$PC{l_5}(g) \rightleftharpoons PC{l_3}(g) + C{l_2}(g)$ is
Total pressure = 16.8 bar
Degree of dissociation, 
The given equilibrium reaction is,
Initially 1 0 0
At equilibrium 
Now we have to calculate the partial pressure of 
The expression of 
Now put all the values in this expression, we get
Therefore, the equilibrium constant for the reaction is, 3.2
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$2P(s) + 3C{l_2}(g)\,\, \to \,\,2PC{l_3}(l);\,\Delta H = - 151.8\,kcal$
$PC{l_3}(l) + C{l_2}(g)\,\, \to \,\,PC{l_5}(s);\,\,\Delta H = - 32.8\,kcal$
......$kcal$
(p_%7BCl_2%7D)%7D%7Bp_%7BPCl_5%7D%7D)
%7D%7B(1%2B%5Calpha)%7D%5Ctimes%20P)(%5Cfrac%7B(%5Calpha)%7D%7B(1%2B%5Calpha)%7D%5Ctimes%20P)%7D%7B%5Cfrac%7B(%5Calpha)%7D%7B(1%2B%5Calpha)%7D%5Ctimes%20P%7D)
