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

CHEMISTRY A LEVEL(FORM SIX) – PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(2)

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Multiplying an equilibrium reaction by a number
When the stoichiometric coefficient of a balanced equation is multiplied by the same factor, the equilibrium constant for the new equation is old equilibrium constant raised to the power of the multiplied factor.
PCl5 (g) PCl3 (g) + Cl (g)
K c (i) =
If the equation (i) is multiplied by ½
PCl5 (g) PCl3 (g) + Cl2 (g)
Note; if an equilibrium reaction is multiplied by then;
K c(ii ) =
K c(ii) =
Example: The K c for the reaction below is
S (s) + O2 (g) SO3 (g); what is the K c for SO3 Equilibriate with S + O2
2 SO3 (g) 2S(s) + 302(g)
S (s) + O2 (g) SO3 (g)………………. (I)
2SO3 (g) 2S(s) + 302(g)…………………. (2)
Equation (2) has been reversed and multiplied by 2
When reversed
K c (i) = = 1.1×1065
K c (ii) =
K c (ii) =
K c (ii) = 9.09 x 10-66
When multiplied
K c (ii) =
3. Adding the equilibrium.
The equilibrium constant for the reaction
i) 2HCl(g) Cl2(g) +H2(g)
K c (i) = 4.17x 10-34 (At 25 0 c)
The equilibrium constant for reaction
ii) I2 (g) + Cl2 (g) 2ICl (g)
K c (ii) = 2.1 x 105 (At 25 0c)
Calculate the equilibrium constant for the reaction
iii) 2HCl (g) + I2 (g) 2ICl (g) + H2 (g)
K c (iii) =?
K c (i) =
K c (ii) =
K c (iii) =
When equation (i) + equation (ii) = equation (iii), Hence
K c (iii) = K c (ii) x K c (i)
= (2.1 x 105) x (4.17 x 10-34)
K c (iii) = 8.757 x 10-29 (mol dm-3)1/2
Question 1:
The following are reactions which occurs at 3500K
i) 2H2 (g) +O2 (g) 2H2O (g) K p (i) = 26.4 atm-1
ii) 2CO (g) +O2 (g) 2CO2 (g) K p (ii) = 0.376 atm-1
Calculate equilibrium constant for reaction
CO 2 (g) + H 2 (g) CO (g) + H2O (g)
K p (iii) =?
Question 2
.Determine the Equilibrium constant for reaction;
M2(g) +O2 (g) +Br2 (g) NOBr (g)
Given that:-
i/ 2NO (g) N2(g) + O2(g) K c = 2.4 x
ii/ NO (g) + Br2 (g) NOBr K c = 1.4
Answers
i) 2H2 (g) + O2 (g) 2H2 O (g) K p (i) =26.4atm-1
ii) 2CO (g) + O2 (g) 2CO2 (g) K p (ii) =0.376 atm-1
iii) CO2 (g) + H2 (g) CO (g) + H2O (g) K p (iii) =?
K p (i) =
K p (ii) =
K p (iii) =
Reverse (ii), multiply (i) and (ii) by ½
i) (2H2 (g) +O2 (g) H2O (g)) K p (i) =
i) H2 (g) + O 2 H2O (g) K p(i) = 5.138 atm-1
Reverse (ii)
ii) 2CO2 (g) 2CO (g) +O2 (g) K p (ii) =
= 2.6595 atm-1
Now multiply by (ii) by ½
(2CO2 (g) 2CO (g) +02(g)) K p (ii) =
ii) CO2 (g) CO (g) + O2 (g) =1.63079
Add equation (i) + (ii), then
K p (iii) = K p (i) x K p (ii)
= 5.138 x 1.63079
=8.379 atm-1
2. i) 2NO(g) N2(g) + O2(g) K c(i) = 2.4 x 1030
ii) NO (g) + Br (g) NOBr (g) K c (ii) =14
iii) N2 (g) + O2 (g) + Br (g) NOBr(g) K c (iii) =?
Reverse ……. (i)
O2 (g) + N2 (g) 2NO (g) K c (i) =
=4.167 x 10-31
Now multiply (i) by ½
O2 (g) + N2 (g) NO (g) K c (i) =-31
=6.455 x 10-16
Equation (i) + (ii) = (iii)
Therefore:

