Question

On heating PCl5 at 523 K, it dissociated to PCl3 and Cl2 and Vl2 in 1.5 litre vessel. Vapour density of equilibrium mixture is 55. Calculate the degree of dissociation at 523 K and the value of Kc.


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Answer 1

Theoretical V.D. of PCl5 = 208.5/2

=104.25 and n=2

$\alpha = \frac{104.25 - 55}{(2 - 1)55} \\ = \frac{49.25}{55} = 0.8954 \: or \: 89.54\%$

Now. PCl5 <-----> PCl3+Cl2

1 -alpha/1.5. alpha/1.5. alpha/1.5

Kc= alpha/1.5×alpha/1.5/1-alpha/1.5

= alpha^2/1.5(1-alpha)

=0.8954× 0.8954/1.5×0.1046

=5.11 mole liyre^-1

Answer 2

Explanation:

$\boxed{Explained\:Answer}$

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