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Chapter 13Chemical Equilibrium

Irreversible ReactionsChemical reactions in which proceed to completion where essentially

all of the reactants are converted into products

Example:

2H2(g) + O2(g) → 2H2O(g)

Such a reaction essentially occurs in one direction, called the “forward” direction as shown by the → symbol

2NO2(g) → N2O4(g)dark brown colorless

After adding pure NO2 to a closed vessel (the syringe), the dark brown color gradually fades but does not disappear completely!

Eventually the color stops changing when it reaches a light brown color!

Clearly the reaction has stopped short of completion but instead has reached a state of equilibrium where the concentration of reactants

and products are constant

Reversible ReactionsIn many chemical reactions there is not a complete conversion of

reactants into products because the reaction not only occurs in the “forward” (→) direction, but also in the “reverse” (←) direction where the products come together to reform the reactants. This is shown by

drawing a symbol between the reactants and products.

Example:

2NO2(g) N2O4(g)

The dark brown color does not completely disappear since as soon as the colorless N2O4 starts to form from the forward (→) reaction the

reverse reaction (←) begins to reform the dark brown NO2

Strictly speaking in a closed system ALL chemical reactions will reach equilibrium

However, in some reactions the products are so favored that the reaction essentially proceeds to completion where the amounts of

reactants at equilibrium is negligible

Example:2H2(g) + O2(g) 2H2O(g)

We say the equilibrium position is shifted far to the right

In other reactions the reactants are so favored that the reaction essentially does not occur so that the amounts of products at

equilibrium is negligible

Example:2CaO(s) 2Ca(s) + O2(g)

Here we say the equilibrium is shifted far to the left

Equilibrium is a Dynamic Condition!

To the eye it appears that a system at equilibrium is static and unchanging since the concentrations of reactants and products are

constant

However, at the molecular level the forward and reverse reactions are still occurring but at the same rate

Analogy:

When traffic is crossing a bridge in opposite directions at a uniform speed, the total number of cars on the bridge at any instant will be

approximately the same even though the traffic is still moving

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Attaining EquilibriumWhen a reversible reaction begins and only the reactants are present,

only the forward reaction occurs

However, as the reactants are used up and the products accumulate, the forward reaction rate decreases and the reverse reaction rate

increases

As the reaction proceeds the forward rate decreases and the reverse rate increases. Eventually, the forward and reverse reaction rates

become equal and the system reaches equilibrium

Similarly, the reactant concentrations decrease and the products concentration increase until equilibrium is reached and the concentrations of

reactants and products stabilize

The Law of Mass ActionAn empirical law relating the concentrations of reactants and products at

equilibrium

For the general equilibrium reaction:

aA + bB cC + dD

The Law of Mass Action is represented by the following equilibrium expression:

K = [C]c[D]d / [A]a[B]b

where the square brackets correspond to the concentrations at equilibrium, the superscripts are the coefficients in the balanced reactions and K is a

constant called the equilibrium constant

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Calculating Equilibrium ConstantsThe value of the equilibrium constant at a given temperature (K

depends on temperature) can be easily calculated in the equilibrium concentrations of the reactants and products at equilibrium are

known

Equilibrium constant values are normally written without units

Manipulating Equilibrium Constant Expressions

Summary

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Equilibrium Constants and Equilibrium PositionsThe equilibrium constant is unique at a given temperature, regardless

of the initial concentrations of gases that are mixed together

While the equilibrium constant, which depends on the ratio of the concentrations will always be the same at a given temperature, the

actual equilibrium concentrations can vary

Each set of possible equilibrium concentrations is called an equilibrium position

There is only one equilibrium constant at a given temperature but there are an infinite number of possible equilibrium positions

The particular equilibrium position attained by a system depends on the initial concentrations but the equilibrium constant does not

Equilibrium Expressions Involving Gases

Equilibria involving gases can also be described using pressures. This comes from the ideal gas equation:

K is used for an equilibrium constant expressed in terms of concentrations while Kp is used for an equilibrium constant expressed

in terms of pressures

Relationship between K and Kp

Here the temperature is in Kelvin and the Universal Gas Constant R = 0.08206 L.atm.mol-1.K-1

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Homogeneous EquilibriaReactants and products are all in the same state, most commonly the

gaseous state:

2O3(g) 3O2(g)

Heterogeneous EquilibriaReactants and products are all in two or more states

Heterogeneous Equilibrium ConstantsThe concentrations of pure solids and liquids in heterogeneous

equilibria are constant so are not included in the equilibrium constant expression

aA(s) + bB(g) cC(l) + dD(g)

K = [D]d / [B]b

This rule does not apply to solutions since their concentrations canchange

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Applications of the Equilibrium Constant

Knowing the equilibrium constant for a reaction we can determine:

1. Whether a set of concentrations represents a system in equilibrium

2. The tendency of a reaction to reach equilibrium (but not the rate it will reach equilibrium)

3. The final equilibrium position that will be attained from a given set of initial concentrations

Interpreting Equilibrium ConstantsThe magnitude of the equilibrium constant can be interpreted as a

measure of the inherent tendency of a reaction to occur

Reactions with large K correspond to equilibria which lie far to the right and therefore essentially go to completion

Reactions with small K correspond to equilibria which lie far to the left and therefore essentially to not occur

Reactions with a K close to 1 correspond to equilibria which contain roughly equal amounts of reactants and products at equilibrium

Equilibria with Large K2SO2(g) + O2(g) 2SO3(g)

K = [SO3]2 / [SO2]2[O2] = 3.4 x 102

Reaction essentially goes to completion

Equilibria with Small KCOCl2(g) CO(g) + Cl2(g)

K = [CO][Cl2] / [COCl2] = 2.2 x 10-10

Little reaction takes place

The size of K and the time to reach equilibrium are not related!