K c (iii) = K c (i) x K c (ii)
= 6.455 x 10-16 x 1.4
= 9.0369 x 10-16(mol dm-3)1/2
3. The equilibrium constants for the reactions which have been determined at 878K are as follows:-
i) COO (s) + H2 (g) CO(s) + H2O (g) K1 = 67
ii) COO(s) + CO (g) CO(s) + CO2 (g) K2 = 490
Using these information, calculate K’s (at the same temperature) for;
iii) CO2 (g) +H2 (g) CO2 (g) + H2O (g) K3 =?
And commercially important water gas reaction
iv) CO (g) + H2O(g) CO2(g)+ H2(g) K4=?
Reverse (ii)
(ii)CO2 (g) + CO(s) CO (g) + COO (g) K2=2.0408 x 10-3
Equation (i) + (ii) = (iii)
K3 = K 1 x K2
= 67 x 2.0408 x 10-3
K3 = 0.1367
To find K4
i) COO (s) +H 2 CO (s) + H2O (g) K1= 67
ii) COO(s) + CO (g) CO (s) + CO2 (g) K2 = 490
iii) CO2 (g) + H2 (g) CO (g) + H2O (g) K3 = 0.1367
iv) CO (g) + H2O (g) CO2 (g) + H2 (g) K4 =?
When equation (iii) is reversed, it is equal to equation (iv)
K4 =
=
= 7.315
4. The heterogeneous equilibrium
i) Fe(s) + H2O FeO(s) + H2 (g)
ii) Fe(s) + CO2 (g) FeO(s) + CO (g)
Have been studied at 800 0c and 1000 0c.Also the rate () is constant = 1.81 at 8000c and 2.48 at 1000 0c.
i) Why are the ratios constant?
ii) Calculate equilibrium constant at two temperatures of the reaction
iii) H2O (g) + CO (g) H2 (g) + CO2 (g)
Answer:
i) The ratios are constant because for any system in equilibrium at a given temperature, the ratio of products of concentration of products to the product of concentration of reactants raised to the point of their mole ratios is constant.
ii) At 800 0C
K p1 =2
K p2 =1.81
K p3 =?
(i) Fe(s) + H2O(l) Fe(s) + H2 (g) K p (i) =2
Reversed (ii) CO (g) + FeO(s) Fe(s) + CO2 (g) K p (ii) =?
Equation (i) + (ii) = (iii)
Kp1 x Kp2=Kp3
Kp3 = Kp1 x Kp2
=2 x 0.55
Kp3 = 1.1
At 1000 0c
Kp1 = 1.49
K p2 =2.48
i) Fe(s) + H2O (g) FeO(s) + H2 (g) K p1 =1.49
Reversed ii) CO (g) + FeO(s) CO2 (g) + Fe(s) Kp2 =
=0.403
Equation (i) + (ii) = (iii)
Kp3 =kp1 x kp2
= 1.49 x 0.403
Kp3 = 0.6
1: When 1 mole of ethanoic acid is maintained at 25 0c with 1 mole of ethanol, 1/3 of ethanoic acid remain when equilibrium is attained. How much would have remained if 3/4 of 1 mole of ethanol had been used instead of 1mole at the same temperature.
CH3CH2OH + CH3COOH CH3COOCH2CH3 + H2O
At start: 1mol 1mol 0 0
At time: 1-x 1-x x x
At equilibrium: 11- x =
K c =
=
=
Kc = 4
CH3CH2OH + CH3COOH CH3COOCH2CH3 +H2O
At start; 1mol 1mol 0 0
At time 1-x 1-x x x
At equilibrium (1 x 3/4)-x 1-x x x
Kc =
4 = =
4 =X2
3- 7x + 4x2 = x2
3- 7X + 4X2 – x2 =0
3x2 + 3 – 7x = 0
3x2 -7x + 3 = 0
a b c
x =
x =
=
=
X= or x =
= 1.76 = 0.566
X cannot be 1.76
X = 0.566
– 0.566 = 0.184 moles or 23/125 moles
0.184 moles of ethanol would have remained
2. The equilibrium constant (K c) for the reaction; 2HI (g) H2 (g) +I2 (g) is 0.02 of 400 0 c. If 2 moles of H2 and 1 mole of I2 were mixed together in a 1.0dm3 at 400 0c, how many moles of HI, I2 and H2 would be present at equilibrium.
2HI (g) H2 (g) + I2 (g) K c = 0.02
Reverse the equation above
H2 (g) + I2 (g) 2HI (g) Kc =
At start 2 1 0
At time 2-x 1-x 2x = = 50
At equilibrium
K c =
50 =
50 =
50 (2 – 3x + x2) = 4x2
100 – 150x + 50x2 = 4x2
100 – 150x +5 0x2 – 4x2 = 0
100 – 150x + 46x2 = 0
x=
=
=
x= or x=
x= 2.33 x = 0.934
x cannot be 2.33
x = 0.934
At equilibrium:
Number of moles of H2 = 2- 0.934
=1.066moles
Number of moles of I2= 1-0.934
=0.066 moles
Number of moles of HI = 2 x 0.934
=1.868 moles
3. The equilibrium constant for the reaction; H2 (g) +Br2 (g) 2HBr (g) at 1024K is 1.6 x 105. Find the equilibrium pressure of all gases if 10 atm of HBr is introduced into a second container at 1024K.
H2 + Br2 2HBr
PREDICTION OF DIRECTION AND EXTENT OF CHEMICAL EQUILIBRIUM
At each point in the progress of a reaction, it is possible to formulate the ratio of concentration having the same form as the equilibrium

