Chemical Equilibrium Reactions Go Both Ways Equilibrium Plural is equilibria (for you Latin fans) Reactions are reversible –they go forward and backward

Chemical Equilibrium

Reactions

Go

Both Ways

Equilibrium

• Plural is equilibria (for you Latin fans)

• Reactions are reversible– they go forward and backward

• There is always some product and reactant

• How much of each is there?– Depends on the reaction

Equilibria is when….

• The rate of the forward reaction and reverse reaction are the same

• Sealed jar example

Add liquid to empty, dry jar

When first added, the air has no liquid vapor.The vapor pressure of the liquid starts to populate the vaporWith molecules

The rate of evaporation does not change,but the rate of condensation continuously increasesUntil the rate of evaporation = the rate of condensation

Reactions Go Both Ways

2 NO2 (g) N2O4 (g)

Brown Colorless

N2O4 (g) 2 NO2 (g)

Colorless Brown

In an Equilibrium

The forward reaction rate is the same as

the reverse reaction

Chemical equilibriumDifferent types of arrows are used in chemical

equations associated with equilibria.

Single arrowSingle arrow

• Assumes that the reaction proceeds to completion as written.

Two single-headed arrowsTwo single-headed arrows

• Used to indicate a system in equilibrium.

Two single-headed arrows of different sizesTwo single-headed arrows of different sizes ..

• May be used to indicate when one side of an equilibrium system is favored.

Chemical equilibrium• Homogeneous equilibriaHomogeneous equilibria - Equilibria that

involve only a single phase.

Examples.Examples.

• All species in the gas phase

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

• All species are in solution.

HC2H2O2 (aq) H+ (aq) + C2H3O2- (aq)

Chemical equilibrium• Basic steps for reaching equilibrium:Basic steps for reaching equilibrium:

• For the general reaction:

A + B C

We can view the reaction as occurring in three steps.

• Initial mixing

• Kinetic region

• Equilibrium region

Chemical equilibriumInitial mixing.Initial mixing.

• When A and B are first brought together, there is no C present.

• The reaction proceeds as

A + B C

• This is just at the very start of the reaction. Things change as soon as some C is produced.

Chemical equilibriumKinetic region.Kinetic region.

• As soon as some C has been produced, the reverse reaction is possible.

A + B C

• Overall, we still see an increase in the net concentration of C.

• As we approach equilibrium, the rate of the forward reaction becomes slower.

Chemical equilibrium

Equilibrium region.Equilibrium region.

• A point is finally reached where the forward and reverse reactions occur at the same rate.

A + B C

• There is no net change in the concentration of any of the species.

Chemical equilibrium•C

on

cen

trati

on

•Time

•C

•B

•A

•Equilibrium•Region

•Kinetic•Region

Equilibrium•A point is ultimately•reached where the•rates of the forward•and reverse changes

•are the same.

• At this point,• equilibrium• is reached.

• Rate

of

Form

ati

on

Rate

of

Form

ati

on

•Time

•Reactants

•Products

Equilibrium•C

on

cen

trati

on

•Time

•Kinetic Equilibrium•Region Region

•Products

•Reactants

•An equilibrium exists when no further change in concentration occurs. Note that the

concentrations of products and reactants do not have to be equal!

•The Equilibrium ConstantThe Equilibrium Constant

•Also note that the Also note that the equilibrium equilibrium

concentrations of concentrations of the reactants and the reactants and products are the products are the

same regardless of same regardless of the whether or not the whether or not you start with only you start with only

the reactants or the reactants or only the products. only the products.

•No productsNo products

•No reactantsNo reactants

Equilibrium Constants• The equilibrium constant is related to the rate laws of

the forward and reverse reactions.• Because the forward and reverse reactions are equal in

rate, the rate laws must be equal to each other.• kf [N2O4] = kr[NO2]2

• Isolating the concentrations and rate constants together…

• [NO2]2/[N2O4] = kf/kr = a constant (Keq), or the equilibrium constant.

