CHEMICAL EQUILIBRIUMCHEMICAL EQUILIBRIUM

THE PROPERTIES OFTHE EQUILIBRIUM

1. Within the natural events, it is the deal point of having the tendency of longing to have either MINIMUM ENTALPHY or MAXIMUM ENTROPY. That is, the equilibrium is the mid-point of two opposite tendencies

2. THE EQUILIBRIUM is DYNAMIC.After being reached to the equilibrium (Temperature and Pressure are held constant and There are no added substances) , the concentrations of the reactants and products do not change with time since the forward and reverse reactions rate remain the same. For this reason, we call it dynamic and since it moves in two directions it is called as reversible.

3. Le CHATELİER PRINCIPLE (ACTION ≡ REACTION): Once a reaction has been reached into the equilibrium, it stays at the equlibrium, unless the conditions are changed. The change in temperature or pressure is effective on the equlibrium . When the net reaction proceeds to one side(left or right) by means of explicit factors, the equilibrium can be reached at a specific time again.(As soon as possible it is reached.)

THE CONDITION OF DYNAMIC EQUILIBIRIUM

1. When a liquid vaporizes into a closed container, after a time vapor molecules condense to the liquid state at the same rate at which liquid molecules vaporize. Even though molecules continiue to pass back and forth between liquid and vapor, the pressure exerted by the vapor remains constant with time. The vapor pressure of a liquid is a property associated with an equilibrium condition.

2. When a solute dissolves in a solvent, a time is reached when the rate of dissolving is just matched by the rate at which dissolved solute precipitates. The solubility of a solute is a property associated with an equilibrium condition.

THE CONDITION OF DYNAMIC EQUILIBIRIUM

3. When an aqueous solution of Iodine is shaken with pure Carbontetrachloride, iodine molecules move into CCl4 layer. When I2 molecules pass between two liquids(water and CCl4) at equal rates, the concentration of I2 in each layer remains constant. The ratio of concentrations of a solute in two immiscible solvents is called the distribution coefficient. The distribution coefficient of a solute between two immiscible solvents is a property associated with an equilibrium condition.

FACTORS AFFECTING EQUILIBRIUM

• The environmental conditions: The change in P,V,n,,T have an effect on it. They create a new equilibirium point.

1. PRESSURE EFFECT: ∑n∑nRTRT ≠ ∑n ≠ ∑nPRPR Δp is EFFECTIVEEFFECTIVE. ∑nRT = ∑nPR ise Δp is not effective , The pressure leads the net reaction to move towards the side with less amount of substance in moles.

2. EFFECT OF VOLUME:

3. MOLARITY:

4. TEMPERATURE EFFECT:

DYNAMIC EQUILIBRIUM

1) H2O(s) ⇄ H2O(g )

2) NaCl(s) ⇄ NaCl(aq)

3) I2(H2O) ⇄ I2(CCl4)

4) CO(g) + 2 H2(g) ⇄ CH3OH(g)

5) HA(aq) ⇄ H+ + A-

EXAMPLES FOR THE DYNAMIC EQUILIBRIUM

THE EQUILIBRIUM CONSTANT EXPRESSION

CO(g) + 2 H2(g) ⇄ CH3OH(g)kf

kr

In the equilibrium f = r

kf[CO][H2]2 = kr[CH3OH][CH3OH]

[CO][H2]2=

kf

kr

Kc =

Forward: CO(g) + 2 H2(g) → CH3OH(g)

Reverse: CH3OH(g) → CO(g) + 2 H2(g)

f= kf[CO][H2]2

r = kr[CH3OH]

kf

kr

CO (g) + 2H2 (g) CH3OH (g)

THREE APPROACHES TO EQUILIBRIUM

CO(g) H2(g) CH3OH(g)

Experiment 1

initial amounts,mol 1,000 1,000 0,000

equil. amounts, mol 0,911 0,822 0,0892

equil. concn, mol/L 0,0911 0,0822 0,00892

Experiment 2

initial amounts,mol 0,000 0,000 1,000

equil. amounts, mol 0,753 1,506 0,247

equil. concn, mol/L 0,0753 0,151 0,0247

Experiment 3

initial amounts,mol 1,000 1,000 1,000

equil. amounts, mol 1,380 1,760 0,620

equil. concn, mol/L 0,138 0,176 0,062

CO (g) + 2H2 (g) CH3OH (g)