constant expression .This generalized ratio is called reaction quotient (Q).
For the reaction; then:-
QC =
Q p =
QC differs from KC in that the concentration in the expression is not necessarily the equilibrium concentration.
When the values of KC and QC are compared, one can predict the direction of the chemical reaction.
If Q c K c the system is not at equilibrium, the reactants must further be converted to products to achieve equilibrium therefore net reaction proceeds from left to right.
If Q c = K c the system is at equilibrium
If Q c > K c the system is not an equilibrium, the products must converted to reactants to achieve equilibrium therefore a net reaction proceeds from right to left.
Example:
1.Consider the reaction
At 2500C, K c = 4.0 x 10-2. If the concentration of Cl2 and PCl3 are both 0.30 M while that of PCl5 is 3.0M, is the system at equilibrium? If not, in which direction does the reaction proceed?
-2
Q c =
=
Q c = 0.03 QC ≠ K c, Q c< K c
The system is not at equilibrium and the reaction proceeds from left to right
2. At 200k, the K p for the formation of NO is 4 x 10-4
N2 (g) + O2 (g) 2NO (g)
If at 200k the partial pressure of N2 is 0.5 atm and that of 02 is 0.25 atm, that of NO is 4.2 x 10-3 atm, decide whether the system is at equilibrium, if not in which direction does the reaction proceed.
K p = 4.0 x 10-4
Q p =
=
Q p = 1.4112 x 10-4 K p ≠Q p, K p > Q p
The system is not at equilibrium, therefore the reaction proceeds from left to right
EQUILIBRIUM CONSTANT WITH DEGREE OF DISSOCIATION ()
-It gives to what extent the reactants are converted to products by dissociation.
-This has to be treated similar to moles.
Example1. 0.01 moles of PCl5was placed in 1L vessel at 210K.It was found to be 52.6% dissociated into PCl3 and Cl2. Calculate the K c at that temperature.
PCl5 (g) PCl3 (g) + Cl2 (g)
At start: 0.01 0 0
At equilibrium: 0.01-(5.26 x10-3) 5.26 x 10-3 5.26×10-3
=
= 5.26 x
PCl5 = 4.74 x 10-3 moles
PCl3 = 5.26 x 10-3 moles
Cl2 = 5.26 x 10-3 moles
K C =
=
K C = 5.837 x 10-3 moles
2. At 1 atm and 85oC, N2O4 is 50% dissociated, Calculate the equilibrium constant in terms of pressure and calculate the degree of dissociation of the gas at 100 c and 550c.
N2O4 2NO2 (g)
Start 1 0
Equilibrium: 1 2
But = 50% = 0.5
1-0.5 2(0.5)
0.5 1
nT = 0.5 + 1
= 1.5 moles
= x 1 = x 1
= 0.33 moles = 0.66moles
K p =
=
K p = 1.32
N2O4 2NO2 (g)
Start: 1 0
Equilibrium: 1- 2
nT = (1- ) + 2
=1+
PT = 10atm
= x 10 = x 10
K p =
=
= x 100 x x
=
1.32 =
1.32 – 1.322 = 402