Example:

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

Hydrogen and oxygen have a strong tendency to form water since water has a lower energy. However, the large activation energy

prevents the reaction from occurring at room temperature

The value of K therefore depends on ∆E while the rate of reaction depends on Ea

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Reaction Quotient, Q

It is possible to determine whether a mixture of reactants and products are at equilibrium and if not the direction the equilibrium

will shift to attain it

To do this we calculate the reaction quotient, Q from the Law of Mass Action using the initial concentrations rather than the

equilibrium concentrations:

Calculating Equilibrium Pressures and Concentrations

Equilibrium problems typically involve finding the concentrations at equilibrium given the equilibrium constant and the initial

concentrations

Problems vary in difficultly depending on the information given

Procedure for Solving Equilibrium Problems

1. Write the balanced equation for the reaction2. Write the equilibrium expression from the law of mass action

3. List the initial concentrations4. Calculate Q and determine the direction of the shift to equilibrium

(not necessary if the initial concentration of one or more of the reactants and products is zero)

5. Construct an ICE table containing the Initial concentrations, the Change needed to reach equilibrium and then the Equilibrium

concentrations6. Substitute the equilibrium concentrations into the equilibrium

expression and solve for the unknown (try to simplify if K is small)

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

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Quadratic Equations

Many equilibrium and acid-base problems require solving quadratic equations so make sure you review this!

If:

then: 02 cbxax

a

acbbx

2

42

Suppose for a synthesis of hydrogen monofluoride from hydrogen and fluorine, 3.000 mol hydrogen and 6.000 mol of fluorine are

mixed in a 3.000 L flask. Given that the equilibrium constant for the reaction at this temperature is 1.15 x 102, calculate the equilibrium

concentration of each component

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Systems with Small Equilibrium Constants

Systems with small equilibrium constants are easier to deal with because we can make approximations which simplify the calculation

Modifying Equilibrium ConditionsIt is possible to shift the position of an equilibrium by modifying the

reaction conditions. Such changes are called “stresses”

Stresses can be applied by changing the concentrations of one or more of the reactants or products as well as changing the

temperature and pressure of the system

Le Châtelier’s PrincipleWhen a stress is placed on a reaction at equilibrium, the position of

the equilibrium shifts in the direction that relieves the stress

Concentration ChangesThe stress of adding a substance to an equilibrium system is relieved by shifting the position of the equilibrium away from that substance

The stress of removing a substance from an equilibrium system is relieved by shifting the position of the equilibrium towards that

substance

These shifts have to occur in order to keep the equilibrium constant Kthe same (assuming T and P or T and V are constant)

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

If we add N2, the equilibrium position shifts to the right forming more NH3. The equilibrium constant, however, remains unchanged

Pressure ChangesThere are three ways to change the pressure of an equilibrium

system containing gaseous components

1. Add or remove a gaseous reactant or product

2. Add an inert gas which is not involved in the reaction

3. Change the volume of the system

#1 has already been considered

#2 increases the total pressure but has no effect on the partial pressures of the reactants or products so there is no shift

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Pressure Changes through Changes in VolumeWhen the volume of an equilibrium system is reduced, the system responds by reducing its own volume. This is achieved by shiftingthe equilibrium position towards the side containing the fewer

number of molecules thereby decreasing the total number of gaseous molecules present in the system

When the volume of an equilibrium system is increased, the system responds by increasing its own volume. This is achieved by shiftingthe equilibrium position towards the side containing the greater

number of molecules thereby increasing the total number of gaseous molecules present in the system

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

If the volume decreases, the system responds by reducing its own volume by shifting the equilibrium position towards the side containing the fewer

number of molecules (the right)

Changing TemperatureWhen we change the temperature of a system at equilibrium, the

equilibrium constant changes

If the temperature of an equilibrium system is increased, the equilibrium position shifts to remove the heat

If the temperature of an equilibrium system is decreased, the equilibrium position shifts to replace the heat

Endothermic Reactions (∆H > 0)Heat is treated as a reactant:

aA(g) + bB(g) + heat cC(g) + dD(g)

Increase Temperature: Equilibrium shifts towards products

K increases

Decrease Temperature: Equilibrium shifts towards reactants

K decreases

Exothermic Reactions (∆H < 0)Heat is treated as a product:

aA(g) + bB(g) cC(g) + dD(g) + heat

Increase Temperature: Equilibrium shifts towards reactants

K decreases

Decrease Temperature: Equilibrium shifts towards products

K increases

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2NO2(g) N2O4(g)

dark brown colorless

100 ºC 0 ºC

exothermic or endothermic?

CatalystsSpeed up the rate of both forward and reverse reactions making the

time required to reach equilibrium shorter

However, have no effect on the equilibrium position

Summary