THREE APPROACHES TO EQUILIBRIUM

[CH3OH]

[CO][H2]2

Kc(1) = 14.5 M-2

Kc(2) = 14.4 M-2

Kc(3) = 14.5 M-2

[CH3OH]

[CO][H2]Kc =

[CH3OH]

[CO].2[H2]

0.596 M-1

1.09 M-1

1.28 M-1

1.19 M-1

2.17 M-1

2.55 M-1

CO(g) + 2 H2(g) ⇄ CH3OH(g)k1

k-1

Relationship of Kc to the Balanced Chemical Equation

a A + b B …. ⇄ g G + h H …. The Equi. constant = Kc= [G] g[H]h [A]a[B]b ….

We must always make certain that the expression for Kc matches the corresponding balanced equation. In doing so,

When we reversereverse an equation, we invertinvert the value of Kc

•When we multiplymultiply the coefficients in a balanced equation by a common factor(2,3…) , we raise the equilibrium constant to the corresponding power power(2,3…)

•When we dividedivide the coefficients in a balanced equation by a common factor, we take the corresponding rootroot of the equilibrium constant(i.e. Square root, cube root,…)

A) N2(g) + ½O2 ⇄ N2O(g) KcA= 2.7x10+18=

B) N2(g) + O2 ⇄ 2 NO(g) KcB= 4.7x10-31

C) N2O(g) + ½O2 ⇄ 2 NO(g) Kc= ?

Kc=

[N2O][O2]½

[NO]2

=[N2][O2]½

[N2O][N2][O2]

[NO]2

=KcB KcA = 1.7x10-13

[N2][O2]

[NO]2

=

[N2][O2]½

[N2O]

EXAMPLES

1. CO (g) + 2 H2 (g) ⇄ CH3OH (g) Kc = 14,5

In the above reaction [CO] = 1,03 M and [CH3 OH] = 1,56 M ; [H2] = ?

[H2] = 0,322 M .

2. The synthesis reaction of NH3(g), and its

Koc value is as follows: N2 (g) + 3 H2 (g) ⇄ 2 NH3 (g) Ko

c=3,5x 108

NH3 (g) ⇄ ½ N2 (g) + 3/2 H2 (g) Koc=?

Solution: 2 NH3 (g) ⇄ N2 (g) + 3 H2 (g) K’c =1/3,5 x 108=2,8 x 10-9 In order to obtain the demanded equation, the coefficients are divived by

2. Since, the square root of K’c is calculated..

K’’c = 5,3 x10-5

[CH3OH]

[CO] [H2]2

1,56

1,03 x [H2]2 Kc = = = 14,5

Equilibria Involving Gases: The Equilibrium Constant, KP

• Mixtures of gases are as much solutions as are mixtures with a liquid solvent.

• We derive an equilibrium constant expression,Kp by using the partial pressurespartial pressures of gases in atm in the equation of ideal gasesideal gases:

2 SO2(g) + O2(g) ⇄ 2 SO3(g) Kc = [SO2]2[O2]

[SO3]2

[SO3]=V

nSO3 = RT

PSO3 [SO2]=

V

nSO2 = RT

PSO2

[O2] =V

nO2 = RT

PO2

The Equilibrium Constant KP = Kc(RT)Δn

2 SO2(g) + O2(g) ⇄ 2 SO3(g)

Kc = [SO2]2[O2]

[SO3] RT

PSO3

2

RT

PSO2

RT

PO2

=

2

= RTPSO3

2PSO2PO2

2

Kc = KP(RT) KP = Kc(RT)-1

KP = Kc(RT)Δngas Δngas=2-(2+1)=-1

Equilibria Involving Pure Liquids and Solids

• Even though solids and liquids participate in a chemical reaction, they remain pure in a heterogenous mixture, that is the concentrations of pure solids and liquids can not vary:

• The concentrations terms for pure solids and liquids The concentrations terms for pure solids and liquids do not appear in equilibrium constant expressions.do not appear in equilibrium constant expressions.

