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

since the atomic number is the number of protons in the nucleus, as we move to the left on the periodic table, the electrostatic pull between the nucleus and the valence electron shell decreases. as we move down a group (vertical column), the valence shell energy level gets further and further from the nucleus.