Share this:


PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM


Chemical reactions which takes place in both directions are called reversible reactions .These reactions do not proceed to completion rather to dynamic equilibrium between products and reactants.
As a result of this, the concentration of reactants and products becomes constant (but not equal)
Reactants EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) products
At equilibrium, the rate of forward reaction is equal to the rate of backward reaction.
Note: The concentration of all species in the equilibrium remains constant since both opposing reactions proceed at the same rate.
The concentration of reactants decreases with time while those of products increases with time until equilibrium is reached where the

concentrations are constant (not equal).
EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
The equilibrium has been attained with high concentration of products than reactants therefore equilibrium lies on product side.

EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
The equilibrium has been attained with high concentration of reactants than products therefore equilibrium lies on reactants side.
The rate of reaction depends on the concentration of reactants. As the concentration of reactants decreases, the rate of forward reaction decreases

too. The rate of backward reaction increases since the concentration of products increases until equilibrium is attained.
EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
Characteristics of chemical equilibrium
1. The equilibrium can be established or attained in a closed system (no part of the reactants or products is allowed to escape out.)
2. The equilibrium can be initiated from either side .The state of equilibrium of a reversible reaction can be approached whether we start from

reactants or products.
Consider the reactions:-
H2(g) + EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) 2HI(g)
EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
3. Constancy of concentration
When a chemical equilibrium is attained, concentration of various species in the reaction mixture becomes constant.
4. A catalyst cannot change the equilibrium point. When a catalyst is added to a system in equilibrium, it speeds up the rate of both forward and

backward reactions to equal extent.Therefore a catalyst cannot change equilibrium point except that it is reached earlier.

Types of chemical equilibrium
1. Homogeneous equilibrium:
This is equilibrium where reactant and products are in the same physical states i.e. all solids, all liquids or all gases.
E.g. H2 (g) + I2 (g) EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) 2HI (g)
N2 (g) + 3H 2(g) EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) 2NH3 (g)
EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
2. Heterogeneous equilibrium;
This is equilibrium when the reactants and product are in the different physical states
CaCO3(s) EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) CaO(s) + CO2 (g)
3Fe (S) + 4H2O (g) EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) Fe3O4(s) + 4H2 (g)
3. Ionic equilibrium:
This is an equilibrium which involves ions.
E.g.
EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
LAW OF MASS ACTION
The law relates the rates of reactions and the concentration of the reacting substances. The law states that, “the rate of a chemical reaction is directly proportional to the product of the molar concentration of the reactants at constant temperature, at any given time”.

The molar concentration means number of moles per litre and is also called active mass
Consider the following simple reactions:
EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
The rate of the reaction, REcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
R = KEcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)Rate equation
Or R = KEcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
Where [A] and [B] are the molar concentrations of the reactant A and B respectively and that K is a constant of proportionality known as rate constant or velocity constant.

If the concentration of each of the reactants involved in the reaction is unity i.e. the concentration of [A] = [B] = 1, thus the rate constant of a

reaction at a given temperature may be defined as a rate of the reaction when the concentration of each of the reactants is unity.
Generally for a reaction
EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
Where a and b are stoichiometric coefficients or mole ratio of the reactants, A and B
r = K [A]a [B]b or R=K [A]a [B]b
LAW OF CHEMICAL EQUILIBRIUM AND EQUILIBRIUM CONSTANT
The law of mass action is applied to reversible reaction to derive a mathematical expression for equilibrium constant known as the law of chemical equilibrium.
Consider a simple chemical reaction (reversible)
A+B EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) C+D
The forward reaction is A+B EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) C+D
Rate of forward (Rf) EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) [A] [B]
Rf = Kf [A] [B]
Kf = Rate constant for forward reaction
Similarly for the backward reaction
EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
Rate of backward (Rb) EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)[C] [D]
Rb = Kb [C] [D]
Kb = Rate constant for backward reaction
At equilibrium both rates are equal
Rforward = Rbackward
Kf [A] [B] = Kb[C] [D]
EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
Ke = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
Generally:-
For the reaction
EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
The equilibrium constant for this expression is given by
K e = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
K e is changed to K c for equilibrium concentration (equilibrium constant) .
K c = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
Where K c is equilibrium concentration (equilibrium constant)
[C], [D] etc are molar concentration of species in mol per litre a, b etc are mole ratios or stoichiometric coefficients.
But for gaseous equilibrium we know that PV= nRT
PV = nRT
EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
Concentration of a gas is directly proportional to partial pressure therefore the equilibrium constant can be represented in terms of both concentration and partial pressure.
For the above equilibrium;
K p = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
Example1.
1.N2 O4 (g) dissociates to give NO2 (g)
EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
K c = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
K p = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
2. H2 (g) + I2 (g) EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) 2HI (g)
K c = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
K p = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
The law of chemical equilibrium states that, “For any system in equilibrium at a given temperature, the ratio of product of concentration of products to the product of the concentration of reactants raised to the power of their mole ratios is constant”.

