1. Apply Le Chatelier's principle for the manufacture of SO3 by contact process and find the conditions for getting maximum yield of SO3.
Contact process involves the equilibrium reaction of oxidation of SO2 gas by gaseous oxygen in air to manufacture large quantities of SO3 gas.
V2O5
2SO2(g) + O2(g) ⇌ 2SO3(g) ΔH0f = -47 kcal / mole
Effect of Pressure:The formation reaction of SO3 involves a decrease in the overall moles of the reactants. According Le Chatlier’s principle, when large pressure is applied, forward reaction is favoured. 700 atm - 1200 atm pressure is maintained on the 2 : 1 mole ratio mixture of pure SO2 and O2 gases in the reaction chamber.
Effect of Temperature:
SO3 production is an exothermic reaction. Hence, increase in temperature favours SO3 dissociation. However, lowering of temperature prolongs the time of attainment of equilibrium. Therefore, an optimum temperature at nearly 400°C to 450°C is maintained.
Effect of Catalyst:
The most widely used catalyst for SO3 production is porous vanadium pentoxide (V2O5). Presence of moisture deactivates the catalyst. Only dry and pure SO2 and O2 gases are used over the catalyst. Since oxidation of SO2 is a slow process, presence of V2O5 speeds up the equilibrium process and high yield of SO3 is achieved in a short period.
Effect of Concentration:
SO3 is the anhydride of H2SO4. Therefore, SO3 from contact process along with steam is used in oleum and H2SO4 manufacturing processes.
The yield of SO3 is nearly 97%.
2. Apply Le Chatelier's principle to Haber's process for the manufacture of Ammonia.
Ammonia formation reaction is an equilibrium reaction.
Fe
N2(g) + 3H2(g) ⇌ 2NH3(g) ΔH0f = –22.0 kcal / mole
Effect of Pressure:
The forward reaction is accompanied by decrease in the number of moles of reactants. According to Le Chatlier’s principle, an increase in pressure favours such a reaction and shifts the equilibrium towards the product formation direction. Therefore, nearly 300-500 atm pressure is applied on 3 : 1 mole ratio of H2 : N2 gas mixture in the reaction chamber for maximum yield of ammonia.
Effect of Temperature:
The ammonia formation reaction is an exothermic reaction. According Le Chatlier’s principle, increase in temperature favours decomposition reaction of ammonia. However, at low temperature the time to reach the equilibrium becomes very long. Hence an optimum temperature close to 500°C-550°C is maintained.
Effect of Catalyst:
Iron catalyst is chosen to speed up the attainment of the equilibrium concentration of ammonia.
Effect of Concentration:
In order to maintain the equilibrium conditions, steam is passed to remove away the ammonia as and when it is formed so that the equilibrium remains shifted towards the product side.
The maximum yield of ammonia is nearly 37%.
3. Derive the expression for Kc and Kp for formation of HI.
The formation of HI from H2 and I2 is an example of gaseous homogeneous equilibrium reaction. It can be represented as
H2(g) + I2(g) ⇌ 2HI(g)
In this equilibrium reaction, the number of moles of the products is equal to the number of moles of the reactants.
Δng = np – nr
= 2 – (1 + 1) = 2 – 2 = 0
∴ Kp = Kc
Derivation of Kc in terms of x:
Let us assume ‘a’ and ‘b’ moles of H2 and I2 gases being present in ‘V’ litres of the reaction vessel. At equilibrium, let x moles each of H2 and I2 react to form 2x moles of HI. Then, the equilibrium concentrations in moles litre of H2, I2 and HI in the reaction mixture will be (a - x) / V, (b - x) / V and 2x / V respectively.
H2(g)
|
I2(g)
|
HI(g)
| |
Initial number of moles
|
a
|
b
|
0
|
Number of moles reacted
|
x
|
x
|
-
|
Number of moles remaining at equilibrium
|
a - x
|
b - x
|
2x
|
Equilibrium concentration
|
(a - x) / V
|
(b - x) / V
|
2x / V
|
According to the law of mass action,
Kc = [HI]2 / [H2] [I2]
= (2x / V)2 / ((a - x ) / V . (b - x) / V)
= (4x2 / V2) x (V2 / (a - x) (b - x))
Kc = 4x2 / (a - x) . (b - x) = Kp
x is also known as the extent of reaction.
Derivation of Kp in terms of x:
Since Δng = 0, Kp = Kc = 4x2 / (a - x) . (b - x)
∴ Kp and Kc are equal in terms of x values.