C(s) + H2O(g) ⇄ CO(g) + H2(g)

Kc = [H2O]

[CO][H2] = PH2O

PCO .PH2 / (RT)1

C(s) does not appear in the equilibrium equation.

Burnt Lime- CaCO3(s) CaO(s) + CO2(g)

CaCO3(s) ⇄ CaO(s) + CO2(g)

Kc = [CO2] KP = PCO2(RT)

Examples1. 2 SO2 (g) + O2 ( g) ⇄ 2SO3 (g) Kc=2,8 x 102 Complete the calculation of Kp =?

T=1000K

Solution: Kp = Kc(RT)-1 = 2,8 x 102 (0,08206 x 1000)-1 = 3,4

2. : H2S (g) + I2 (k) ⇄ 2 HI (g) + S (k) at 60o C the partial pressures of the gases in the

equilibrium are PHI=0,00365 atm and PH2S =0,996 atm . Calculate the Kp value of the reaction?

Solution: Pure solids do not appear in the equilibrium constant expression

Kp = = = 1,34 x 10-5

3 : The steam-iron process is used to generate H2(g) mostly for use in hydrogenating oils. Iron metal and steam (H2O(g) ) react to produce Fe3O4(s) and H2. Write expressions for Kc, Kp for this reversible reaction.

Solution: 3Fe(s) + 4H2O(g) ⇄ Fe3O4(s) + 4H2(g)

Kc=[H2]4 / [H2O]4 Δn=0

Kp=P4H2 / P4H2O= Kc (RT)0=Kc

(PHI)2

(PH2S)

(3,65 x 10-3)2

9,96 x 10-1

SIGNIFICANCE OF THE MAGNITUDE OF AN EQUILIBRIUM CONSTANT

The Reaction Quotient, Q: Predicting the Direction of A Net Reaction

• A very large numerical value of Kc or Kp signifies that the forward reaction goes to completion or very nearly so.

• A very small numerical value of Kc or Kp, signifies that the forward reaction does not occur to any significant extent.

• If the numerical value of Kc or Kp in the range from 10-10 to 1010 , a reaction is most likely to reach a state of equilibrium in which significant quantities of both reactants and products are present.

• .• In some equilibrium calculations it is helpful to determine the direction of net

change as a first step:• For any set of initial concentrations in a reaction we can set up a ratio of

concentrations called the reaction quotient, Qreaction quotient, Qcc

• Qc = Kc

CO(g) + 2 H2(g) ⇄ CH3OH(g)k1

k-1

Qc = [A]t

m[B]tn

[G]tg[H]t

h

Example for predicting the Direction of a Net Reaction

To increase the yield of H2(g) in the reaction of C(s) and H2O(g), an additional reaction called « water-gas shift reaction» is generally used. For the water-gas shift reaction at about 1100 K Kc =1,00. The following amount of substances are brought together and followed to react at 1100 K: 1,00 mol CO, 1,00 mol H2O , 2,00 mol CO2 and 2,00 mol H2. Compared to their initial amounts, which of the substances will be present in a greater amount and which , in a lesser amount, when equilibrium is established?

CO (g) + H2O (g) ⇄ CO2 (g) + H2 (g)

Solution:

Qc = = = 4,00[CO2] [H2] (2,00/V)(2,00/V)

[CO] [H2O] (1,00/V)(2,00/V)

Because Qc > Kc , a net reaction occurs to the left. When equilibrium is established, the amounts of H2O and CO will be greater than

initially and the amounts of CO2 and H2 will be less.

ALTERING EQUILIBRIUM CONDITIONS: LE CHÂTELIER PRINCIPLE

When an equilibrium system is subjected to a change in temperature, pressure, or concentration of a reacting species, the system reacts in a way that partially offsets the change while reaching a new state of equilibrium.

