Question

If 8g mol of $PC{l_5}$ heated in a closed vessel of \[{\mathbf{10L}}\] capacity and \[{\mathbf{25}}\% \] dissociation into $PC{l_3}$ and $C{l_2}$ at the equilibrium then value of ${K_p}$ will be equal to
A) $\dfrac{P}{{30}}$
B) $\dfrac{P}{{15}}$
C) $\dfrac{2}{{3P}}$
D) $\dfrac{3}{{2P}}$

Answer 1

Hint: Degree of dissociation of a substance is the fraction of the total number of molecules dissociated into simpler molecules at a particular temperature.
$Degree\;of\;dissociation\;\left( \alpha \right) = \dfrac{{number\;of\;moles\;dissociated}}{{total\;number\;of\;moles\;taken}}$

Complete step-by-step solution:
Let us look into our reaction first
$PC{l_5} \rightleftharpoons PC{l_3} + C{l_2}$
By stoichiometry $1$ mole of $PC{l_5}$ gives $1$ mole of $PC{l_3}$ and $1$ mole of $C{l_2}$. It is given that \[25\% \] of $PC{l_5}$ dissociated to give $PC{l_3}$ and $C{l_2}$. Also, given that \[8g\] mol of $PC{l_5}$ is used.
When \[25\% \] of \[8g\] mol of $PC{l_5}$ is dissociated we get \[6g{\text{ }}mol\] of $PC{l_5}$ remaining.
Let us see how that is
$PC{l_5} \rightleftharpoons PC{l_3} + C{l_2}$

Initial mol	$8$	$0$	$0$
Moles after Dissociation	\[8 - \alpha \]	$\alpha $	$\alpha $

Here $\alpha $ is given in $\% $ so we can say that $8 \times \dfrac{{25}}{{100}} = 2$ so, $2\;mol$ is dissociated and we get $2\;mol$ of $PC{l_3}$ and $C{l_2}$. We can see now how the reaction looks like
$PC{l_5} \rightleftharpoons PC{l_3} + C{l_2}$.

Initial mol	$8$	$0$	$0$
Moles after Dissociation	\[8 - 2\]	$2$	$2$

i.e., \[6g{\text{ }}mol\] of $PC{l_5}$and $2\;mol$ of $PC{l_3}$ and $C{l_2}$ each.

In case of gaseous reaction, the equilibrium constant is calculated from partial pressure of gaseous reactants and products.
${K_p} = \dfrac{{{P_{PC{l_3}}} \times {P_{C{l_2}}}}}{{{P_{PC{l_5}}}}}$
We can also write this in terms of mole fraction
${{\text{P}}_{\text{A}}} = {P_T}{{\text{X}}_{\text{A}}}$,

Applying this to equation
${K_p} = \dfrac{{P{X_{PC{l_3}}} \times P{X_{C{l_2}}}}}{{P{X_{PC{l_5}}}}}$, where P be the total pressure
Where, X represents the mole fraction
Also the total moles at equilibrium is equal to $6 + 2 + 2 = 10$ (from the reaction after dissociation)
$\Rightarrow {X_{PC{l_3}}} = \dfrac{{number\;of\;moles\;of\;PC{l_3}}}{{total\;number\;of\;moles}} = \dfrac{2}{{10}}$
$\Rightarrow {X_{C{l_2}}} = \dfrac{{number\;of\;moles\;of\;C{l_2}}}{{total\;number\;of\;moles}} = \dfrac{2}{{10}}$
$\Rightarrow {X_{PC{l_5}}} = \dfrac{{number\;of\;moles\;of\;PC{l_5}}}{{total\;number\;of\;moles}} = \dfrac{6}{{10}}$
Applying the value to the equation we get,
$\Rightarrow {K_p} = \dfrac{{\dfrac{{2P}}{{10}} \times \dfrac{{2P}}{{10}}}}{{\dfrac{{6P}}{{10}}}} = \dfrac{{4P}}{{60}}$
$\Rightarrow {K_p} = \dfrac{P}{{15}}$

Hence the correct answer is option (B).

Note: We can also find the equilibrium constant in case of gaseous reaction using the equilibrium constant in terms of mole fraction,
For the reaction,
$aA + bB \rightleftharpoons xX + yY$
Then,
${K_x} = \dfrac{{\chi _X^x\;\chi _Y^y}}{{\chi _{A\;}^a\chi _B^a}}$
Where $\chi _X^x,\;\;\chi _Y^y,\;\chi _{A\;}^a,\;\chi _B^a$ are the mole fractions of X,Y, A and B respectively, ${K_x}$depends upon temperature as well as pressure and volume of the chemical system. It is found that
${K_p} = {K_x}{\left( P \right)^{\vartriangle {n_g}}}$
Where P is external pressure and $\vartriangle {n_g}$ is change in the number of gaseous moles.