Question

For the reaction AB--><--A + B is 33% dissociated at a total pressure P. Therefore, P is related to Kp by:
Kp=2P
Kp=4P
Kp=6P
Kp=8P

Answer 1

Answer : The relation between P and $K_p$ is, $P=8K_p$

Solution : Given,

Total pressure = P

Degree of dissociation = 30% = 0.33

The given equilibrium reaction is,

                            $AB\rightleftharpoons A+B$

Initially                  1         0     0

At equilibrium   $(1-\alpha)$     $\alpha$   $\alpha$

$\text{ Total number of moles}=1-\alpha+\alpha+\alpha=1+\alpha$

$\text{ Partial pressure of AB}=(\frac{1-\alpha}{1+\alpha})P$

$\text{ Partial pressure of A}=(\frac{\alpha}{1+\alpha})P$

$\text{ Partial pressure of B}=(\frac{\alpha}{1+\alpha})P$

The relation between the $K_p$ and the total pressure is,

$K_p=\frac{p_A\times p_B}{p_{AB}}$

Now put all the values of partial pressure, we get

$K_p=\frac{\alpha^2P}{(1-\alpha^2)}$

where,

$K_p$ = equilibrium constant

P = total pressure

$\alpha$ = degree of dissociation

Now put all the given values in the above formula, we get

$K_p=\frac{(0.33)^2P}{[1-(0.33)^2]}$

$K_p=0.122P$

$P=8.19K_p$

$P=8K_p$

Therefore, the relation between P and $K_p$ is, $P=8K_p$