calculate the value of kp and kc when pcl5 is dissociated in 47.1% at 18°c at 1atm pressure
Answers
Answer:
PCl
5
⇌PCl
3
+Cl
2
1 0 0 Moles before dissociation
(1−α) α α Moles after dissociation
Given α=0.2 at 1 atm pressure
K
P
=
n
PCl
5
n
PCl
3
×n
Cl
2
×[
∑n
P
]
△n
=
(1−α)
α.α
[
1+α
P
]=
1−α
2
Pα
2
=
1−(0.2)
2
1×(0.2)
2
K
P
=0.0416atm
Again when α is desired at 0.5, K
P
remains constant and thus,
K
P
=
1−α
2
Pα
2
0.0416=
1−(0.5)
2
P×(0.5)
2
∴P=0.1248atm
Answer:
PCl5 is 47.1% dissociated at 18°C and one atmospheric pressure. The value of Kp is 0.253
The equilibrium constant Kp is equal to the partial pressures of the products divided by the partial products of the reactants and the partial products are to be raised by the coefficient of that substance in the balanced reaction form.
So, Kp = (Partial pressure of product)^(coefficient in balanced equation) / (Partial pressure of reactant)^(coefficient in balanced equation)
The reaction when PCl5 is dissociated is:
PCl5 ⇄ PCl3 + Cl2
Let degree of dissociation be 'α'.
PCl5 ⇄ PCl3 + Cl2
Initially (at t=0): 1 0 0
At equilibrium: 1-α α α moles
∴ Total moles in the equilibrium mixture = 1-α + α + α = 1+α
Kp = [(p-PCl3)^2 X (p-Cl2)^1] / [(p-PCl5)^2]
Partial pressure of a substance = Mole fraction of that substance X Pressure
So, partial pressure of PCl3 = (α/1+α) X P [α/1+α is the mole fraction of PCl3 at equilibrium as number of moles is α divided by total number of moles 1+α and P is the pressure = 1 atm]. = (α/1+α)
Similarly, partial pressure of Cl2 = (α/1+α) X P = (α/1+α)
partial pressure of PCl5 = (1-α/1+α) X P = (1-α/1+α)
In the question, α = 0.471, So,
partial pressure of PCl3 = (0.471/1+0.471) = 0.32
partial pressure of Cl2 = (0.471/1+0.471) = 0.32
partial pressure of PCl5 = (1-0.471 / 1+0.471) = 0.36
Putting these values in equation (1), we get,
Kp = (0.32)^2 X (0.32)^1 / (0.36)^2
= 512/2025 = 0.253
So, value of Kp is 0.253.
Explanation: