Chemical equilibrium

ByBy Dr. Hisham Ezzat Abdellatef Dr. Hisham Ezzat Abdellatef

Professor of Pharmaceutical Analytical Professor of Pharmaceutical Analytical ChemistryChemistry

First Year 2011-2012

Classification of reaction

Homogeneousonly one phase

Heterogeneousmixture is not uniform

e.g: In the gas phase:H2(g) + I2 → 2 HI(g)

In the liquid phase: CH3COCl + CH3OH → CH3COOCH3 + HCIAcetyl Chloride methyl alcohol methylacetate

irreversible reaction or complete reaction

reversible reaction →

A + B → AB N2(g) + 3 H2(g) 2 NH3(g)

at equilibrium ----the rate of forward reaction equal the rate of backward reaction

Factors influencing equilibrium:

1. forward and reverse rates2. partial pressures3. concentrations4. temperature

Law of mass action

A2(g) + B2(g) 2 AB(g)

rate of the forward reaction = Kf. [A2] [B2]rate of the reverse reaction is = Kr [AB]2

At equilibrium ratef = rater

Kf [A2] [B2] = Kr [AB] 2

]][B[A

[AB]K22

2

K which is called the equilibrium constant

aA + bB eE + fF eE +fF aA + bB

ba

fe

c [B] [A][F][E] K

In general, for any reversible reaction

fe

ba\

[F][E][B] [A] K

cK1 \ K

Mechanisms of more than one step 1

2 NO2CI(g) 2 NO2(g) + CI2(g)

22

22

2

Cl][NO][Cl][NO Kc

mechanism consisting of two steps:

1- NO2CI NO2 + Cl

2. NO2CI NO2 + CI2

Cl][NO[Cl] ][NOK

2

21

[Cl] ][NO][Cl ][NOK

2

222

[Cl] ][NO

][Cl ][NOCl][NO[Cl] ][NO= K K = K

2

22

2

221c 2

2

22

2

Cl][NO][Cl][NO

• For reactions involving gasesP α C

using partial pressures instead of concentration.

N2(g) + 3 H2(g) 2 NH3(g) 3H2N2

2NH3

p .PPp K

322

23

c ][H ][N][NHK

3H2N2

2NH3

p .PPp K

(Kc ≠ Kp)

The Relationship Between Kp and KC:

aA + bB eE + fF

bB

aA

fF

eE

p P .PP .P K ba

fe

c [B] [A][F][E]K

Assuming ideal gas PV= nRT

concentration of a gas X RTP

Vn = [X] XX Px is its partial pressure

Px = [X] RT

bB

aA

fF

eE

P .PP .P Kp bbaa

ffee

(RT)[B] (RT)[A](RT)[F] (RT)[E]

Kp = b)(af)(e

ba

fe

(RT)[B] [A][F][E]

Kp = Kc. (RT)n(g)

Example 1:For the reaction N2O4(g) 2 NO2(g)

The concentrations of the substances present in an equilibrium mixture at 25°C are[N2O4] = 4.27 x 10-2 mol/L [NO2] = 1.41 x 10-2 mol/L

what is the value of Kc for this temperature.

Solution:

Kc = mol/l4.66x10mol/l)(4.27X10mol/l)(1.41x10

]O[N][NO 3

2

22

42

2

2

Example 2:At 500 K. 1.0 mol of ONCI(g) is introduced into a one - liter container. At equilibrium the ONCI(g) is 9.0% dissociated:

2 ONCI(g) 2 NO(g) + CI2(g)

Calculate the value of Kc for equilibrium at 500 K.

Solution: [ONCI(g) ] = 1 mol/L since 9.0% dissociated,Number of moles dissociated =

at equilibrium, [ONCI] = 1.0 mol/L - 0.09 mol/L = 0.91 mol/Lamounts of CI2 :

2 ONCI 2NO + CI2 2 x x

0.09 mol 0.045 mol

ONCI mol 0.09mol x1.0100 9 = [X]

Therefore, at Kc = 22

2

[ONCl]][Cl[NO]

= mol/l 4.4x10

mol/l) (0.9l)(0.045mol/mol/l) (0.09 4

2

2

2ONCI 2NO + CI2

at start 1.0 mol/L ----- ------ Change - 0.09 mol/L + 0.09 + 0.045at equilibrium 0.91 0.09 0.045

Example 3:

For the reaction2 SO3(g) 2 SO2(g) + O2(g) at 1100 K

Kc is 0.0271 mol/L. what is Kp at same temperature.

Solution:n = 3-2 =1 Kp = Kc (RT)+1 = 0.0271 mol/L x (0.0821 L. atm / K. mol) (1100 K) = 2.45 atm

Example 4:What is Kc for the reaction?

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

At 500°C if Kp is 1.5 x 10-5 / atm-2 at this temperature.

Solution:Kp = Kc (RT)n

Kp = Kc (RT)-2

Kc = Kp (RT)2 = x [0.0821.atm/K.mol x 773K]2 = (1.5 x I0-5/ atm2) (4.03 x 103 L2. atm2 / mol2) = 6.04 x 10-2 L2 / mol2

n = 2 - 4 = -2T = 273 + 500 = 773 K

Try

At 127oC, K = 2.6 x 10-5 mol2/L2 for the reaction2NH3(g) N2(g) + 2H2(g)

Calculate Kp at this temperature

Reaction quotient (Q).

Predicting the Direction of a Relation:• For the reaction

PCl2(g) PCl3(g) + Cl2(g) at 250oC

• Suppose that a mixture of 1.00 mol of PCI5(g)/ 0.05 mol of PCI3(g)/ and 0.03 mol of CI2(g) is placed in 1.0 L container. Is this an equilibrium system, or will a net reaction occur in one direction or the other?

Kc = l0.0415mol/][PCl

]][Cl[PCl

5

23

1- Q < Kc from left to right (the forward direction) to approach equilibrium.

2- Q = Kc The system is in equilibrium.

3- Q > Kc from right to left (the reversible direction) to approach equilibrium.

Q = mol/l 0.015 0.1

03.005.0][PCl

]][Cl[PCl

5

23 x

Q (0.015 mol/L) < kc (0.0415 mol/L). The system is not at equilibrium. The reaction will proceed from left to right.

Example 4:• For the reaction 2 SO2(g) + O2(g) 2 SO3(g)

at 827°C, kc is 36.9 L / mol. If 0.05 mol of SO2(g), 0.03 mol of O2(g), and 0.125 mol of SO3(g) are mixed in a 1.0 L container, in what direction will the reaction proceed?

• Solution:

Since Q (208 L/mol) > kc (36.9 L/mol), the reaction will proceed from right to left (SO3 will dissociate).

L/moL 208][0.03mol/lmol/l] [0.05

l][0.125mol/][O][SO

][SOQ 2

2

22

2

23

Heterogeneous Equilibria:

• The concentration of a pure solid or a pure liquid is constant and do not appear in the expression for the equilibrium constant.

• For exampleCaCO3(S) CaO(S) + CO2(g)

Kc = [CO2]

3 Fe(s) + 4 H2O(g) Fe3O4(s) + 4 H2(g)

Example 5:

Kc for the HI equilibrium at 425°C is 54.5:

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

A quantity of HI(g) is placed in a 1.01 container and allowed to come to equilibrium at 425°C What are the concentrations of H2(g) and I2(g) in equilibrium with 0.5 mol/L of HI(g)

Solution:at equilibrium [H2] = [I2] [H2] = [I2] = x [HI] = 0.5 mol/L

x = 0.068 mol/L The equilibrium concentration is: [HI] = 0.5 mol/ L [H2] = [I2] = 0.068 mol/ L

54.5][I ][H

[HI]K 422

2

c

54.5Xmol/l] [0.5

2

2

222

2 /Lmol 0.0045654.5(0.5)X

Example 6:

For the reaction

H2(g) + CO2(g) H2O(g) + CO(g)

kc is 0.771 at 750°C. If 0.01 mol of H2 and 0.01 mol of CO2 are mixed in 1 liter container at 750°C, what are the concentrations of all substances present at equilibrium?

