At 800 ° C , the following equilibrium is established as F _ 2 (g)⇌...

According to Graham's law of effusion
$r ∝ \dfrac{1}{M}$

$\dfrac{{r}_{mix}}{{r}_{\text{SO} 2}} = \sqrt{\dfrac{{M}_{\text{SO} 2}}{{M}_{mix}}}$

$⇒ \text{2.56} = \dfrac{6 4}{{M}_{mix}}$

$⇒ {M}_{mixture} = 2 5$

Let mole fraction of ${F}_{2}$ is x. Therefore, mole fraction of $F$ will be \left(1 - x\right)$<br /><br />$2 5 = \dfrac{3 8 \times \text{x} + \left(1 - \text{x}\right) \times 1 9}{1}$<br /><br />$\text{x} = 0.315$;<br /><br />$⇒ {K}_{p} = \dfrac{{P}_F^{2}}{{P}_{\text{F} 2}} = \dfrac{{\left(\text{0.685P}\right)}^{2}}{\left(\text{0.315P}\right)} ≃ \text{1.49 atm}