UNITS OF EQUILIBRIUM CONSTANT
The units of equilibrium constants, K c and K p depends on the number of moles of reactants and products involved in the reaction
1. N2 (g) + O2 (g) EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) 2NO (g)
K p = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
K p = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) = Unit less
2. N2 (g) + 3H2(g) EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) 2NH3 (g)
K p = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
K p =EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
K p = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
K c = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
= EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
K c = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
A large value
of K p or K c means the equilibrium lies on the sides of the product and a small value of K c and K p means equilibrium lies on the sides of the reactants thus the equilibrium constant shows to what extent the reactant are converted to the product.
If K c is greater than 103, products pre- dominate over reactants equilibrium therefore the reaction proceeds nearly to completion.
If K c is less than 10-3, reactants dominate over products, the reaction proceeds to very small extent.
If K c is in the range of 10-3 and 103, appreciable concentrations of both reactants and products are present.
SOLIDS AND PURE LIQUIDS IN EQUILIBRIUM EXPRESSION
The concentration of a solid or pure liquid (but not a solution) is proportional to its density. Therefore, their densities are not affected by any gas pressure hence remain constant hence their concentration never appear in the equilibrium expression.

Example
1. What is the equilibrium expression for the following reactions?
i) 3Fe (g) + 4H2O (g) EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) Fe3O4 + 4H2 (g)
Solution
K c =EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) OR K p = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
ii) PCl5 (g) EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) PCl3 (g) + Cl 2 (g)
K c = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
K p = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
iii) HCl (g) + Li H(S) EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) H2 (g) +LiCl(s)
K c = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
K p = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
iv) Cu (OH) 2(s) EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) Cu2+ (aq) + 2OH (aq)
K c = [Cu 2+] [OH ] 2
v) CaCO3(s) EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) CaO(s) + CO 2(g)
K c = [CO2]
vi) [Cu(NH3)4]2+(aq) EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) Cu2+(aq) + 4NH3(aq)
EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
2. The equilibrium constant for the synthesis of HCl, HBr and HI are given below;
H2 (g) +Cl2 (g) EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) 2HCl (g) KC =1017
H2 (g) + Br2 (g) EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) 2HBr (g) KC=109
H2 (g) +I 2(g) EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) 2HI (g) KC=10
3. a) What do the value of Kc tell you about the extent of each reaction?
b) Which of these reactions would you regard as complete conversion? Why?
Answers
a) For all the 3 reactions, the equilibrium lies on the product side (i.e. RHS) and the extent of formation of HCl is greater than HBr which in turn is greater than HI.
b) For the 1st and 2nd reaction, it can be regarded as complete conversion because products pre-dominates over reactants at equilibrium.
Characteristics of equilibrium constant
1. Equilibrium constant is applicable only when the concentrations of reactants and products have attained their equilibrium.
2. The value of equilibrium concentration is independent of the original concentration of reactants.
3. The equilibrium constant has a definite value for every reaction at a particular temperature.
4. For a reversible reaction the equilibrium constant for the forwarded reaction is the inverse of the equilibrium constant for the backward reaction.
5. The value of equilibrium constant tells the extent to which reaction proceeds in the forward or reverse direction.
6. Equilibrium constant is independent of the presence of catalyst. This is because the catalyst affects the rate of forward and backward reaction equally.

Relationship between K c and K p for a gaseous equilibrium
For any given reaction, Kc or Kp is a function of the reaction itself and temperature.
Consider the following gaseous equilibrium
aA (g) + bB(g) EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) cC (g) + dD (g)
K c = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) —————— (1)
K p = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) ————– (2)
From PV= nRT
[X] = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
I.e. [A] = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) , [B] = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
Substitute these concentration in terms of partial pressures in equation (1)
K c = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
= EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) . EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
= K p EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
= K p EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
But; (c+d) = total number of moles in the products.
(a +b) = total number of moles in the reactants.
(c +d) – (a + b) = The difference between moles of products and reactants
So, (c
+d)-(a + b) = Δn

EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
Thus, the numerical value of K p and K c are equal when there is the same number of moles on products and reactants side numerically.
Example:
1. What is the relationship between K p and K c in the following reactions?
2H2 (g) + O2 (g) EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) 2H2 O (g)
From; K p = K c (RT) Δn
= K c (RT) (2- (1+2))
K p = Kc (RT)-1
ii) Derive the relationship between K p and k c for the particular reaction.
2H2 (g) + O2 (g) EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) 2H 2O (g)
K c = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) …………… (1)
K p = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) ……….. (2)
From PV = nRT
P = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)RT
P = [X] RT
EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)= [H 2] RT , EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)= [ EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)] RT………etc
Substitute the partial pressure in equation (2)
EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
K p =EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) . EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
K p = K c (RT) (2-(1+2))
K p = K c (RT)-1
2. a) Derive the relationship between K p and K c for ammonic synthesis.
N2 (g) + 3H2 (g) EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) 2NH3 (g)
K c = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) ………… (1)
K P = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) …….. (2)
From PV = nRT
P = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)RT
P = [X] RT
EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) = [EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)] RT, EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)= [EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)] RT, EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) = [EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)] RT
Substitute the partial pressure in equation (2)
K p = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
K P =EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
K p = KC x EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
K p = K C.(RT)-2
b) If Kc = 0.105 mol-2dm6 at 4720C. Calculate K p
[R=8.31 dm3 KPa mol-1 K-1]
Solution:
K p = K c x (RT)-2
= 0.105 x (8.31x 745)-2
= 0.105 x (6190.95)-2
= 2.7395x 10-9(KPa)-2
DETERMINATION OF EQUILIBRIUM CONSTANT
Consider the reaction:-
CH3CH2OH + CH3COOH EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) CH 3COOCH2 CH3 + H2O
EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
Example:
1. An equilibrium system for the reaction between H2 and I2, to form HI at 670K in 5l flask contains 0.4 moles of H2, 0.4 moles of I2 and 2.4 moles of HI. Calculate the equilibrium constant K c.
H2 (g) +I2 (g) EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) 2HI (g)
K c = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
[HI] = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)= 0.48 EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
[H2] = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)= 0.08 EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
[I2] =EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) = 0.08 EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)K c = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
K c = 36
2. A mixture of 1.0 x10-3 moldm-3 H 2 and 2.0 x 10-3 moldm-3 I2 are placed into a container at 4500C. After equilibrium was reached the HI concentration was found to be 1.87 x 10EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1). Calculate the equilibrium constant.
H2 (g) + I2 (g) EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) 2HI (g)
At t=0 a b 0
Equilibrium: a-x b-x 2x
EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)2x = 1.87 x 10-3
X= 9.35 x 10-4moldm -3
a=1.0 x 10-3
a-x = 1.0 x 10-3 – 9.35x 10-4
= 6.5 x 10-5moldm-3
b = 2.0 x 10-3
b -x=2.0EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)10-3– 9.35EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)10-4
=1.065EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)10-3moldm-3
K c = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
= EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
K c = 50.51
3. a) It was found that if 1 mole of acetic acid and half a mole of ethanol react to equilibrium at certain temperature, 0.422 moles of ethyl acetate are produced. Show that the equilibrium constant for this reaction is about 4.
EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
Start: – 1mol 0.5mol 0 0
Equilibrium: 1-0.422 0.5-0.422 0.422 x
0.578 0.078 0.422 0.422
Equilibrium EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
Concentration
K c = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
= EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
K c = 3.95
K c ≈ 4 shown
b) From the same reaction above, 3moles of acetic acid and 5 moles of ethanol reacted. Find the amounts which will be present equilibrium.
CH3COOH (l) + C2H5OH (l) EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) CH3COOCH2CH3 (I) + H2O (I)

t= 0 3 5 0 0
t = t (3-x) (5-x) x x
Equilibrium
Concentration EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
K c = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
4 = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
4 = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
4 = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
X2 = 60 – 32x + 4x2
3x2 – 32x + 60 = 0
X = 8.24 or x = 2.43
Logically; x = 2.43
At equilibrium CH3COOH EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) 3 – 2.43 = 0.57 moles
C2 H5 OH EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) 5 – 2.43 = 2.57 moles
4 .For the reaction;
CO2 (g) + H2 (g) EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) CO (g) + H2 O (g)
The value of K c at 5520C is 0.137. If 5moles of CO 2, 5moles of H2, 1 mole of CO and 1 mole of H2 O are initially present, what is the actual concentration of CO2, H2, CO and H2O at equilibrium?
CO2 (g) + H2 (g) EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) CO2 (g) + H2 O (g)
At start; 5 1 1 1
At equilibrium 5-x 1-x 1x 1x
K c = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
0.137 = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
0.137 = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
3.425 – 1.37x + 0.137x2 = X2
3.425 – 1.37x +
0.137 x2–x2 =0
3.425 -1.37x – 0.863x2 =0
x =EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
x = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
x =EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) = x =1.349
COMBINING EQUILIBRIUM REACTIONS
1. Reversing an equilibrium reaction
Consider the reaction
PCl5 (g) EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) PCl3 (g) +Cl2 (g)………………. (i)
K c (i) =EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) …………………….. (i)
Reversing the reaction;
PCl3 (g) + Cl2 (g) EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) PCl5 (g)
K c (ii) = EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1) ………………. (2)
When reaction equation is reversed, the equilibrium constant is reciprocated i.e. from equation (1) and (2)
K c(ii) =EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)
2.




Share this:

EcoleBooks | CHEMISTRY A LEVEL(FORM SIX) NOTES - PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)

subscriber

Leave a Reply

Your email address will not be published. Required fields are marked *

Accept Our Privacy Terms.*