Equilibrium Constant

• The position of the equilibrium can be described mathematically

• We use the balanced chemical equation to develop a mathematical expression

• Use molarity to describe concentration– Remember, that’s moles per liter of solution

Law of Mass Action

pA + qB rC + tD

K = [C]r • [D]t

[A]p • [B]q

[ ] = molarity

K = equilibrium constant

Products “over”

reactants

Law of Mass Action

4 NH3(g) + 7O2 (g) 4NO2 (g) + 6H2O (g)

K = [C]r • [D]t = [NO2]4 • [H2O]6

[A]p • [B]q [NH3]4 • [O2]7

Equilibrium Constant

Keq = [products]x

[reactants]y

• More products means a bigger K– Reaction lies to the right

• More reactants means smaller K– Reaction lies to the left

Y Reactant X Product (g)

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

K = [NH3]2

[N2] • [H2]3

K = 6.02 x 10-2

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

K = [NH3]2

[N2] • [H2]3

K = 6.02 x 10-2

K = 6.02 x 10-2

N2(g) + 3H2(g) 2NH3(g)K = [NH3]2

[N2] • [H2]3

K = 6.02 x 10-2

K = 6.02 x 10-2

K = 6.02 x 10-2

Equilibria

• It doesn’t matter where you start

• The “ratio” of products to reactants is the same– Because the equilibria is established by the rate of the

forward reaction vs the rate of the reverse reaction

Equilibrium Constant, Keq

• The equilibrium constant, Keq, is the ratio

of the concentrations of the products compared to the concentrations of the reactants.

• Keq > 1 indicate that forming products is

favored. Keq < 1 indicate that reactants

don’t react easily.

•Equilibrium Constant, KEquilibrium Constant, Keqeq

• 3H2(g)+N2(g) 2NH3(g)• The equilibrium constant, Keq, for the

reaction above is determined by:

• Keq = [ NH3 ] 2

[ H2 ] 3 [ N2 ]

• At 472C, Keq = 0.105


2NH3(g) 3H2(g)+N2(g)•Reversing a reaction will result in the new Reversing a reaction will result in the new equilibrium constant, Kequilibrium constant, Keq neweq new, that is equal , that is equal

to 1 / Kto 1 / Keq oldeq old

•Keq new = [ H2 ] 3 [ N2 ] = 1 / 0.105

[ NH3 ] 2 or 9.52


NH3(g) 1.5H2(g)+0.5N2(g)•Changing the number of moles of reactants Changing the number of moles of reactants and products will exponentially change the and products will exponentially change the

equilibrium constant:equilibrium constant: KKeq neweq new= (K= (Keq oldeq old)^)^nn

Keq new = [ H2 ]1.5 [ N2 ]0.5 = (9.52)^0.5

[ NH3 ] or 3.09

Partial pressure equilibrium constantsAt constant temperature, the pressure of a gas is

proportional to its molarity.

Remember, for an ideal gas: PVPV = = nRTnRT

and molarity is: M = mole / liter or n/V

so: P = MR T

where R is the gas law constant

T is the temperature, K.

Partial pressure equilibrium constantsFor equilibria that involves gases, partial pressures can be used

instead of concentrations.

aA (g) + bB (g) eE (g) + fF (g)

Kp =

Kp is used when the partial pressures are expressed in units of atmospheres.

•pEe pF

f

•pAa pB

b

Partial pressure equilibrium constantsIn general, Kp ≠ Kc, instead

KKpp = = KKcc ( (RTRT))nngg

ng is the number of moles of gaseous products minus the number of moles of gaseous reactants.

ng = (e + f) - (a + b)

Partial pressure equilibrium constantsFor the following equilibrium, Kc = 1.10 x 107 at

700. oC. What is the Kp?

•atm L•mol K

• 2H2 (g) + S2 (g) 2H2S (g)

Kp = Kc (RT)ng

T = 700 + 273 = 973 K

R = 0.08206

ng = ( 2 ) - ( 2 + 1) = -1

Partial pressureequilibrium constants

Kp = Kc (RT)ng

= 1.10 x 107 (0.08206 ) (973 K)

= 1.378 x105

•atm L•mol K[ ]

-1

Equilibrium constants and expressions

Heterogeneous equilibriaHeterogeneous equilibria

• Equilibria that involve more than one phase.

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

• Equilibrium expressions for these types of systems do do notnot include the concentrations of the pure solids (or liquids).

Kc = [CO2]

Equilibrium constants and expressionsHeterogeneous equilibriaHeterogeneous equilibria - We don’t include the

pure solids and liquids because their concentrations do not vary. These values end up being included in the K value.

•CO2

•CaO &

•CaCO3As long as the temperature is constant and some solid is present, the amount of

solid present has no effect on the equilibrium.

Writing an equilibrium expression• Write a balanced equation for the

equilibrium.

• Put the products in the numerator and the reactants in the denominator.

• Omit pure solids and liquids from the expression

• Omit solvents if your solutes are dilute (<0.1M).

• The exponent of each concentration should be the same as the coefficient for the species in the equation.

Writing an equilibrium expressionExample.Example.

• What would be the equilibrium expression for the following?

(NH4)2CO3(s) 2 NH3(g) + CO2(g) + H2O(g)

•Kc = [NH3]2 [CO2] [H2O]

•(NH4)2CO3 is a pure solid so is not included Kc

•We use [NH3]2 because the coefficient for NH3(g) in the equation is 2.

Equilibrium constant alphabet soup

•KKcc

•KKpp

•The equilibrium The equilibrium constant expressed inconstant expressed in concentrationconcentration unitsunits

•The equilibrium The equilibrium constant expressed constant expressed inin pressurepressure unitsunits


•KKspsp

•For the limited For the limited dissociation of dissociation of

insoluble solids.insoluble solids.

KKspsp is called theis called the

•solubility product solubility product constant.constant.


•KKaa

•KKbb

•KKww

For the dissociation of For the dissociation of weakweak acidsacids

For the dissociation For the dissociation of weakof weak basesbases

For the dissociation ofFor the dissociation of waterwater into Hinto H++ and OH and OH--

Equilibrium and rate of reactionChemical reactions tend to go to equilibrium providing

that reaction takes place at a significant rate.

There is no relationship between the magnitude of the equilibrium constant and the rate of a reaction.

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

Kc = 2.9 x 1031 = [H2O]2

[H2]2 [O2]

However, the reaction will take years to reach equilibrium at room temperature.

Determining equilibrium constantsEquilibrium constants can be found by

experiment.

If you know the initial concentrations of all of the reactants, you only need to measure the concentration of a single species at equilibrium to determine the Kc value.

Lets consider the following equilibrium:

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

Determining equilibrium constantsH2 (g) + I2 (g) 2HI (g)

Assume that we started with the following initial concentrations at 425.4oC.

H2 (g) 0.00500 M

I2 (g) 0.01250 MHI (g) 0.00000 M

At equilibrium, we determine that the concentration of iodine is 0.00772 M

Determining equilibrium constantsThe equilibrium expression for our system is:

Kc =

Based on the chemical equation, we know the equilibrium concentrations of each species.

I2 (g) = the measured amount

= 0.00772 M

That means that 0.00478 mol I2 reacts (0.01250M - 0.00772M) to produce HI in 1.00 L of solution.

[HI]2

[H2] [I2]

Determining equilibrium constantsI2 (g) = 0.00772 M

HI (g) = 0.00478 M

= 0.00956 M

H2 (g) = 0.00500 M - 0.00478 M

= 0.00022 M

Kc = = = 54

2 mol HI

1 mol I2

•[HI]2

•[H2] [I2]

(0.00956)2

(0.00022)(0.00772)

Equilibrium calculationsWe can predict the direction of a reaction by

calculating the reaction quotient.

Reaction quotient, Reaction quotient, QQ

For the reaction: aA + bB eE + fF

Q has the same form as Kc with one important difference. Q can be for any set of concentrations, not just at equilibrium.

Q =[E]e [F]f

[A]a [B]b

Reaction quotientAny set of concentrations can be given and a Q

calculated. By comparing Q to the Kc value, we can predict the direction for the reaction.

QQ < < KKcc Net forward reaction will occur.

QQ = = KKcc No change, at equilibrium.

