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

Chemical Equilibrium

Many reactions do not go to completion - under the given conditions it is possible that not all of the reactants are consumed.

Instead the extent of the reaction is determined by the equilibrium point.

A + B --> C + D

As the concentrations of C and D increase, C and D could react to form A and B - the REVERSE reaction.

A + B C + D

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

The forward and reverse directions “oppose” one another.

At some point in time, the rate of the forward reaction will equal the rate of the reverse reaction - this point corresponds to EQUILIBRIUM.

Hence, when equilibrium has been reached, the concentration of A, B, C and D stay constant, as long as the conditions are held the same.

At the equilibrium point: A and B combine to form C and D; C and D combine to form A and B; but both occur at the same rate.

There is no NET change in the concentrations of A, B, C & D.

Equilibrium

When opposing forces acting on a system are equal in magnitude, the system is said to be in a state of equilibrium.

A dynamic equilibrium is one at which changes to the system do occur at the microscopic level, but at the macroscopic level these changes are not observed.

In general: Processes not at equilibrium will act or react to reach equilibrium.

A B

Characteristics of equilibrium

1) The attainment of equilibrium is spontaneous; i.e. it is a natural tendency

2) At equilibrium there is no macroscopic evidence of any changes in the system

3) A dynamic balance is established between opposing forces

4) Equilibrium is reached from either direction

The Equilibrium Expression

the ratio:

(concentration C)c (concentration D)d

(concentration A)a (concentration B)b= K (constant)

K is called the EQUILIBRIUM CONSTANT

Note: K has a fixed value for a particular reaction and varies with temperature

a A + b B c C + d D

For a general reaction:

If all reactants and products are gases, the relationship between the partial pressures of all gases at equilibrium is:

PCc PD

d

= KPA

a PBb

If all reactants and products are in solution, the relationship between the concentrations of all species at equilibrium is:

[C]c [D]d

= K[A]a [B]b

Where [X] is the concentration (example molarity) of species X at equilibrium

Homogenous reactions; reactants and products in the same phase

Heterogeneous reaction: reactants and products are not in the same phase

a A(aq) + b B(aq) c C(aq) + d D(g)

[C]c PDd

= K[A]a [B]b

K is a dimensionless quantity.

2A(g) B(g)

K = PB/P2A

is actually

K= PB/Pref

(PA/Pref)2

Pref is set to 1 atm

K is dimensionless

For solutions, if concentration is M; [A]ref = 1M

HCl(aq) H+(aq) + Cl-(aq)

[H+] [Cl-]= K

[HCl]

CH3COOH(aq) H+(aq) + CH3COO-(aq)

[H+] [CH3COO-]= K

[CH3COOH]

K ~ 107 at 25oC

K ~ 10-5 at 25oC

NH3(g) H2(g) + N2(g)

The reaction of SO2(g) and O2(g) forming SO3(g)

2 SO2(g)+ O2(g) 2SO3(g)

P2SO3

= K

PSO2

2PO2

Equilibrium can be reached for different partial pressures of SO2, O2, and SO3, depending on the starting conditions, but at 25oC, the value of K is the same.

The Magnitude of the Equilibrium Constant

The magnitude of the equilibrium constant reveals the extent to which the reaction will proceed in the desired direction.

a A + b B c C + d D

Reactions that have K values > 1 are favored in the direction written; i.e. forward direction.

Reactions that have K values < 1 are favored in the reverse direction

Reactions for which K is near I have substantial amounts of both reactants and products when equilibrium is established.

Applying the Equilibrium Expression to Gas Phase Reactions

PCl5(g) PCl3(g) + Cl2 (g) K = 2.15

What are the equilibrium partial pressures of all three gases in a closed container containing only PCl5 at 0.100 atm and held at 250oC?

According to the ideal gas laws, the partial pressures of gases is proportional to the number of moles of each gas, as long as the volume and temperature are kept fixed.

The stoichiometry of this reaction is 1 : 1 : 1

PCl5(g) PCl3(g) + Cl2 (g)

Initial P (atm)

Change in P (atm)

Equilibrium P (atm)

-x x x

0.100 0 0

0.100-x x x

If the partial pressure of PCl5 decreases by x at equilibrium, the partial pressures of PCl3 and Cl2 increases by x at equilibrium.

At equilibrium:

PPCl3 PCl2 = K

PPCl5

This is a quadratic equation of the form

ax2+ bx + c = 0

and the solution of this equation is of the form

= 2.15(0.100-x)

(x) (x)

x2 = 2.15 (0.100-x)

x2 + 2.15x - 0.215 = 0

x = (-b ± √b2 - 4ac)

2a

Using this expression and solving for x, the roots of the equation are x = 0.0957 and -2.25 atm.

At equilibrium, the partial pressures of Cl2 and PCl3 are 0.0957 atm, and that of PCl5 is (0.100 - 0.0957) = 0.004atm

4 NO2(g) N2O(g) + 3 O2 (g)

The three gases are introduced into a container at partial pressures of 3.6 atm NO2, 5.1 atm N2O and 8.0atm O2 and react to reach equilibrium at a fixed temperature. The equilibrium partial pressure of NO2 is measured to be 2.4 atm. Calculate the equilibrium constant of the reaction at this temperature, assuming that no competing reactions occur.

