IB Chemistry on Equilibrium Constant Calculation and Le Chatelier's Principle

Equilibrium Law applies to system at equilibrium

• Kc is constant at constant temperature

• Kc unaffected by Concentration, Pressure and Catalyst

• Addition of reactants and products will only shift position of equilibrium to right or left changing to a new equilibrium

concentration but Kc remains the same

• Magnitude of Kc indicate the extend or how far the reaction forms product/reactants but not how fast

• Kc High – High product but rate can be very slow

• Kc Low – Low product but rate can be very fast

Reaction between H2 + I2 ↔ 2HI Kc = 46.4 at 730K Kc = [HI]2 = 46.4 [H2][I2]

Effect of Concentration on equilibrium constant, Kc

• Increase in Conc ↑ - position of equilibrium shift to right/left - Conc will be Reduced ↓ • Decrease in Conc ↓ – position of equilibrium shift to right/left - Conc will be Increased ↑

Reaction contain different Initial Conc of H2 and I2 and HI but at equilibium Kc remains the same regardless of initial conc

H2 + I2 ↔ 2HI Initial Conc Equilibrium Conc

Kc = [HI]2 = 46.4 [H2][I2]

Equilibrium Law

How dynamic equilibrium is achieved in a closed system?

As reaction proceeds

Conc of H2 decrease ↓over time

As reaction proceeds

Conc of NH3 increase ↑over time

Rate of forward rxn, Rf decrease ↓ over time Rate of reverse rxn, Rr increase ↑over time

At dynamic equilibrium

• Forward Rate = Reverse Rate

• Conc of reactant/product remain

constant

N2(g) + 3H2(g) ↔ 2NH3 (g)

H2

H2

NH3

NH3

How dynamic equilibrium is shifted when H2 is added ?

N2(g) + 3H2(g) ↔ 2NH3 (g)

• When more H2 added, position equilibrium will shift to right, to reduce the H2 conc and increase the NH3 conc • Rate of forward and reverse will increase • New equilibrium concentration will be achieved when Rate of forward kf = Rate of reverse kr • More products NH3 and less reactants N2 but Kc value remains unchanged

H2 added New equilibrium Conc

At equilibrium

Rate forward kf = Rate reverse kr

At equilibrium again

Rate forward kf = Rate reverse kr

Equilibrium shifted to right

Rate forward kf > Rate reverse kr

Kc = 4.07 Qc = 2.24 Kc = 4.07

How dynamic equilibrium is shifted when H2 is added ?

N2(g) + 3H2(g) ↔ 2NH3 (g)

• When more H2 added, position equilibrium will shift to right, to reduce the H2 conc and increase the NH3 conc • Rate of forward and reverse will increase • New equilibrium concentration will be achieved when Rate of forward kf = Rate of reverse kr • More products NH3 and less reactants N2 but Kc value remains unchanged

H2 added New equilibrium Conc

N2(g) + 3H2(g) ↔ 2NH3 (g)

• Kc = (NH3)2 (N2)(H2)3

• Kc = (0.67)2

(0.20)(0.82)3

• Kc = 4.07

Equilibrium Conc for H2 = 0.82M

Equilibrium Conc for N2 = 0.20M

Equilibrium Conc for NH3 = 0.67M

New Conc for H2 = 1.00M

Equilibrium Conc for N2 = 0.20M

Equilibrium Conc for NH3 = 0.67M

N2(g) + 3H2(g) ↔ 2NH3 (g)

• Qc = (NH3)2 (N2)(H2)3

• Qc = (0.67)2

(0.20)(1.00)3

• Qc = 2.24

N2(g) + 3H2(g) ↔ 2NH3 (g)

• Kc = (NH3)2 (N2)(H2)3

• Kc = (0.75)2

(0.19)(0.90)3

• Kc = 4.07

New Equilibrium Conc for H2 = 0.90M

New Equilibrium Conc for N2 = 0.19M

New Equilibrium Conc for NH3 = 0.75M

Qc < Kc

2.24 < 4.07

• Kc must remain constant

• Shift to the right

• Increase products and decrease reactants

• New equilibrium Conc is achieved

• Qc = Kc again

N2O4 (g) ↔ 2NO2 (g)

Rf (rate forward) = Rr (rate reverse) kf [N2O4] = kr [NO2]2

Kc = (NO2)2 (N2O4)

• Kc is a ratio of rate constant and is temperature dependent

• Kc is a ratio of products conc to reactants conc

• Magnitude of Kc indicate how far/extend of the reaction proceeds towards a product at a given temperature

Small Kc : N2(g) + O2(g) ↔ 2NO(g) Kc = 1 x 10 -30 small ↓

Kc = (NO)2

(N2)1(O2)