QQ > > KKcc Net reverse reaction will occur.

Reaction quotient exampleFor an earlier example

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

we determined the Kc to be 54 at 425.4 oC.

If we had a mixture that contained the following, predict the direction of the reaction.

[H2] = 4.25 x 10-3 M

[I2] = 3.97 x 10-1 M

[HI] = 9.83 x 10-2 M

Reaction quotient example

Q =

=

= 5.73

Since Q is < Kc, the system is not in equilibrium and will proceed in the forward direction.

[ HI ]2

[ H2 ] [ I2 ]

(9.83 x 10-2)2

(4.25 x 10-3)(3.97 x 10-1)

Calculating equilibrium concentrations

If the stoichiometry and Kc for a reaction is known, calculating the equilibrium concentrations of all species is possible.

Commonly, the initial concentrations are known.

One of the concentrations is expressed as the variable x.

All others are then expressed in terms of x.

Equilibrium calculation exampleA sample of COCl2 is allowed to decompose. The

value of Kc for the equilibrium

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

is 2.2 x 10-10 at 100 oC.

If the initial concentration of COCl2 is 0.095M, what will be the equilibrium concentrations for each of the species involved?

Equilibrium calculation exampleCOCl2 (g) CO (g) Cl2 (g)

Initial conc., M 0.095 0.000 0.000

Change in conc. - X + X + Xdue to reaction

EquilibriumConcentration, M(0.095 -X) X X

Kc = =[CO ] [Cl2 ]

[ COCl2 ]

X2

(0.095 - X)

Equilibrium calculation exampleX2

(0.095 - X)Kc = 2.2 x 10-10 =

•Rearrangement gives

X2 + 2.2 x 10-10 X - 2.09 x 10-11 = 0

This is a quadratic equation. Fortunately, there•is a straightforward equation for their solution

Quadratic equationsAn equation of the form

a X2 + b X + c = 0

Can be solved by using the following

x =

Only the positive root is meaningful in equilibrium problems.

-b + b2 - 4ac

2a

Equilibrium calculation example

-b + b2 - 4ac

2a

X2 + 2.2 x 102.2 x 10-10-10 X - 2.09 x 102.09 x 10-11-11 = 0

a bb cc

X =

- 2.2 x 10-10 + [(2.2 x 10-10)2 - (4)(1)(- 2.09 x 10-11)]1/2

2

•X = 9.1 x 10-6 M

Equilibrium calculation exampleNow that we know X, we can solve for the concentration of all of

the species.

COCl2 = 0.095 - X = 0.095 M

CO = X = 9.1 x 10-6 M

Cl2 = X = 9.1 x 10-6 M

In this case, the change in the concentration of is COCl2 negligible.

Summary of method of calculating equilibrium concentrations

Write an equation for the equilibrium.

Write an equilibrium constant expression.

Express all unknown concentrations in terms of a single variable such as x.

Substitute the equilibrium concentrations in terms of the single variable in the equilibrium constant expression.

Solve for x.

Use the value of x to calculate equilibrium concentrations.

You Try!• A 40.0g sample of solid (NH4)2CO3 is placed in a closed

evacuated 3L flask and heated to 400oC. It decomposes to produce NH3, H2O, and CO2 :

• (NH4)2CO3 ↔ 2 NH3 + H2O + CO2

• The equilibrium constant, Kp , is 0.295

• Write the Kp expression.

• Calculate Kc

• Calculate the partial pressure of NH3

• Calculate the total pressure in the flask at equilibrium.• Calculate the number of grams of solid at equilibrium.• What is the minimum amount of solid needed to establish

equilibrium?

Answers• Kp = PNH3

2 PH2O PCO2

• 3.17x10-8

• 1.04 atm

• 2.08 atm

• 37.3g

• More than 2.72g

Predicting Shifts: LeChatelierAny stress placed on an equilibrium system will Any stress placed on an equilibrium system will

cause the system to shift to minimize the effect cause the system to shift to minimize the effect of the stress.of the stress.

•How can you cause the color to change from pink to blue?

• You can put ssttressress on a system by adding or removing something from one side of a

reaction.

