EQUILIBRIUM
Sheet 2
I. Given the following balanced equation
CH3COOH + C2H5OH
acid alcohol
CH3-COOC2H5 + H2O
ester
The initial concentrations are as follows: acid 10 g/l and alcohol 8 g/l and equilibrium
concentrations of ester 9.68 g/l and H2O is 2 g/l.
a.
Calculate Kc.
b.
If 1 mole of each reactant (acid and alcohol) had been present initially, what
weight of the products (ester and water) would have been produced.
II. Given I2 + H2
2HI
If at equilibrium, half of the reacting substance is present in equal amounts of I 2 and H2
and half is HI, calculate Kc.
III. Kc for A + B
C + D is 4
Calculate the moles of C and D at equilibrium if 5 m/l of A are added to 5 m/l of B.
.
KEY TO SHEET 2 ON EQUILIBRIUM
I. a. Change concentration of each substance from g/l to moles/l:
acid 10/60 = .17
alcohol 8/46 = .17
ester 9/68/88 = .11
water 2/18 = .11
Since. 11 m/l of the ester and water were formed from the acid and the alcohol, then
each of the concentrations of the acid and the alcohol is equal to .17 - .11 = .06.
Kc =
.11.11 = 3.36
.06.06
b. .17 moles of acid or alcohol gave .11 moles of ester and water. 1 mole of acid or
alcohol gave
1.11
= .647 m/l of ester and water or
.17
.647
m 88 g

= 56.9 g/l of ester
1 mole
m 18 g

= 11.6 g/l of water
1 mole
.647
II. I2 + H2
2 HI
Assume that total concentration of reacting substance to be 1 M; hence, concentration of
HI = .5 concentration of I2 + concentration H2 = .5 or each concentration of I2 = H2 = .25
Kc =
=
 HI2
I 2 H 2 
.52  4
.25.25
III. Let x m/l of A react with x m/l of B to give x m/l of C and x m/l of D.
Then the equilibrium concentrations are
A = (5-x) m/l
C = x m/l
B = (5-x) m/l
D = x m/l
Substitute these values in the equilibrium expression
C D
A B
Kc =
4=
 X X
5  X5  X
4 (25-10x + x2) = x2
3x2 - 40x + 100 = 0
Use quad. formula to find x
x=
 b  b 2  4ac
2a
=
40  1600  4  3  100
6
=
40  20
6
x = 10
or x = 3.33
x = 10 impossible solution