4. Derive the expressions for Kc and Kp for decomposition of PCl5.
The dissociation equilibrium of PCl5 in gaseous state is written as
PCl5(g) ⇌ PCl3(g)+ Cl2(g)
For this reaction Δng = np – nr= 1∴ Kp = Kc (RT)
= (1 + 1) – 1 = 2 – 1
Derivation of Kc in terms of x:
Let ‘a’ moles of PCl5 vapour be present in ‘V’ litres initially. If x moles of PCl5 dissociate to PCl3 and Cl2 gases at equilibrium at constant ‘V’ litres, then molar concentrations of PCl5, PCl3 and Cl2 gases at equilibrium will be a - x / V , x / V and x / V respectively.
PCl5(g)
|
PCl3(g)
|
Cl2(g)
| |
Initial number of moles
|
a
|
0
|
0
|
Number of moles reacted
|
x
|
-
|
-
|
Number of moles remaining at equilibrium
|
a - x
|
x
|
x
|
Equilibrium concentration
|
(a - x) / V
|
x / V
|
x / V
|
According to law of mass action,
Kc = [PCl3] [Cl2] / [PCl5]
= (x / V) (x / V) / (a - x) / V
= (x2 / V2) x (V / (a - x))
Kc = x2 / (a - x)V
x is also known as twhich represents the fraction of total moles of reactant dissociated.
x = Number of moles dissociated / Total number of moles present initially
If initially 1 mole of PCl5 is present then
Kc = x2 / (1 - x)V = x2P / (1 - x)RT
If the degree of dissociation is small compared to unity, then (1 – x) is approximately equal to 1.0.
Kc = x2 / V
Or
x2 = K . V
x a √V
But V a 1 / P
x a √1 / P
Where x is small, degree of dissociation varies inversely as the square root of pressure (or) varies directly as the square root of volume of the system.
Derivation of Kp in terms of x:
PCl5(g)
|
PCl3(g)
|
Cl2(g)
| |
Initial number of moles
|
a
|
0
|
0
|
Number of moles reacted
|
x
|
-
|
-
|
Number of moles remaining at equilibrium
|
a - x
|
x
|
x
|
Equilibrium concentration
|
(a - x) / V
|
x / V
|
x / V
|
∴ Total number of moles at equilibrium
|
= 1 - x + x + x = (1 + x)
| ||
We know that,
Partial pressure is the product of mole fraction and the total pressure.
Mole fraction is the number of moles of that component divided by the total number of moles in the mixture.
Therefore
pPCl5 = (1 - x / 1 + x) . P
pPCl3= (x / 1 + x) . P
pCl2 = (x / 1 + x) . P
In terms of partial pressures of PCl5, PCl3 and Cl2 then Kp = pPCl3. pCl2 / pPCl5
Substituting the values of partial pressures in the above equation, we get
Kp = (x / 1+x)P . (x / 1 + x)P / (1-x / 1+x)P
= x2 . P2 . (1+x)2 x (1+ x / 1 - x) x 1 / P
Kp = x2P / (1 - x2)
When x < < 1, x2 value can be neglected when compared to one i.e., (1 - x2) = 1.
∴ Kp ≈ x2P
5. Derive the relation Kp = Kc(RT)Δng for a general chemical equilibrium reaction. Or Derive a relation between (equilibrium constant) Kp and Kc.
Consider a general chemical equilibrium reaction in which the reactants and products are in gaseous phases,
aA + bB + cC + .... ⇌ lL + mM + nN + ...
Then,
Kp = pLl pMm pNn ... / pAa pBb pCc ...
Where,
p is the partial pressure of the respective gases.
In terms of molar concentrations of reactants and products
Kc = [L]l [M]m [N]n ... / [A]a [B]b [C]c ...
For any gaseous component ‘i’ in a mixture, its partial pressure ‘pi’ is related to its molar concentration ‘Ci’ as Ci = pi / RT
Since
pi = (ni / V) RT
Where,
(ni / V) = Ci = Number of moles of i per litre. V = Volume in litres.
Substituting concentration terms by partial pressures,
Kc = (pL/RT)l (pM/RT)m (pN/RT)n / (pA/RT)a (pB/RT)b (pC/RT)c
= pLl x pMm x pNn / pAa x pBb x pCc X (1/RT)(l + m + n + …) – (a + b + c + …)
Kc = Kp / (RT)Δng
Or
Where,
Δng = (l + m + n + …) – (a + b + c + …)
Δng = Total number of stoichiometric moles of gaseous products, np – (Minus)
Total number of stoichiometric moles of gaseous reactants, nr.