1 Kc small → low product ↓, more reactants ↑, close to no reaction at all

Large Kc : 2COg) + O2 ↔ 2CO2(g) Kc = 2.2 x 10 22 high ↑

Kc = (CO2)2

(CO)2(O2)1 Kc large → high product ↑, small reactants ↓ , close to completion

Intermediate Kc : 2HI(g) ↔H2(g) + I2(g) Kc = 0.02

Kc = (H2)(I2)

(HI)2 Kc intermediate → significant amount of reactants and products

Equilibrium Law

Kc = (C)c(D)d

(A)a(B)b

At equilibrium:

N2O4 (g) ↔ 2NO2 (g)

Rf (rate forward) = Rr (rate reverse) kf [N2O4] = kr [NO2]2

kf/kr = (NO2)2 (N2O4)

Kc = kf / kr

aA + bB ↔ cC and dD

When a reversible reaction achieved dynamic equilibrium - aA + bB ↔ cC and dD

• Equilibrium constant Kc = ratio of molar conc of product (raised to power of their respective stoichiometry coefficient)

to molar conc of reactants (raised to power of their respective stoichiometry coefficient)

• Kc is constant at constant temperature

A reaction quotient Q is a ratio made up of product conc multiplied together and divided by reactant conc with

each term raised to the power of its stoichiometric coefficient

aA + bB ↔ cC and dD

Qc = (C)c(D)d

(A)a(B)b Qc = Kc at equilibrium

If overall reaction is the sum of two or more reactions, the overall reaction quotient. Kc is the product of reaction quotients for single reactions.

N2(g) + O2(g) ↔ 2NO Kc1 = 4.3 x 10 -25

2NO(g) + O2(g) ↔ 2NO2(g) Kc2 = 6.4 x 10 9

A reaction quotient Kc for a forward reaction is the reciprocal of reaction quotient for reverse reaction. Kc forward = 1/Kc reverse

H2 + I2 ↔ 2HI Kc = (HI)2

(H2) (I2)

Kc = 50

2HI ↔ H2 + I2

Kc1 = (H2) (I2) (HI)2

Kc1 = 1/Kc

= 1/50 = 0.02

If all coefficient are multiplied by some factor, that factor becomes the exponent for relating the reaction quotients

Equilibrium Law

Kc = (C)c(D)d

(A)a(B)b

N2(g) + O2(g) ↔ 2NO Kc1 = 4.3 x 10 -25

2NO(g) + O2(g) ↔ 2NO2(g) Kc2 = 6.4 x 10 9

N2(g) + 2O2(g) ↔ 2NO2(g) Kc (overall) = Kc1 x Kc2

Kc (overall) = 4.3 x 10 -25 x 6.4 x 10 9

2SO2 + O2 ↔ 2SO3 Kc = (SO3)2

(SO2)2(O2)

Kc = 200

SO2 + 1/2O2 ↔ SO3

Kc1 = (SO3)

(SO2)(O2)1/2

Kc1 = (SO3)2 x exponent ½

(SO2)2(O2) Kc1 = Kc x exponent ½ Kc1 = (200)1/2

4SO2 + 2O2 (↔ 4SO3

Kc2 = (SO3)4

(SO2)4(O2)2

Kc2 = (SO3)2 x exponent 2

(SO2)2(O2) Kc2 = Kc x exponent 2 Kc2 = (200)2

Initial Conc Equilibrium Conc

Kc forward = 1/Kc reverse

X factor of 2 X factor of 1/2

Equilibrium Law

For gaseous partial pressure , Kp can be used

PV = n R T

P = (n/V) R T

P = Conc R T

N2O4 (g) ↔ 2NO2 (g)

Kp = (pNO2)2

(pN2O4)

= (NO2)2(RT)2

(N2O4)(RT) Kp = (NO2)2RT Kc = (NO2)2

(N2O4) (N2O4)

Relationship between Kp and Kc

• Kp = Kc(RT)Δn • Δn = Total number of moles of product – Total number of reactants

p = Conc RT (pNO2) = Conc (NO2)RT (pN2O4) = Conc(N2O4)RT (pNO2) = (NO2)RT (pN2O4) = (N2O4) RT

2SO2 + O2 ↔ 2SO3 Partial pressure for SO2 = 0.05atm Partial pressure for O2 = 0.025atm Partial pressure for SO3 = 1.00atm Kp = p(SO3)2

p(SO2)2p(O2) = (1.0)2

(0.05)2(0.025) = 1.6 x104

2SO2 + O2 ↔ 2SO3 Kp = 11.57atm-1 at 873K, R = 0.0821dm3 atm Kp = Kc(RT)Δn

11.57 = Kc (0.0821 x 873)-1

Kc = 11.57 x (0.0821 x 873)

