Respuesta :
Kp is 0.55
Explanation:
A(g) + B(g) C(g), K = 3.53 - 1
2 A(g) + D(g) C(g), K = 6.98 -2
C(g) + D(g) 2 B(g), K = ? -3
To obtain the third reaction, multiply reaction 1 with 2 and then reverse and add with second reaction.
2A(g) + 2B(g) ⇒ 2C(g), K = 3.53
2 C(g) ⇒ 2A(g) + 2B(g), K = 6.98
2A(g) + D(g) ⇒ C(g)
C(g) + D(g) ⇒ 2B(g)
The equilibrium constant follow the same changes as the reaction occur.
Therefore, Kreverse = 1/(K)²
Kreverse = 1/(3.53)²
= 0.08
The equilibrium constant for the second reaction is represented by K2
K = Kreverse X K2
K = 0.08 X 6.98
= 0.56
We know,
Kp = Kc (RT)Δⁿ
Kp = equilibrium constant in terms of partial pressure
Kc = equilibrium constant in terms of molar concentration
Δn = difference in number of moles of product to the number of moles of reactant
Kp = 0.55 X (RT)⁰
= 0.55 X 1
= 0.55
Therefore, Kp is 0.55
The mole fraction of B is 0.29
Explanation:
Given K= 3.53 at 45^o for the reaction A(g) + B(g)------ C(g)
k= 6.98 at 45^o for the reaction 2 A(g) + D(g)-----C(g)
to find out the value of K for the reaction C(g) + D(g)-------- 2 B(g)
for first equation
K=[C]/ [A] [B]
3.53=[A][B]/[C]
3.53C= [A][B]
for the second equation:
K= [C]/[A]^2[D]
D= 3.50 [A][B]/7.10[A]^2
putting the values in third reaction
K=[B]^2/[C][D]
= [B]^2/3.50 [A][B].[C]7.10[A]^2
= 0.58
Kp= (2x)^2/(1.70-x)^2
= 0. 70 atm
the equilibrium constant for the reverse reaction
k (reverse)=1/k^2
k(reverse)=1/3.5
=0.0186
The equilibrium constant of the second reaction is K2
for calculating the equilibrium constant of third reaction
K=Kreverse*K2
= 0.0186*7.10
kc = 0.58
The equation for molar concentration in terms of partial pressure is given as:
Kp=Kc [RT]^n
from the reaction 3 it is found that mole difference between reactant and product is zero.
when n=o
RT becomes 1
Kp= 0.58
from the third reaction mole fraction is 0.58/2
= 0.29