• Co(HCo(H22O)O)662+2+ + 4Cl1- CoClCoCl44

2-2- + 6H2O

Predicting shifts in equilibria• Equilibrium concentrations are based on:

– The specific equilibrium

– The starting concentrations

– Other factors such as:• Temperature

• Pressure

• Reaction specific conditions

• Altering conditions will stress a system, resulting in an equilibrium shift.

Le Chatelier’s principleCo(HCo(H22O)O)66

2+2+ + 4Cl1- CoClCoCl442- 2- + 6H2O

Note: ∆H = + valueNote: ∆H = + value

•What other stresses could be placed on this system to change the color back and forth

between pink and blue?

• Add heatAdd heat

• (removes (removes HH22O)O)• Add Add

AgAg1+1+ • (removes (removes ClCl1-1- ) ) • Add Add iceice

• Add acetoneAdd acetone

• Add HCl Add HCl

(adds (adds ClCl1-1-) )

Le Chatelier’s principle

Any stress placed on an equilibrium Any stress placed on an equilibrium system will cause the system to shift to system will cause the system to shift to minimize the effect of the stress.minimize the effect of the stress.

• You can put stress on a system by adding or removing something from one side of a reaction.

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

•What effect will there be if you added more•ammonia? How about more nitrogen?

Changes in concentrationChanges in concentration do not change the value

of the equilibrium constant at constant temperature.

When a material is added to a system in equilibrium, the equilibrium will shift away from that side of the equation.

When a material is removed from a system in equilibrium, the equilibrium will shift towards that sid of the equation.

•Le Châtelier’s PrincipleLe Châtelier’s Principle• Change in Reactant or Product ConcentrationsChange in Reactant or Product Concentrations

• Consider the Haber processN2(g) + 3H2(g) 2NH3(g)

• If HIf H22 is added while the system is at equilibrium, the is added while the system is at equilibrium, the

system must respond to counteract the added Hsystem must respond to counteract the added H22 (by (by

Le Châtelier).Le Châtelier).

• That is, the system must consume the HThat is, the system must consume the H22 and produce and produce

products until a new equilibrium is established.products until a new equilibrium is established.

• Therefore, [H2] and [N2] will decrease and [NH3] increases.



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

• Adding a reactant or product shifts the equilibrium Adding a reactant or product shifts the equilibrium away from the increase.away from the increase.

• Removing a reactant or product shifts the equilibrium Removing a reactant or product shifts the equilibrium towards the decrease.towards the decrease.

• To optimize the amount of product at equilibrium, we need to flood the reaction vessel with reactant and

continuously remove product (Le Châtelier).• We illustrate the concept with the industrial

preparation of ammonia


• N2 and H2 are pumped into a

chamber.• The pre-heated

gases are passed through a heating coil to the catalyst

bed.

• The catalyst bed is kept at 460 - 550 C under high

pressure.

• The product gas stream

(containing N2, H2 and NH3) is passed over a cooler to a refrigeration unit.

• In the refrigeration unit, ammonia liquefies but not N2 or H2.


• The unreacted nitrogen and hydrogen are recycled with the new N2 and H2 feed gas.

• The equilibrium amount of ammonia is optimized because the product (NH3) is continually removed and the reactants (N2 and H2) are continually being added.

• Effects of Volume and PressureEffects of Volume and Pressure• As volume is decreased pressure increases.

• Le Châtelier’s Principle:Le Châtelier’s Principle: if pressure is if pressure is increased the system will shift to counteract the increased the system will shift to counteract the

increase by producing fewer moles of gas.increase by producing fewer moles of gas.

Changes in pressure• Changing the pressure does not change the value of

the equilibrium constant at constant temperature.

• Solids and liquids are not effected by pressure changes.

• Changing pressure by introducing an inert gas will not shift an equilibrium.

• Pressure changes only effect gases that are a portion of an equilibrium.

Changes in pressureIn general, increasing the pressure by decreasing

volume shifts equilibria towards the side that has the smaller number of moles of gas.