= 829.3 dm3mol -1

Relationship between Kp and Kc

N2 + 3H2 ↔ 2NH3 Kp = Kc (RT)Δn

Kp = Kc(RT)2-4

Kp = Kc(RT)-2

Relationship between Kc and Kp for gaseous system

Conc = n/V

Ex 1

Ex 2

Reaction CO + Br 2 ↔ COBr2. Equilibrium conc of the gas are shown below

Calculate Kc. (Volume vessel is 1dm3 )

Equilibrium Conc : [CO] = 8.78 x 10-3 M, [Br2] = 4.90 x 10-3M, [COBr2] = 3.40 x 10-3M

CO + Br2 ↔ COBr2 Equilibrium/Conc 8.78 x 10-3 4.90 x 10-3 3.40 x 10-3

Reaction PCI5 ↔ PCI3 + CI2. Calculate equilibrium conc for CI2, Kc = 0.19M. (Volume vessel is 1dm3 )

Equilibrium Conc : [PCI5] = 0.20M, [PCI3]= 0.01M

PCI5 ↔ PCI3 + CI2

Equilibrium/Conc 0.20 0.01 x

Equilibrium Conc = Moles /Volume


Equilibrium Law

Kc = (COBr2 ) (Br2)(CO)

= (3.40 x 10-3 ) (8.78 x 10-3)(4.90 x 10-3) Kc = 79 dm3 mol-1

Kc = (PCI3)(CI2) (PCI5)

0.19 = (0.01) (x) (0.20) x = 3.8M

Reaction N2O4 ↔ 2NO2. Calculate equilibrium conc for N2O4, Kc = 1.06 x 10-5M. (Volume vessel is 1dm3 )

Equilibrium Conc : [NO2] = 1.85 x 10-3 M

N2O4 ↔ 2NO2

Equilibrium/Conc x 1.85 x 10-3

Kc = (NO2)2

(N2O4)

1.06 x10-5 = (1.85 x10-3)2

(x) x = 0.323M


Ex 1

Ex 3

Ex 2

Equilibrium Law (Using ICE Method)

1 mol of ethanoic acid added to 1 mol of ethanol and at equilibrium, 0.67mol ester was produced.

Calculate Kc . (Reaction volume is 1dm3)

CH3COOH + C2H5OH ↔ CH3COOC2H5 + H2O

Initial mole 1.0 1.0 0 0

Change (1-0.67) (1-0.67) 0.67 0.67

Equilibrium/Conc 0.33/1 0.33/1 0.67/1 0.67/1

In esterification, 2 mol of ethanoic acid, 3 mol of ethanol and 2 mol H2O were mixed .

Calculate the equilibrium concentration if Kc for reaction is 4.0 (Reaction volume is 1dm3 )

CH3COOH + C2H5OH ↔ CH3COOC2H5 + H2O

Initial mole 2 3 0 2

Change (2- x) (3-x) x 2 + x

Equilibrium/Conc (2-x)/1 (3-x)/1 x/1 (2 + x)/1

Conc CH3COOC2H5 = 1.33M Conc H2O = 3.33M Conc CH3COOH = 0.67M Conc C2H5OH = 1.67M

Kc = (CH3COOC2H5 ) (H2O ) (CH3COOH ) (C2H5OH ) Kc = (0.67) (0.67) (0.33) (0.33) Kc = 4.12

Kc = (CH3COOC2H5) (H2O) (CH3COOH) (C2H5OH ) Kc = (x) (2+x) (2-x) (3-x) 4 = (x) (2+x) (2-x) (3-x) x = 1.33

Change = Reacted and Produced




Ex 4

Ex 5

Reaction X + Y ↔ Z. Initial amount of X (0.02) and Y (0.01). At equilibrium 9.00 x 10-3 Z was produced

Calculate equilibrium constant Kc. (Volume vessel is 600cm3 )

X + Y ↔ Z

Initial mole 0.02 0.01 0 Change (0.02 – 9.00 x 10-3) (0.01 - 9.00 x 10-3) 9.00 x 10-3 Equilibrium/Conc 0.011/0.6 0.001/0.6 9.00 x 10-3 /0.6

Mixture of 1.9 mol H2 and 1.9 mol I2 was added to vessel . At equilibrium concentration 3mol of HI was formed

Calculate Kc at 400C. (Volume vessel is 1dm3 )

H2 + I2 ↔ 2HI

Initial mole 1.9 1.9 0 Change (1.9 – 1.5) (1.9 – 1.5) 3 Equilibrium/Conc 0.4/1 0.4/1 3/1


Kc = (Z) (X)(Y)

Kc = (0.015) (0.0183) (0.0016)

Kc = 491M-1

Kc = (HI)2

(H2)(I2)

= (3)2

(0.4)(04)

Kc = 56

Reaction SO2CI2 ↔ SO2 + CI2 . 0.05 mol of SO2CI2 was used and at equilibrium 0.0345 mol of CI2 was formed

Calculate equilibrium constant Kc. (Volume vessel is 2dm3 )

SO2CI2 ↔ SO2 + CI2

Initial mole 0.05 0 0 Change (0.05 – 0.0345) 0.0345 0.0345 Equilibrium/Conc 0.0155/2 0.0345/2 0.0345/2

Kc = (SO2) (CI2) (SO2CI2)

Kc = (0.0173) (0.0173) (0.00775)

Kc = 0.0386M







Ex 6

Ex 7

Ex 8

Reaction CO + H2O ↔ CO2 + H2. Initial amount of 3 mol CO and 3 mol H2O was used.

Calculate the equilibrium concentration for H2 if Kc for reaction is 4.0 (Volume vessel is 1dm3 )

CO + H2O ↔ CO2 + H2

Initial mole 3 3 0 0

Changes (3- x) (3-x) x x

Equilibrium/Conc (3-x)/1 (3-x)/1 x/1 x/1

Reaction PCI5 ↔ PCI3 + CI2. Initial amount of PCI5 is 0.01 mol. Kc = 0.19M. (Volume vessel is 0.5dm3 )

Calculate equilibrium conc for PCI5, PCI3 and CI2

PCI5 ↔ PCI3 + CI2

Initial mole 0.01 0 0

Changes 0.01 - x x x

Equilibrium/Conc (0.01 – x)/0.5 x/0.5 x/0.5


Kc = (CO2) (H2) (CO) (H2O ) Kc = (x) (x) (3-x) (3-x) 4 = (x)(x) (3-x) (3-x) x = 2M

Kc = (PCI3)(CI2) (PCI5)

0.19 = (x/0.5)2

(0.01 –x)/0.5

x = 0.0091M [PCI5] = (0.01 – 0.0091)/0.5 = 1.8 x 10-3M [PCI3] = 0.0091/0.5 = 1.8 x 10-2M [CI2] = 0.0091/0.5 = 1.8 x 10-2M





Ex 9

Ex 10

Equilibrium formed during this reaction was investigated in two experiment carried out at different temperature.

The result are shown in table below

For each experiment, deduce the concentrations of other species present at equilibrium.

Calculate the values of Kc for the forward reaction for each experiment.

2HI ↔ H2 + I2 Initial mole 0.06 0 0

Change (0.06 – 0.02) 0.01 0.01

Equilibrium/Conc 0.04/1 0.01/1 0.01/1

2HI ↔ H2 + I2 Initial mole 0 0.04 0.04

Change 0.04 0.02 0.02


2HI ↔ H2 + I2

EXPT 1

EXPT 2

Questions on IB Equilibrium Law

Kc = (H2) (I2) (HI)2

Kc = (0.01) (0.01) (0.04)2

Kc = 6.25 x10-2

Kc = (H2)(I2) (HI)2

Kc = (0.02) (0.02) (0.04)2

Kc = 0.25





Ex 11

Mixture of 1 mol H2 and 2 mol I2 was added to a reaction vessel .

Calculate the equilibrium concentration for HI, H2 and I2 at 400C. (Volume vessel is 1dm3 )

2HI ↔ H2 + I2

Initial mole 1 0 0

Change (1 -0.22) 0.11 0.11


1 mol of HI was heated at 400C and 0.78 mol of HI remains at equilibrium.

Calculate Kc. (Volume vessel is 1dm3 )

H2 + I2 ↔ 2HI

Initial mole 1 2 0

Change (1 –x) (2-x) 2x

Equilibrium/Conc (1-x)/1 (2-x)/1 2x/1

Conc HI = 2x = 1.87M Conc H2 = 1 - x = 0.065M Conc I2 = 2 - x = 1.065M

Kc1 = 1/Kc

= 1/0.02 Kc1 = 50


Kc = (H2)(I2) (HI)2

Kc = (0.11) (0.11) (0.78)2

Kc = 0.02



Kc1 = (HI)2

(H2)(I2)

50 = (2x)2

(1-x)(2-x)

x = 0.935M



Ex 12

Ex 13

2NO (g) + O2 ↔ 2NO2 (g) ΔH = -ve

4NH3(g) + 5O2(g) ↔ 4NO(g) + 6H2O(g) ΔH = -ve

Effect of Concentration, Pressure, Temperature and catalyst on equilibrium system

What are the effect of Increasing Conc, Pressure, Temp and Catalyst on the rate, rate constant, Kc and position of equilibrium

What are the effect of Increasing Conc, Pressure, Temp and Catalyst on the rate, rate constant, Kc and position of equilibrium

Ex 1

Ex 2

Education

IB Chemistry on Equilibrium Constant Calculation and Le Chatelier's Principle