Given K = 3.53 at 45°C for the reaction A(g) + B(g) C(g) and K = 6.98 at 45°C for the reaction 2 A(g) + D(g) C(g) what is the value of K at the temperature for the following reaction? C(g) + D(g) 2 B(g) What is the value of Kp at 45°C for the same reaction? Starting with 1.70 atm partial pressures of both C and D, what is the mole fraction of B once equilibrium is reached?

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