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

N2O2 (g) 2NO2 (g)

•Unaffected by pressure•Unaffected by pressure

•Increased pressure, shift to left•Increased pressure, shift to left

•Le Châtelier’s PrincipleLe Châtelier’s Principle• Effects of Volume and PressureEffects of Volume and Pressure

• That is, the system shifts to remove gases and decrease pressure.

• An increase in pressure favors the direction that has An increase in pressure favors the direction that has fewer moles of gas.fewer moles of gas.

• In a reaction with the same number of product and reactant moles of gas, pressure has no effect.

• Consider N2O4(g) 2NO2(g)

•Le Châtelier’s PrincipleLe Châtelier’s Principle• Effects of Volume and PressureEffects of Volume and Pressure

• An increase in pressure (by decreasing the volume) favors the formation of colorless N2O4.

• The instant the pressure increases, the system is not at The instant the pressure increases, the system is not at equilibrium and the concentration of both gases has equilibrium and the concentration of both gases has

increased.increased.• The system moves to reduce the number moles of gas The system moves to reduce the number moles of gas

(i.e. the forward reaction is favored).(i.e. the forward reaction is favored).• A new equilibrium is established in which the mixture

is lighter because colorless N2O4 is favored.

•Le Châtelier’s PrincipleLe Châtelier’s Principle• Effect of Temperature ChangesEffect of Temperature Changes

• The equilibrium constant is temperature dependent.

• For an endothermic reaction, H > 0 and heat can be considered as a reactant.

• For an exothermic reaction, For an exothermic reaction, HH < 0 and heat can be < 0 and heat can be considered as a product.considered as a product.

• Adding heat (i.e. heating the vessel) favors away from the increase:

– if H > 0, adding heat favors the forward reaction,

– if H < 0, adding heat favors the reverse reaction.


• Removing heat (i.e. cooling the vessel), favors towards the decrease:

if H > 0, cooling favors the reverse reaction,

if H < 0, cooling favors the forward reaction.

for which H > 0.

–Co(H2O)62+ is pale pink and CoCl4

2- is blue.

Cr(H2O)6(aq) + 4Cl-(aq) CoCl42-(aq) + 6H2O(l)



If a light purple room temperature equilibrium mixture is placed in a beaker of warm water, the mixture turns deep blue.

Since H > 0 (endothermic), adding heat favors the forward reaction, i.e. the formation of blue CoCl4

2-.

If the room temperature equilibrium mixture is placed in a beaker of ice water, the mixture turns bright pink.

Since H > 0, removing heat favors the reverse reaction which is the formation of pink Co(H2O)6

2+.

Cr(H2O)6(aq) + 4Cl-(aq) CoCl42-(aq) + 6H2O(l)• Co

•Le Châtelier’s PrincipleLe Châtelier’s Principle

• The Effect of CatalystsThe Effect of Catalysts

• A catalyst lowers the activation energy barrier for the reaction.

• Therefore, a catalyst will decrease the time a catalyst will decrease the time taken to reach equilibriumtaken to reach equilibrium.

• A catalyst A catalyst does notdoes not effect the effect the composition of the equilibrium composition of the equilibrium

mixture.mixture.

You try!• H2(g) + S(s) → H2S(g) ∆Hrxn = -20.17kJ/mol

• An amount of solid S and an amount of gaseous H2

are placed in an evacuated container at 25oC. At equilibrium, some solid S remains in the container. Predict and explain the following:

• Effect on partial pressure of H2S when S is added.

• Effect on partial pressure of H2 when H2S is added

• Effect on mass of S when volume is increased.• Effect on S when the temperature increased.• Effect of adding a catalyst to initial concentrations.

Answers• Solids have no effect on equilibrium

• P H2 will increase.

• S remains the same, equal numbers of moles of gases on both sides means no change in equilibrium.

• S decreases. Equilibrium is not effected by solids, but solid amounts can be changed by equilibrium. If more gases are produced, solid must go down.

• Catalysts make them reach equilibrium faster, but do not change equilibrium positions.

Documents

Chemical Equilibrium Reactions Go Both Ways Equilibrium Plural is equilibria (for you Latin fans) Reactions are reversible –they go forward and backward