Chemistry Chemistry A Level(Form Six) Chemistry Notes Form Six (A Level)

CHEMISTRY A LEVEL(FORM SIX) – PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(1)

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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 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).

The equilibrium has been attained with high concentration of products than reactants therefore equilibrium lies on product side.

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.


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) + 2HI(g)
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) 2HI (g)
N2 (g) + 3H 2(g) 2NH3 (g)
2. Heterogeneous equilibrium;
This is equilibrium when the reactants and product are in the different physical states
CaCO3(s) CaO(s) + CO2 (g)
3Fe (S) + 4H2O (g) Fe3O4(s) + 4H2 (g)
3. Ionic equilibrium:
This is an equilibrium which involves ions.
E.g.
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:
The rate of the reaction, R
R = K Rate equation
Or R = K
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
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 C+D
The forward reaction is A+B C+D
Rate of forward (Rf) [A] [B]
Rf = Kf [A] [B]
Kf = Rate constant for forward reaction
Similarly for the backward reaction
Rate of backward (Rb) [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]
=
Ke =
Generally:-
For the reaction
The equilibrium constant for this expression is given by
K e =
K e is changed to K c for equilibrium concentration (equilibrium constant) .
K c =
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
=
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 =
Example1.
1.N2 O4 (g) dissociates to give NO2 (g)
K c =
K p =
2. H2 (g) + I2 (g) 2HI (g)
K c =
K p =
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) 2NO (g)
K p =
K p = = Unit less
2. N2 (g) + 3H2(g) 2NH3 (g)
K p =
K p =
K p =
K c =
=
K c =
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) Fe3O4 + 4H2 (g)
Solution
K c = OR K p =
ii) PCl5 (g) PCl3 (g) + Cl 2 (g)
K c =
K p =
iii) HCl (g) + Li H(S) H2 (g) +LiCl(s)
K c =
K p =
iv) Cu (OH) 2(s) Cu2+ (aq) + 2OH (aq)
K c = [Cu 2+] [OH ] 2
v) CaCO3(s) CaO(s) + CO 2(g)
K c = [CO2]
vi) [Cu(NH3)4]2+(aq) Cu2+(aq) + 4NH3(aq)

2. The equilibrium constant for the synthesis of HCl, HBr and HI are given below;
H2 (g) +Cl2 (g) 2HCl (g) KC =1017
H2 (g) + Br2 (g) 2HBr (g) KC=109
H2 (g) +I 2(g) 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) cC (g) + dD (g)
K c = —————— (1)
K p = ————– (2)
From PV= nRT
[X] = =
I.e. [A] = , [B] =
Substitute these concentration in terms of partial pressures in equation (1)
K c =
= .
= K p
= K p
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

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) 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) 2H 2O (g)
K c = …………… (1)
K p = ……….. (2)
From PV = nRT
P = RT
P = [X] RT
= [H 2] RT , = [ ] RT………etc
Substitute the partial pressure in equation (2)
K p = .
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) 2NH3 (g)
K c = ………… (1)
K P = …….. (2)
From PV = nRT
P = RT
P = [X] RT
= [] RT, = [] RT, = [] RT
Substitute the partial pressure in equation (2)
K p =
K P =
K p = KC x
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 CH 3COOCH2 CH3 + H2O

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) 2HI (g)
K c =
[HI] = = 0.48
[H2] = = 0.08
[I2] = = 0.08
K c =
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 10. Calculate the equilibrium constant.
H2 (g) + I2 (g) 2HI (g)
At t=0 a b 0
Equilibrium: a-x b-x 2x
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=2.010-3– 9.3510-4
=1.06510-3moldm-3
K c =
=
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.
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
Concentration
K c =
=
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) CH3COOCH2CH3 (I) + H2O (I)

t= 0 3 5 0 0
t = t (3-x) (5-x) x x
Equilibrium
Concentration
K c =
4 =
4 =
4 =
X2 = 60 – 32x + 4x2
3x2 – 32x + 60 = 0
X = 8.24 or x = 2.43
Logically; x = 2.43
At equilibrium CH3COOH 3 – 2.43 = 0.57 moles
C2 H5 OH 5 – 2.43 = 2.57 moles
4 .For the reaction;
CO2 (g) + H2 (g) 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) CO2 (g) + H2 O (g)
At start; 5 1 1 1
At equilibrium 5-x 1-x 1x 1x
K c =
0.137 =
0.137 =
3.425 – 1.37x + 0.137x2 = X2
3.425 – 1.37x + 0.137 x2–x2 =0
3.425 -1.37x – 0.863x2 =0
x =
x =
x = = x =1.349
COMBINING EQUILIBRIUM REACTIONS
1. Reversing an equilibrium reaction
Consider the reaction
PCl5 (g) PCl3 (g) +Cl2 (g)………………. (i)
K c (i) = …………………….. (i)
Reversing the reaction;
PCl3 (g) + Cl2 (g) PCl5 (g)
K c (ii) = ………………. (2)
When reaction equation is reversed, the equilibrium constant is reciprocated i.e. from equation (1) and (2)
K c(ii) =
2.

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