At equilibrium, the partial pressure of NO2 is 2.4 atm

3.6 - 4x = 2.4 => x = 0.3 atm

Hence PN2O at equilibrium = 5.7 atm; PO2 = 8.9 atm

4 NO2(g) 2 N2O(g) + 3 O2 (g)

Initial P (atm)

Change in P (atm)

Equilibrium P (atm)

- x 2x/4 3x/4

3.6 5.1 8.0

2.4 5.1+2x 8.0+3x

K = (5.7)2 (8.9)3

(2.4)4= 6.9 x 102

Change in P (atm) - 4x 2x 3x

In applying the equilibrium expression the following must be considered.

1) The equilibrium constant for a reverse reaction is the reciprocal of the equilibrium constant for the corresponding forward reaction.

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

= K1

PH2O

2

PH2 PO2

2

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

PH2O

2 = K2

PH2 PO2

2

K1 = K2

-1

2) When the coefficients in a balanced chemical equation are multiplied by a constant factor, the corresponding equilibrium constant is raised to the power equal to that factor.

= K2

PH2O

PH2 PO2

1/2

H2(g) + O2 (g) H2O(g) 12

K2 = √K1

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

= K1

PH2O

2

PH2 PO2

2

3) When chemical equations are added or subtracted to obtain a net equation, the corresponding equilibrium constants are multiplied or divided to obtain the equilibrium constant of the net equation.

2 BrCl(g) Cl2 (g) + Br2(g)

= K1 = 0.45 at 25oCPBrCl

2

PCl2 PBr2

Br2(g) + I2 (g) 2 IBr(g)

= K2 = 0.051 at 25oC2

PBr2 PI2

PIBr

Adding the two chemical equations gives:

2 BrCl(g) + Br2 (g) + I2 (g) 2 IBr(g) + Cl2 (g) + Br2(g)

2 BrCl(g) + I2 (g) 2 IBr(g) + Cl2 (g)

= K3

PBrCl

2

PCl2 PIBr

2

PI2

Looking at the expressions for K1, K2 and K3

K1 K2 = K3

Hence, K3 = 0.023 at 25oC

The Ideal Gas Equation and Chemical Equilibrium

For gaseous reactants or products, the concentration may be in moles/liter.

The concentrations of the gases in moles/liter must be converted to partial pressures.

Concentration of a species A in moles/lit = [A]

[A] = nV

PA

R T=

This equation can be used to convert concentration of a gas in moles/lit to partial pressure of the gas.

where Pref is the reference pressure = 1 atm and ensures that K is unitless.

a A(g) + b B(g) c C(g) + d D(g)

Hence, for a general gas phase reaction

[C]c [D]d

= K[A]a [B]b

R TPref

( )(a+b-c-d)

For the general reaction:

PCc PD

d

= QPA

a PBb

a A(g) + b B(g) c C(g) + d D(g)

Define the reaction quotient, Q:

where P is the partial pressure of a species at any point in time.

Reaction Quotient

If Q = K, the reaction is at equilibrium

If Q ≠ K, the reaction is not at equilibrium.

If Q > K , reaction proceeds from right to left

If Q < K, reaction proceeds from left to right

a A(g) + b B(g) c C(g) + d D(g)

The equilibrium constant for the reaction

CH4(g) + H2O(g) CO(g) + 3 H2(g)

Equals 0.172 at 900K. The concentrations of H2(g), CO(g), and H2O(g) in an equilibrium mixture of gases all equal 0.00642 mol/L. Calculate the concentration of CH4(g) in the mixture, assuming that this is the only reaction taking place.

[CO] [H2 ]3

= K[CH4] [H2O]

R TPref

( )-2

[0.00642 mol/L] [0.00642 mol/L ]3

= 0.172

[CH4] [0.00642 mol/L]

[(0.08206 L atm mol-1) 900 K]-2

[CH4] = 0.00839 mol/L

What happens if a system at equilibrium undergoes a change in conditions?

The tendency of a system to achieve equilibrium is spontaneous.

Once a system is at equilibrium it will remain at equilibrium.

However, if conditions change the system will respond to this change in a way to achieve equilibrium again.

Note: the concentrations of species when equilibrium is re-established need not be the same as the ones established at the previous equilibrium.

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

LeChatelier’s Principle

If a stress is applied to a system at equilibrium, the system tends to react so that the stress is minimized

PCc PD

d

= QPA

a PBb

a A(g) + b B(g) c C(g) + d D(g)

A) Changing the concentration of a reactant or product

If reactant added: Q < K, reaction proceeds from left to rightIf product added: Q > K, reaction proceeds from right to left

If a product is removed from the equilibrium mixture, Q also decreases, and the reaction once again proceeds to the right to increase the concentration of the product.

