Question

A quantity of $PC{l_5}$ was heated in a 2L vessel at $525K$. It dissociates as:
$PC{l_5}\left( g \right) \rightleftharpoons PC{l_3}\left( g \right) + C{l_2}\left( g \right)$
At equilibrium 0.2mol, each of $PC{l_5}$, $PC{l_3}$ and $C{l_2}$ is found in the reaction mixture. The equilibrium constant ${K_c}$ for the reaction is
A. 0.2
B. 0.5
C. 0.1
D. 0.05

Answer 1

Hint: We know that in a chemical reaction, chemical equilibrium is the state where the rates of forward reaction and the reverse reaction are the same. The result is that the concentration of the reactants and the products remains unchanged. The rate constant of the forward reaction divided by the rate constant of the reverse reaction gives the equilibrium constant of a reaction. We can calculate the value of equilibrium constant by multiplying the concentration of the products and dividing it by the concentration of reactants. We can calculate the concentration of the reactants and products, their moles and volume.

Complete step by step answer:
Given data contains,
Mass of $PC{l_5}$ is 0.2mol.
Mass of $PC{l_3}$ is 0.2mol.
Mass of $C{l_2}$ is 0.2mol.
Volume is 2L.
We must remember that the given equation is the decomposition reaction of phosphorus pentachloride. It decomposes into phosphorus trichloride and chlorine. The balanced equilibrium reaction is,
$PC{l_5}\left( g \right) \rightleftharpoons PC{l_3}\left( g \right) + C{l_2}\left( g \right)$
We can give the expression for equilibrium constant as,
${{\text{K}}_{\text{c}}}{\text{ = }}\dfrac{{{\text{Products}}}}{{{\text{Reactants}}}}$
${K_c} = \dfrac{{\left[ {PC{l_3}} \right]\left[ {C{l_2}} \right]}}{{\left[ {PC{l_5}} \right]}}$
Let us now substitute the values of moles and volume in the expression for the equilibrium constant.
$ \Rightarrow {K_c} = \dfrac{{\left[ {\dfrac{{0.2}}{2}} \right]\left[ {\dfrac{{0.2}}{2}} \right]}}{{\left[ {\dfrac{{0.2}}{2}} \right]}}$
$ \Rightarrow {K_c} = \dfrac{{\left[ {0.1} \right]\left[ {0.1} \right]}}{{\left[ {0.1} \right]}}$
On simplifying we get,
$ \Rightarrow {K_c} = 0.1$
The value of equilibrium constant ${K_c}$ for the reaction is calculated as $0.1$.

So, the correct answer is Option C .

Note:
We can determine the composition of the system at equilibrium using the initial composition of the system and known values of equilibrium constant.
The reciprocal of forward reaction gives the equilibrium expression of the reverse reaction.
The different types of equilibrium constants are stability constants, binding constants, formation constants, dissociation constants, and association constants.
Factors such as temperature, ionic strength, and solvents affect the value of equilibrium constant.