Question

a)The numerical value of d/D when α=0 for $PC{l_5} \rightleftharpoons PC{l_3} + C{l_2}$ ; where density of equilibrium mixture and D is vapour density of $PC{l_5}$
b) The value of ${K_c}$ , when $\Delta {G^ \circ } = 0$ for a reaction in equilibrium is:

Answer 1

Hint:We know that the degree of dissociation is given by α and one can find density measurement from the degree of association. The following method is only valid for the reactions having at least one gas and having no solution.

Complete answer:
a)The above reaction of dissociation of phosphorus pentachloride is one such reaction for which we can derive the relation between density measurement and degree of dissociation where we know that d is the vapour density at equilibrium, D is the initial vapour density of $PC{l_5}$ . vapour density is inversely proportional to the moles. Therefore for this reaction if initially we have C moles of $PC{l_5}$ , then at equilibrium $PC{l_5}$ moles will decrease to C(1-α) and the moles of $PC{l_3}$ and $C{l_2}$ formed respectively will be Cα each. Therefore
$\Rightarrow \dfrac{D}{d} = \dfrac{{C(1 - \alpha )}}{C} = 1 - \alpha $
In the question we have been given α=0 therefore substituting we get,
$\Rightarrow \dfrac{D}{d} = 1 - 0 = 1$
Hence similarly 1 is the numerical value for $\dfrac{d}{D}$.
b) ${K_c}$ is the value of equilibrium constant, it is a function of stoichiometric coefficient used for any balanced chemical equation like the above dissociation of $PC{l_5}$
according to the question we have been given gibbs energy as zero we know that it means that the reaction is at equilibrium and we have the relation of equilibrium constant and gibbs energy as
$ - \Delta {G^ \circ } = 2.303RT\log {K_c}$
And we have
$
  \Delta {G^ \circ } = 0 \\
\Rightarrow - 0 = 2.303RT\log {K_c} \\
\Rightarrow \log {K_c} = 0 \\
\Rightarrow {K_c} = 1
$

Note:
We already know that the concentration of both reactants and products remain constant at equilibrium which is dynamic in nature. Hence the rate of conversion from reactant to product or vice versa at equilibrium is equal just like the moles of $PC{l_3}$ and $C{l_2}$ in the above reaction.