1.32 = 402 + 1.322
1.32 = 41.322
2 =0.0319
= 0.1787
Degree of dissociation = 17.87%
DEGREE OF DISSOCIATION BY DENSITY MEASUREMENT
This method is used for the determination of degree of dissociation of gases in which 1 molecule produces 2 or more molecules.
i.e. PCl5 (g) PCl3 (g) + Cl2 (g)
Thus at constant temperature and pressure, the volume increases. The density at constant pressure decreases.
The degree of dissociation can be calculated from the difference in density between the undissociated gas and that of partially dissociated gas at equilibrium.
If we start with 1 mole of the gas (PCl5) and the degree of dissociation ( ), Then
PCl5 (g) PCl3 (g) + Cl2 (g)
Moles at equilibrium: 1-
Total moles: 1 –
= 1+
Note:
The density of an ideal gas at constant temperature and pressure is inversely proportional to the number of moles for a given weight.
Hence the ratio of density
=
When = Density of gas mixture of equilibrium
= Density of gas before dissociation
Degree of dissociation
Examples1. When PCl5 is heated, it gasifies and dissociates into PCl3 and Cl2.The density of the gas mixture at 2000C is 70.2 Find the degree of dissociation of PCl5 at 2000C
Solution
Observed density, ρ2 =70.2
ρ1 =?
V.D =
=
=104.25
=
= 48.5%
2. At 900C, the V.D of N2O4 is 24.8. Calculate the % dissociation into NO2 molecules at this temperature.
N2O4 NO2 + NO2
Density at equilibrium, ρ2 = 24.8
= V.D = = = 46
?
=
=
= 85.48%
DETERMINATION OF DEGREE OF DISSOCIATION BY MOLECULAR MASS
Molecular masses are proportional at constant temperature and pressure to the density of their gases; therefore we can substitute the molecular masses for the density in the degree of dissociation.
Where:-
= Molecular mass of undissociated gas
= Average Molecular mass of gases at equilibrium
Example1:1.588g of N2O4 gives a total pressure of 1atm when partially dissociated at equilibrium in a 500cm3 glass vessel at 250C.What is the degree of dissociation at this temperature?
N2O4 2NO2
M1 = (14x 2) + (16 x4)
= 92gmol-1
M2 =?
From PV= nRT, W here n =
M2 =
=
= 77.7gmol-1
=
= 0.184
FACTORS AFFECTING EQUILIBRIUM REACTION
These factors are as follows:-
i) Temperature
ii) Concentration
iii) Pressure
The first three affects both rates and position of chemical equilibrium (i, ii and iii)
The other three affects the rate of chemical equilibrium
1) Temperature
a) Increasing the temperature, increases the rate of reaction because usually at high temperature the collision factor increases also the number

