobutane]Initial 0.5 +1.50 1.25Change x +x

Equilibrium








Equilibrium 2.00 x 1.25 + x

Using the value of K, determine the change in concentration, “x”.



Butane-Isobutane EquilibriumButane Isobutane K = 2.50

Using the value of K, determine the change in concentration, “x”.




X = 1.07

The new concentrations for butane and isobutane are:

[butane] = 0.93 M [isobutane] = 2.32 M

Equilibrium is shifted to the right, forming more products.




Effect of Volume Changes on Gas-Phase EquilibriaConsider the following reaction:

At Equilibrium, the concentrations are:

[N2O4] = 0.0280 M & [NO2] = 0.0128 M

What would happen if the total volume of the system was suddenly doubled?

22 2 42NO (g) N O (g) K = 1.70 10 @298K


Effect of Volume Changes on Gas-Phase EquilibriaWhat would happen if the total volume of the system was suddenly doubled?

½ [N2O4] = ½ 0.0280M = 0.0140M

½ [NO2] = ½ 0.0128M = 0.00640M

22 2 42NO (g) N O (g) K = 1.70 10 @298K



½ [N2O4] = ½ 0.0280M = 0.0140M

½ [NO2] = ½ 0.0128M = 0.00640M

Calculate Q:

22 2 42NO (g) N O (g) K = 1.70 10 @298K



Q = 342 > K: Therefore some products must shift to reactants (left) to reestablish equilibrium.

22 2 42NO (g) N O (g) K = 1.70 10 @298K


Effect of Volume Changes on Gas-Phase Equilibria

Conclusion:• Increasing the volume of a container favors the side

of equilibrium with the greatest number of gas moles.

• Decreasing the volume favors the side with the least number of moles.

• When the number of gas moles on either side is the same, there is no effect.


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Chemical Kinetics and Chemical Equilibrium

A + 2B AB2

kf

kr

ratef = kf [A][B]2

rater = kr [AB2]

Equilibriumratef = rater

kf [A][B]2 = kr [AB2]

kf

kr

[AB2][A][B]2

= Kc =

95

Modifying the Chemical Equation (cont’d)

What will be the equilibrium constant K"c for the new reaction?

Consider the reaction: 2 NO(g) + O2(g) 2 NO2(g)

[NO2]2

Kc = ––––––––– = 4.67 x 1013 (at 298 K) [NO]2 [O2]

Now consider the reaction: NO2(g) NO(g) + ½ O2(g)

[NO] [O2]1/2 1 1/2

K"c = ––––––––– = ––– [NO2]2 Kc

= 2.14 x 10–14 = 1.46 x 10–7

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If the coef in the reaction is: Then K is:Doubled SquaredHalved Square rootReversed in sign InvertedMultiplied by a constant n Raised to the nth power

•Calculate K of reversed reaction, ½ of a reaction or doubled

Example: 2SO2(g) + O2(g) 2SO3(g) 4SO2(g) + 2O2(g) 4SO3(g) SO2(g) + ½ O2(g) SO3(g) 2SO3(g) O2(g) + 2SO2(g)

Example: 2CO(g) + O2(g) 2CO2(g) K = 2.75 x 1020 @1000KCO2(g) CO(g) + ½ O2(g) K = 6.03 x 10-11 @1000K

Summarize:

chapter 16 principles of chemical reactivity: equilibria

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