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

B) Changing the Volume

Decreasing the volume, increases the total pressure of the reaction mixture.

The reaction will then proceed in the direction which reduces the total pressure.

2 NO2(g) N2O4(g)

If the volume is decreased, the above reaction will move to the right, to decrease the total number of molecules.

C) Changing Temperature

The effect of changing the temperature of a system at equilibrium depends on whether the reaction proceeds by absorbing energy (endothermic) or by releasing energy (exothermic).

An endothermic reaction lowers the temperature of the system and an exothermic reaction raises the temperature of the system.

If a reaction is exothermic, raising the temperature causes the equilibrium to shift to the left.

a A(g) + b B(g) c C(g) + d D(g) + heat

Forward reaction is exothermic; reverse reaction is endothermic.

Heterogeneous Reactions

CaCO3(s) + SO3 (g) CaSO3 (s) + CO2(g)

[CaSO3] PCO2 = K[CaCO3] PSO3

In all equilibrium expressions, the concentrations of all pure solids and liquids are set to 1.

PCO2 = KPSO3

Hence,

In general, to write the equilibrium expression for a reaction

1) Concentration of gases are expressed as partial pressures

2) Concentration of dissolved species in solution are expressed as moles/liter

3) Concentrations of pure solids and pure liquids are set to 1

(for a solvent taking part in a reaction, its concentration is also set to 1 providing the solution is dilute)

Problem: The reaction between Ni(s) and CO(g) to form Ni(CO)4(g) is as follows:

Ni(s) + 4CO(g) Ni(CO)4(g)

A quantity of Ni(s) is added to a vessel containing CO at a partial pressure of 1.282 atm and 354 K. At the equilibrium point of this reaction, the partial pressure of CO is 0.709 atm. Calculate the equilibrium constant of this reaction at 354 K.

Initial P (atm)

Change in P (atm)

Equilibrium P (atm)

0.573 0.143

1.282 0

Ni(s) + 4CO(g) Ni(CO)4(g)

0.709 0.143

PNiCO4K=PCO

4

PNiCO4K=PCO

4

0.143K=

(0.709)4= 0.567

Product yields can be increased by adjusting conditions

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

Gas phase reaction: 4 moles of gas --> 2 moles of gas

Exothermic reaction

Reaction conditions that favor NH3(g) production

high pressure (~ 250 atm)

low temperature (use a catalyst)

Applications of Chemical Equilibria

Hemoglobin (Hb) carries oxygen from the lungs to the body tissue, transporting oxygen from a region of high concentration to low concentration.

The oxygen-hemoglobin complex, oxyhemoglobin (HBO2) transports O2

Hb(aq) + O2(g) HbO2 (aq)

KO2=

[HbO2]

[Hb]PO2

Hemoglobin/O2 Equilibrium

Level of O2 in blood is increased by ~ 70 times because of hemoglobin

Because of the formation of the HbO2 complex, the amount of O2 in blood is increased by a factor of 70.

Hb(aq) + O2(g) HbO2 (aq)

LeChatelier’s principle predicts that in regions of high O2 partial pressure, the Hb-HbO2 equilibrium is shifted to the right, which is the case in the lungs

In regions of low O2 partial pressure, the equilibrium shifts to the left, resulting in a breakup of the HbO2 complex, releasing O2 to the body’s tissues.

Why is CO lethal?

Hb(aq) + CO(g) HbCO (aq)

KCO=[HbCO]

[Hb]PCO

KO2=

[HbO2]

[Hb]PO2

When Hb is exposed to both O2 and CO, there is competition for the Hb, and the following reaction takes place:

HbO2 (aq) + CO(g) HbCO(aq) + O2 (g)

The Kcompetition for this reaction is:

Kcompetition =[HbO2]

[HbCO] PO2

PCO

=KCO

KO2

PO2

PCO=

[HbO2]

[HbCO]Kcompetition =

KCO

KO2

HbO2 (aq) + CO(g) HbCO(aq) + O2 (g)

Since KCO > KO2 Kcompetition is >1

At 38oC, the value of Kcompetition is 210 strongly favoring the formation of the HbCO complex

CO displaces O2 from the HbO2 complex, resulting in asphyxiation.

The process is reversible - from LeChatelier’s principle, a large partial pressure of O2 will shift the reaction above from right to left.

Extraction and Separation

A solute dissolved in a solvent A can be extracted using another solvent B.

Condition: The solute must dissolve in both solvents A & B and solvent B must be immiscible with the solvent A.

CCl4 and H2O are immiscible.

I2(s) dissolves in both solvents.

If to an aqueous solution of I2(aq) CCl4 is added, and the

flask shaken, some of the I2 in the water layer will be extracted into the CCl4 layer

I2(aq) I2(CCl4)

The following equilibrium is established:

K=[I2]CCl4

[I2]H2O H2O layer

CCl4 layer

K: partition coefficient

Chromatography

time

sig

nal

kA =time in mobile phase

time in stationary phase

kB =time in mobile phase

time in stationary phase