of molecules having necessary activation energy is large.
b) Effect on the position of equilibrium is explained by using Le-Chateliers principle which states that “when a system at equilibrium is

subjected to a change, processes occur which tend to counteract the change” (If a system in equilibrium is disturbed (change in temperature and pressure) the system adjusts itself so as to oppose the disturbance).
Consider the reaction
2SO2 (g) + O2 (g) 2SO3 (g) + Heat (negative)
If temperature is increased in the system, the equilibrium moves in a direction where there is a absorption of heat and if the temperature is decreased in the system, the equilibrium moves in a direction where there is release of heat.
Effect of temperature, on the position of equilibrium can be explored by Vant Hoff`s law of mobile chemical equilibrium which states that

“For any system in equilibrium high temperature favours endothermic reactions and low temperature favours exothermic reactions.
The way in which equilibrium constant changes with temperature is found both theoretically and experimentally governed by the following

relationship;
=
∆ Hm = change in molar heat
K = Equilibrium constant
On intergrating the equation above;
Where c = constant
If K1 and K2 are equilibrium constants corresponding to T1 and T2 ,the constant term can be eliminated from the equation above so as to give Vant Hoff`s equation i.e.
lnK1 = …………………. (1)
lnK2 = …………………. (2)
Subtracting equation (1) from (2) i.e.
lnK2 – lnK1 =))
= ……………………1
But 1 ln = 2.303log
= ……………………2
But =
= ……………………3
Where 1 = 2 = 3
Example1: For the reaction;
N2O4 2NO2 ∆H = 61.5KJ mol-1
KP = 0.113 at 298K
i) What is the value of KP at 00C?
ii) At what temperature will KP =1?
Answers:
i) From =
KP2 = 0.113
KP1 =?
T 1 = 273K
T 2 = 298K
∆Hm = 61.5Kjmol-1
R=8.314
=
= 3211.9676 (3.0729896 x 10-4)
= 0.987
To remove ‘log’ make 100.987
ii) From log =
KP2 = 0.113 T1 =?
KP1 = 1 T2=298K
∆Hm = 61.5KJ mol-1
R = 8.314
=
-0.9469 = 3211.9676
-2.948 x 10-4 =
-0.0878T 1 = 298 – T 1
-0.0878T 1 + T1 =298
0.912T1 = 298
T1 = 326.7k
VARIATION OF EQUILIBRIUM CONSTANT WITH TEMPERATURE
According to Vant Hoff`s law of chemical equilibrium
i) Exothermic reaction
-Increasing to temperature decreases the KC value as the equilibrium shifts to the left hand side to decrease the concentration of products and increase the concentration of reactants.
-Decreasing the temperature increase the KC value
ii) Endothermic reaction
-Increasing the temperature increases the KC value
-Decrease the temperature decreases the KC value
2) Concentration
a) Effects on rate of reaction
-Increase in concentration of reactants in the same amount of space/volume increases the number of molecules thus increases the chances

of collision between the molecules and hence this increases the rate of reaction.
-Decreases in concentration result to decrease in rate of reaction as it decreases the number of molecules per unit volume, hence less

collision.
b) Effect on the position of equilibrium
-This depends on either the concentration of products or reactants have been increased or decreased
-It also depends on Le-Chateliers principle
Example
1. Consider, 2SO2 (g) + O2 (g) 2SO3 (g)
State what happens when:
i) [SO2] is increased at the same temperature
ii) [O2] is increased at same temperature
iii) [SO3] is increased at same temperature
According to Le-Chateliers principle the system will adjust itself so as to cancel out the effect by shifting the equilibrium to the right side i.e.

increase the number of products hence decreasing the amount of SO2
2. Using;
CH3COOH (g) + CH3CH2OH (g) CH3COOCH2 (g) + H2O (l)
What happens when;
i) H2O is added
ii) CH3CH2OH is added
iii) NaOH is added
iv) Anhydrous copper (II) sulphate is added
v) CH3COOCH2 is added
Answers:
i) H2O will react with CH3COOH and concentration of CH3COOH will decrease hence forward reaction, equilibrium will lie on the products side.
ii) Forward reaction since it will get on converted to products
iv) Reverse reaction since H2O will react with CUSO4 to form CUSO4.5H2O
3. The following reaction occurs in human body
Hb(s) + O2 (g) HbO2 (g)
i) What happens in the tissues?
ii) What happens in the lung?
Answers:
i) In the tissue the amount of O2 is less and hence according to Le-Chateliers principle the system will adjust itself so as to increase the amount of O2 by favouring the decomposition of HbO2 (backward reaction). Hence the equilibrium shifts to left hand side.
ii) In the lung, the amount of O2 is a lot more and hence according to Le-Chateliers principle the system will adjust itself so as to decrease the amount of O2 by favouring the forward reaction hence, the equilibrium shifts to the right hand side.
4. Consider the following equilibrium

Orange yellow
What would you expect to see it:
i) Dilute NaOH is added to the equilibrium mixture
ii) Dilute HCl is added to the equilibrium mixture
Answers:
i) When Dilute NaOH is added to the mixture, it will react with H+ to form H2O and Na+. This will decrease the concentration of H+ ions

hence equilibrium will shift to left hand side. The colour will change yellow to orange.
ii) Dilute HCl dissociated to form H+ and Cl therefore when added to the mixture, the concentration of H + ions increases. According to