Q = = Kc [SO2]2[O2][SO3]2 Q > Kc

2 SO2(g) + O2(g) ⇄ 2 SO3(g)k1

k-1

Kc = 2,8x102 1000 K

EXAMPLES1 .Predict the effect of adding more H2 (g) to a constant volume equilibrium mixture of N2, H2 and NH3

N2 (g) + 3 H2 (g) ⇄ 2 NH3 (g)

Solution:Solution: The addition of H2 stimulates a shift of the equilibrium condition to

the“right” However, only a portion of the added H2 is consumed in this reaction . When

equilibrium is reestablished, there will be more H2 then present originally . The amount

of NH3 will be also greater , but the amount of N2 will be smaller. Some of the original

N2 , must be consumed in converting some of the added H2 to NH3 .

• 2CO(g) + O2(g) ⇄ 2CO2 (g) Predict the effect of adding of O2(g) , while the system is in equilibrium?

• Solution Solution :Addition of O2(g) stimulates the reaction to the right

• 2. EXAMPLE : CaCO3(s) ⇄ CaO (s) + CO2(g) When equilibrium is established, in which way does the addition of the compounds CaO(s), CaCO3(s), CO2(g) influence the system? Explain for each compound the effects briefly

• Solution Solution : Addition of CaO(s) and CaCO3(s) does not change the equilibrium, however Addition ofCO2(g) leads the net reaction to proceed to the left.

EFFECT OF CHANGES IN PRESSURE OR VOLUME IN EQUILIBRIUM

• We can add a gaseous reactant or product to an equilibrium mixture in which we change the equilibrium conditions(either to the left or to the right ) .

• We can add an inert gas to the constant volume reaction mixture. This has the effect of increasing the total pressure but the partial pressures of the reacting species are all unchanged. As a result the inert gas added in this way has no effectno effect on the equilibrium condition

• We can change the pressure of a system by changing its volumevolume..

Kc = [SO2]2[O2]

[SO3] V

nSO3

2

V

nSO2

V

nO2

=

2

= VnSO3

2nSO2nO2

2

EFFECT OF CHANGES IN VOLUME ON EQUILIBRIUM

nDnCd

Kc = [C]c[D]d

[G]g[H]h

= V(c+d)-(g+h)nG

c

g nHh

• When the volume of an equilibrium mixture of gases is reduced, a net reaction occurs in the direction producing a smaller number of moles of gases. When the volume is increased, a net reaction occurs in the direction producing a larger number of moles of gases.

V-ΔnnG

g nHh

=nDnC

dc

EXAMPLE

An equilibrium mixture of gases N2(g), H2(g) NH3(g) is transfered from a 1,50 L flask to a 5,00 L flask. In which direction does a net reaction occur to restore equilibrium?

N2 (g) + 3 H2 (g) ⇄ 2 NH3 (g)

Solution: When the gaseous mixture is transferred to the larger flask, the partial pressure of each gas and the total pressure drop. Some of the NH3 decomposes back to N2 and H2. A net reaction occurs to the left(reverse direction) in restoring equilibrium. We would come to the same conclusion if we decreased the total pressure instead of increasing the volume.

EFFECT OF TEMPERATURE AND CATALYST ON EQUILIBRIUM

Effect of Catalyst• A catalyst in a reaction mixture speeds up both the forward and reverse reactions.

Equilibrium is achieved more rapidly but .– THE EQUILIBRIUM AMOUNTS are UNCHANGED by the catalyst.

Raising the temperature of an equilibrium mixture shifts the equilibrium condition in the direction of the ENDOTHERMIC reaction.Lowering the temperature causes a shift in the direction of the EXOTHERMIC reaction

Van’t Hoff equation

cmxy

BTR

HK

eBK

p

RT

H

p

ln1

)ln(

EXAMPLE

2 SO2 (g) + O2 (g) ⇄ 2 SO3 (g) ∆H= -180kJ

Will the amount of SO3(g) formed from given amounts of SO2(g) and O2(g) be greater at high or low temperatures?

Solution: Raising the temperature favors the endothermic reaction. To favor the exothermic reaction requires that we lower the temperature. Therefore, an equilibrium mixture would have a higher concentration of SO3 at lower temperatures.

