Question

In the reaction A gives B+2C equilibrium is established by heating 4 moles of A in a closed container at 2 atm pressure. If degree of dissociation of A is 0.25 then Kp is? Please answer if you know....

Answer 1

Answer:

$K_p= \frac{1}{3}   atm^2$

Explanation:

For a reaction

$aA + bB$ ⇄ $cC + dD$

Kp is defined as the ratio of  Partial pressure of the product raised to their stoichiometric coefficient and partial pressure of reactant raised to the stoichiometric coefficient of the respective reactant.

$K_p= \frac{(P_C)^c*(P_D)^d}{(P_A)^a*(P_B)^b}$

A ⇄ B +2C

initially we have 4 moles of A and and pressure of 2 atm,

degree of dissociation is given 0.25

After equilibration number of moles of A decomposed and B and C formed are

                                           A                   ⇄   B       +     2C              Total

moles after equilibrium   4-4*0.25               4*0.25   2*4*0.25       6

Lets say P is reduced

and P is formed

                                          2-P                     P               2P                 2+2P

so partial pressure of A is 2-P

and mole fraction of A is 3/6=1/2

We know mole fraction of A * Total pressure = Partial Pressure of A

1/2*(2+2P)= 2-P

we find P is 0.5

So partial pressure of A is 1.5 atm, B is 0.5 and pressure of C is 1.0 Atm

Putting the values in$K_p = \frac{0.5^1*1^2}{1.5} = 1/3 atm^2$ the Kp equation.