Le Chateliers principle the equilibrium will adjust itself in such a way that the concentration of H+ ions decreases hence backward

reaction is favoured
3) Pressure
a) Effect on the rate of reaction
Increase in pressure, increases the rate of reaction. This is because increases in pressure decreases the concentration per unit volume

hence increases effective collision thus increase the rate of reaction, vice versa is also true.
b) Effect on position of equilibrium
Effect if number of moles of reactants and products are differ
Homogeneous gaseous equilibrium
Consider the reaction
N2O4 2NO2
What will be the effect of equilibrium, when;
i) Pressure is increased?
ii) Pressure decreased?
Answers:
i) When the pressure is increased according to Le Chateliers principles the equilibrium will adjust itself in such a way that the backward

reaction is favoured
ii) According to Le Chateliers principle when pressure is decreased the equilibrium will adjust itself in such a way that the forward

reaction is favoured
N.B:
Pressure has no effect on position of equilibrium if the number of moles in reactants side is equal to number of moles in products side.
Heterogeneous equilibrium
Position of equilibrium is affected by changing the partial pressure of the gases only
E.g.
If partial pressure of CO2 is decreased, the equilibrium shifts to the right hand side to increase the partial pressure of CO2. Vice versa is true.

4) Catalyst
The catalyst speeds up the rate of both forward and backward reaction to the some extent since it powers the activation energy of the reaction.
Therefore the equilibrium is not altered when catalyst is added. It only changes the rates at which the reaction approaches the equilibrium.

5) Physical state and sunlight
They do not affect position of equilibrium, only on rate of reaction.
Assignment
Read and write on the important industrial applications of chemical equilibrium (specifically on harber process and contact process.)

MANUFACTURE OF AMMONIA (HARBER PROCESS)
Harber’s process for the manufacture of ammonia involve direct combination (synthesis) of nitrogen and hydrogen
1mol 3mol 2mol
1vol 3vol 2vol
This reaction is;
i) Reversible
ii) Exothermic
iii) Proceeds with change in volume
According to Le Chateliers principle, the favourable conditions for formation of ammonia are:
The temperature should be kept as low as possible but at very low temperature, the rate of reaction becomes slow. It has been found that the yield of NH3 is maximum at about 5000C which is the optimum temperature of the reaction.
b) High pressure
High pressure favours reaction which is accompanied by a decrease in volume. In actual practice, a pressure of 200-900atm is employed in this process.
c) Catalyst
To increase the speed of the reaction, a catalyst should be finely divided iron containing molybdenum or alumina is used as a catalyst. Molybdenum or alumina (Al2O3) acts as a promoter and increases the efficiency of the catalyst. A mixture of iron oxide and potassium aluminate has been found to work more effectively. A catalyst iron oxide containing Al2O5 and K2O is also used in the process.
A diagram to show the manufacture of ammonia by Born Harber process
MANUFACTURE OF SULPHURIC ACID BY CONTACT PROCESS
In contact process sulphur dioxide is oxidized by air in the presence of catalyst Vanadium pentaoxide. Sulphur trioxide produced is absorbed in concentration H2SO4 to produce oleum (H2S2O7). Oleum is then reacted with calculated amount of H2O to form H2SO4 of the desired concentration.
The chemistry involved in the contact process is described as follows:

i) Production of So2
Sulphur dioxide ( is obtained by burning sulphur or iron pyrites
ii) Catalytic oxidation of SO2 to SO3
SO2 is oxidized by air in the presence of a catalyst to give SO3

The reaction is exothermic and proceeds with a decrease in volume. Therefore according to Le Chateliers principle, the favourable conditions for the maximum yield of SO3 are:-
a) Air or oxygen required for oxidation of SO2 must be in excess
b) The temperature must be low, a temperature between 350 – 4500C gives maximum yield of the products.
c) The pressure must be high, 2atm.
d) Platinised asbestos was used as a catalyst previously, but now days it is replaced by much cheaper Vanadium pentaoxide (V2O5).

A V2O5 remains unaffected the impurities while platinised arbestos is poised by the impurities in the gases.
e) The gases used (SO2 and O2) must be free of impurities, viz, dust particles, arsenious oxide e.t.c, to prevent catalyst poisoning
iii) Conversion of So3 to Oleum
SO3 is dissolved in concentratedH2SO4 to produce oleum or fumingH2SO4.
iv) Conversion of oleum to sulphuric acid
Oleum is diluted with a calculated amount of H2O to get H2SO4 of desired concentration.
The great advantage of the contact process is that it produces a pure acid of any desired concentration and especially the fuming acid which is

of great value in chemical industry.


THE CONTACT PROCESS


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