Example

Dinitrogentetroxide, N2O4(l), is an important component of rocket fuels. For example as an oxidizer of liquid hydrazine in the Titan rocket. At 25˚C N2O4 exists as a colorless gas, which partially dissociates into NO2 , a red brown gas. The color of an equilibrium mixture of these two gases depends on their relative proportions

Equilibrium is established in the reaction N2O4 (g) ⇄ 2 NO2 (g) ,the quantities of the two gases present in a 3L vessel are 7,64 g N2O4 and 1,56 g NO2 . What is the value of Kc

o for this reaction?

Solution:For both N2O4 and NO2 ,we need the conversions g mol mol/L [N2O4] = 7,64 g /92,01 g/molx3L= 0,0277 M

[NO2] = 1,56 g / 46,01 g/molx3L=0,0113 M

Kc = [NO2]2/ [N2O4] = (0,0113)2 / 0,0277 = 4,61 x 10-3

EXERCISEEquilibrium involving SO2(g), O2(g) and SO3(g) is important in sulfuric acid production. When a 0,0200 mol sample of SO3(g) is introduced into an evacuated 1,52 L vessel at 900 K, 0,0142 mol SO3 is found to be present at equilibrium. What is the value of Kp for the decomposition of SO3 (g) at 900 K?

2 SO3 (g) ⇄ 2 SO2 (g) + O2 (g)

Solution:Let us first determine Kc and then convert to Kp by using the equation

2 SO3 (g) ⇄ 2 SO2 (g) + O2 (g)

initial amounts 0,02 mol 0,00 mol 0,00 mol change -0,0058 mol + 0,0058 mol +0,0029 mol equil. amounts 0,0142 mol 0,0058 mol 0,0029 mol equil. conc. 0,0142mol/1,52 L 0,0058mol/1,52 L 0,0029mol/1,52L

[SO3]=9,34x10-3 [SO2]=3,8x10-3 [O2]=1,9x10-3

Kc =[SO2]2*[O2] / [SO3]2= 0,0038)2*(0,0019) / (0,00934)2 = 3,1 x 10-4

Kp=Kc(RT) ∆n =3,1 x 10-4(0,0821 x 900)(2+1)-2 = 2,3 x 10-2

EXERCISEAmmonium hydrogen sulfide, NH4HS(s), (used as a photographic developer) is unstable and dissociates at room temperature..

NH4HS (s) ⇄ NH3 (g) + H2S (g) Kp(atm) = 0,108 (25oC de)

A sample of NH4HS(s), is introduced into an evacuated flask at 25oC ? What is the total gas pressure at equilibrium?

Solution:

Kp =(PNH3)(PH2S) = (PNH3

)(PNH3) = (PNH3

)2 = 0,108

PNH3 = 0,329 atm PH2S = PNH3

= 0,329 atm

Ptoplam = PNH3 + PH2S = 0,329 + 0,329 = 0,658 atm

EXERCISE

An 0,024 mol sample of N2O4 (g) is allowed to come to equilibrium with NO2 (g) in an 0,372 L flask at 25˚C. Calculate (a) the amount of N2O4 (g) present at equilibrium and (b) the percent dissociation of the N2O4

N2O4 (g) ⇄ 2 NO2 (g) Kc = 4,61 x 10-3 (25oC)Solution: a) χ = the number of moles of N2O4 that dissociate.

N2O4 (g) ⇄ 2 NO2 (g)

initial amounts 0,0240 mol 0,00 mol change -χ mol + 2χ mol equil. amounts (0,0240-χ) mol 2χ mol equil. conc. [N2O4]= (0,0240-χ)/0,372 [NO2]=2χ/0,372

Kc== 4,61 x 10-3=[NO2]2/[N2O4]= (2χ / 0,372)2: (0,0240-χ/0,372)=4χ2 : 0,372 (0,0240-χ)

4χ2 = 4,12 x 10-5 - (1,71 x 10-3) χ χ2 + (4,28 x 10-4) χ - 1,03 x 10-5=0 χ = 0,003 mol N2O4

The amount of N2O4 at equilibrium =(0,0240-χ) = 0,024-0,003=0,0210 mol N2O4

b) the percent dissociation % of N2O4 = 0,003*100/0,024= 12,5