• Solution:If x mol of H2 reacts with x mol of CO2 out of the

total amount supplied, x mol H2O and x mol CO will be produced. Hence

H2(g) + CO2(g) H2O(g) + CO(g)

At start

 

0.01 mol/L

 

0.01 mol/L

 

----

 

-----

 

Change

 

- x

 

- x

 

+ x

 

+ x

 

at equilibrium

 

0.01 -x

 

0.01 -x

 

X

 

X

 

0.771][CO ][H

[CO] O][HK22

2c

2

2

c X) - (0.01XK square root 878.0

X - 0.01X

X = 0.0878 – 0.878 XX= 0.00468 mol/l

At equilibrium, therefore

[H2] = [CO2] = 0.01 mol/L - 0.00468 mol/L = 0.0053 mol/L [H2O] = [CO] = 0.00468 mol/L

Example 7: •For the reaction

C(s) + CO2(g) 2 CO(g)

Kp is 167.5 atm at 1000°C. What is the partial pressure of CO(g) in an equilibrium system in which the partial pressure ofCO2(g) is 0.1atm?

•Solution:

•PCO2= 16.8•PCO= 4.10 atm

atm 167.5PPK

CO2

2CO

p

atm 167.50.1P2

CO

Example 8:

Kp for the equilibrium:

FeO(s) + CO(g) Fe(s) + CO2(g)

at 1000°C is 0.403. If CO(g) at a pressure of 1.0 atm, and excess FeO(s) are placed in a container at 1000°C, what are the pressures of CO(g) and CO2(g) when equilibrium is attained?

FeO(s) + CO(g) Fe(s) + CO2(g)

At start

Change

1.0 atm

- x

---

+ x

At equilibrium 1.0 - x atm x

403.0PPK

CO

CO2p

403.0X - atm 1.0

atm X

Solution: Let x equal the partial pressure of CO2 when equilibrium is attained

X = PCO2 = 0.287 atm1.0 – X = Pco = .713

• If a change is made to an equilibrium, the equilibrium shifts in the direction that consumes the change – Case 1: Changing the amounts of reactants /

products.– Case 2: Changing the volume by changing

pressure. – Case 3: Changing the temperature.

• If the concentration of substance is increased, the equilibrium will shift in a way that will decrease the concentration of the substance that was added.

e.g: H2(g) + I2(9) 2 HI(g)•

Increase H2 or I2 → shift to to formation of HI• Removal of H2 or I2 ← Reaction shift to

decomposition of HI.

• Increasing the pressure causes a shift in the direction that will decrease the number of moles of gas.

2 SO2(g) + O2(g) 2 SO3(g)3 moles 2 moles

• When the pressure on an equilibrium mixture is increased (or the volume of the system decreased), the position of equilibrium shifts to the right., and vice versa.

• For reactions in which n = 0, pressure changes have no effect on the position equilibrium.

e.g: N2(g) + O2(g) 2 NO(g)

For the reactionN2(g) + 3 H2(g) 2 NH3(g) H = - 92.4 KJ

Since H is -ve, the reaction to the right evolves heat

N2(g) + 3 H2(g) 2 NH3(g) + 92.4 KJ

The highest yields of NH3 will be obtained at the lowest temperatures and high pressures.

Also consider the reactionCO2(g) + H2(g) CO(g) + H2O(g) H = + 41.2KJ

Since H is + ve , we can write the equation

41.2 KJ + CO2(g) + H2(g) —— CO(g) + H2O(g)

Increasing the temperature always favors the endothermic change, and decreasing the temperature always favors the exothermic change.