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Chapter three
Concept of chemical equilibrium
If two reactants A and B are mixed together a chemical reaction
can take place to yield the product C&D
The chemical reaction can be presented as follows :aA + bB
cC + dD …………….(1)
a, b, c, d : stoichiometric coefficients
in many chemical reaction the reverse reaction i.e reaction
between c with d to form A&B the reaction is also possible
hence the reaction is presented by:
aA + bB
cC + dD
aA + bB
cC + dD
when forward and reverse reaction become equal, the system
has reacted to the condition know as " Chemical equilibrium "
the reaction appear to stop when equilibrium condition is
actually reached since there is no change in the concentration of
all substances
The equilibrium constant
the equilibrium condition is often expressrd by a factor or
known as equilibrium constant
Chemical Equilibrium Calculations
Equilibrium constants are useful for calculating the
concentration of varies ions- in equilibrium the following
examples give an idea for calculations using equilibrium
constants
Example |(1)
the chemical substances (A) and (B) are reacted to produce (C)
and (D) according to the reaction A + B C + D if 0.2 mol of
(A) and 0.5mol of (B) are dissolved in 1 liter solution and the
reaction takes place with equilibrium constant K = 0.3. calculate
the concentrations of reactants and products at equilibrium
let x mol/l of A and B that is reacted, equilibrium constant of
[A] = 0.2-x mol/l
let x mol/l of A and B that is reacted, equilibrium constant of
[B] = 0.5 –x mol/l
[C] = x mol/l
[D] = x mol/l
[ ][ ]
[ ][ ]
K=
=
= 0.3
x2 = 0.3 (0.2-x) (0.5-x)
0.7x2 + 0.21 x – 0.03 = 0
This is a quadratic equation and can be solved by general
quadratic formula
x=
√
x= -0.21±
√
= 0.11M
[A] = 0.09 , [B] = 0.39 , [C] = 0.11 , [D] = 0.11M
Example (2)
Calculate the equilibrium concentration of A and B in a 0.10 M
solution of weak electrolyte AB with an equilibrium constant of
3× 10-6 AB
A+B
K=
[ ][ ]
[
]
Let x = equil. Conc. of A and B , AB = 0.1- x
K= quite small compare to the initial conc. Hence [AB]=0.1-x 0.1
3×10-6 =
=√
= 5.5× 10-4 M = A = B
Note : this simplification will generally hold if the initial conc.
is ≥ 100k , other wise , a quadratic equation will have to be
solved factors. Affecting on chemical equilibrium
1.
2.
3.
4.
Temp.
Pressure
Conc.
Catalysts
Example (3)
A solution was prepared by dissolving 0.1 mol of trichloroacetic
Acid (CCl3COOH) in enough water to make 1.0liter of solution at
25 the dissociation constant Ka for trichloroacetic acid = 0.129
calculate :a) The percent dissociation of trichloroacetic acid
b) The percent dissociation of trichloroacetic if 0.1mole of
soluble salt of trichloroacelate (CCl3COO-) is added to the
solution (neglect any reaction of trichloroacetate with
water)
The chemical reaction CCl3COOH↔ CCl3COO- + H+
Ka =
[
][
[
]
= 0.129
]
Let x = moles of the acid dissociates during the reaction
= (0.1- x) M
[CCl3COOH] =
=(x) M
[CCl3COO-] = [H+] =
= 0.129
x2 + 0.129x – 0.0129 = 0
→
√
X=
= 0.066 mole
% dissoiciation of trichloroacetic acid =
× 100 = 66%
b) if 0.1 moles of trichloroacetate is added to the solution
hence
[CCl3COOH] = (0.1 – x) M
[H+] = x M
[CCl3COO-] =(
= 0.129
)M
→
= 0.129
X2 + 0.229 X – 0.0129 = 0 then
X=
√
X = 0.0465 mole
% dissociation of trichloroacetic acid = 46.5 %
Hence
It is concluded that the addition of one of the products
decreases the dissociation of the acid from 66% to 46.5%
which shows the effect of concentation