Question

At {{227}^{o}}C,$$$$30% of $2$ moles of $PC{{l}_{5}}$ gets dissociated in a $2-litre$ container. The value of ${{K}_{p}}$will be
A.$64.28\text{ }R$
B.$400\text{ }R$
C.$50025\text{ }R$
D.$100\text{ }R$

Answer 1

We know that the reaction given is reversible reaction and hence equilibrium can be attained at one process, so we have to find equilibrium constant. For a given degree of dissociation; by creating an initial and by change in equilibrium table which we also know as ICE table, and by using this equation we can easily formulate expressions for equilibrium constant. For that we need to find a variable multiplied by R to which will be equal to equilibrium constant and solve a simple equation add.

Complete step-by-step answer: At ${{227}^{o}}C$, 30% of 2 moles of $PC{{l}_{5}}$ gets dissociated in a 2-litre container and the chemical reaction is given: $PC{{l}_{5}}(g)\rightleftharpoons PC{{l}_{3}}(g)+C{{l}_{2}}(g)$
The given reversible gaseous reaction and its ice table can be represented as:

| $PC{{l}_{5}}(g)$| $PC{{l}_{3}}(g)$| $C{{l}_{2}}(g)$
---|---|---|---
I| $2mol$| $0mol$| $0mol$
C| $-2\alpha mol$| $2\alpha mol$| $2\alpha mol$
E| $2(1-\alpha )mol$| $2\alpha mol$| $2\alpha mol$

Here we have degree of dissociation about $30$ by which we can easily determine the volume of equilibrium, $\alpha =30$
Also we know that the, volume of equilibrium mixture which is given as $V=2\text{ }L$
For $PC{{l}_{5}}$we have ice table value $2(1-\alpha )mol$from which we can easily divide volume of equilibrium and which is given by:
$[PC{{l}_{5}}(g)]=\dfrac{2(1-\alpha )}{V}=\dfrac{2(1-0.3)}{2}=0.7\dfrac{mol}{L}$
Similarly for $PC{{l}_{3}}$we have ice table value $2\alpha mol$from which we can easily divide volume of equilibrium and which is given by:
$[PC{{l}_{3}}(g)]=\dfrac{2(\alpha )}{V}=\dfrac{2(0.3)}{2}=0.3\text{ }\dfrac{mol}{L}$
Likewise for $C{{l}_{2}}$ we have ice table value $2\alpha mol$from which we can easily divide volume of equilibrium and which is given by:
$[C{{l}_{2}}(g)]=\dfrac{2(\alpha )}{V}=\dfrac{2(0.3)}{2}=0.3\text{ }\dfrac{mol}{L}$
Now, that we have acquired the values of $PC{{l}_{5}}(g)$ , $PC{{l}_{3}}(g)$ , $C{{l}_{2}}(g)$and by these values we can calculate the value of${{K}_{p}}$ which is given by, Product divided by reactant.
${{K}_{p}}=\dfrac{[PC{{l}_{3}}(g)][C{{l}_{2}}(g)]}{[PC{{l}_{5}}(g)]}$
$=\dfrac{0.3\times 0.3}{0.7}\dfrac{mol}{L}$
$\Rightarrow 0.128\dfrac{mol}{L}$
Also we have the value of R in atmospheric pressure which is in standard atmospheric pressure is $R\text{ }=\text{ }0.082057\dfrac{Latm}{Kmol}$
Hence we acquire the value which is given by $100R$

Therefore, the correct answer is D i.e. the value of ${{K}_{p}}$ will be $100R$.

Note: Note that the equilibrium constant of a chemical reaction is the value of its reaction quotient at chemical equilibrium, a state approached by a dynamic chemical system after sufficient time has elapsed at which its composition has no measurable tendency towards further change. For a given set of reaction conditions, the equilibrium constant is independent of the initial analytical concentrations of the reactant and product species in the mixture. Thus, given the initial composition of a system, known equilibrium constant values can be used to determine the composition of the system at equilibrium. However, reaction parameters like temperature, solvent, and ionic strength may all influence the value of the equilibrium constant.