6. Discuss the effect of temperature and pressure on the following equilibrium:
Effect of change of temperature
A chemical equilibrium actually involves two opposing reactions. One favouring the formation of products and the other favouring the formation of reactants.
If the forward reaction in a chemical equilibrium is endothermic (accompanied by absorption of heat) then the reverse reaction is exothermic (accompanied by evolution of heat).
Let us consider the example
N2O4(g) ⇌ 2NO2(g) ; ΔH = +59.0 kJ / mole
In this equilibrium, the reaction of the product formation (NO2) is endothermic in nature and therefore, the reverse reaction of reactant formation (N2O4) should be exothermic.
If the above equilibrium reaction mixture is heated then its temperature will be raised. According to Le Chatelier’s principle, the equilibrium will shift in the direction which tends to undo the effect of heat. Therefore, the equilibrium will shift towards the formation of NO2 and subsequently dissociation of N2O4 increases. Therefore, generally, when the temperature is raised in a chemical equilibrium, among the forward and reverse reactions, the more endothermic reaction will be favoured.
Similarly, if the temperature of the equilibrium is decreased i.e., cooled, then the exothermic reaction among the forward and reverse reaction of the equilibrium will be favoured.
Effect of change of pressure
If a system in equilibrium consists of reactants and products in gaseous state, then the concentrations of all components can be altered by changing the total pressure of the system. Consider the equilibrium in the gaseous state such as
N2O4(g) ⇌ 2NO2(g)
Increase in the total pressure of the system in equilibrium will decrease the volume proportionately. According to Le Chatlier’s principle, the change can be counteracted by shifting the equilibrium towards decreasing the moles of products. Hence, the reaction of combination of NO2 molecules to N2O4 formation will be favoured.
7. The dissociation equilibrium constant of HI is 2.06 x 10 –2 at 458 K. At equilibrium the concentrations of HI and I2 are 0.36 M and 0.15M respectively. What is the equilibrium concentration of H2 at 458 K?
Dissociation equilibrium of HI, 2HI(g) ⇌ H2(g) + I2(g)
Dissociation equilibrium constant of HI, Kc = 2.06 x 10 –2
Equilibrium concentration of I2, [I2] = 0.15M
Equilibrium concentration of HI, [HI] = 0.36 M
Equilibrium concentration of H2, [H2] = ?
Kc = [H2] [I2] / [HI]2
2.06 x 10 –2 = [H2].0.15 / (0.36)2
[H2] = 2.06 x 10 –2 x (0.36)2 / 0.15
Equilibrium concentration of H2, [H2] = 1.78 x 10-2 M
8. Apply Le Chatelier’s principle for the following reaction N2(g) + O2(g) ⇌ 2NO(g) and discuss the effect of pressure and concentration on it.
ONE MARKS THREE MARKS8. Apply Le Chatelier’s principle for the following reaction N2(g) + O2(g) ⇌ 2NO(g) and discuss the effect of pressure and concentration on it.
Effect of
pressure
If a system in
equilibrium consists of reactants and products in gaseous state, then the
concentrations of all components can be altered by changing the total pressure
of the system. Consider the equilibrium in the gaseous state such as N2(g)
+ O2(g) ⇌
2NO(g)
Increase in the
total pressure of the system in equilibrium will decrease the volume
proportionately. According to Le Chatlier’s principle, the change can be
counteracted by shifting the equilibrium towards decreasing the moles of
products.
In this equilibrium reaction, the number of moles of the products is
equal to the number of moles of the reactants.
Pressure
has therefore no effect on the equilibrium
Effect of
concentration
At the
equilibrium conditions the reaction mixture contains both the reactant and
product molecules, that is, N2, O2 and NO molecules. The concentrations of
reactant and product molecules are constant and remain the same as long as the
equilibrium conditions are maintained the same.
If a change is
imposed on the system by purposely adding
NO into the reaction mixture then the product concentration is raised.
Since the system possesses equilibrium concentrations of reactants and
products, the excess amount of NO react in the reverse direction to produce back the reactants and this results
in the increase in concentrations of N2 and O2.
Similarly if the
concentration of reactants such as N2 and O2 are purposely raised when the system is
already in the state of equilibrium, the excess concentrations of N2 and O2 favour
forward reaction. Concentration of
NO is raised in the reaction mixture.
In
general, in a chemical equilibrium increasing the concentrations of the reactants results in shifting the equilibrium in favour of the
products
while increasing
the concentrations of the products results in shifting the equilibrium in favour of the reactants.
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