Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

How many moles per litre of ${ PCl }_{ 5 }$ has to be taken to obtain 0.1 mole ${ Cl }_{ 2 }$, if the value of the equilibrium constant ${ K }_{ c }$ is 0.04?
(a) 0.15
(b) 0.25
(c) 0.35
(d) 0.05

seo-qna
Last updated date: 20th Jun 2024
Total views: 404.7k
Views today: 4.04k
Answer
VerifiedVerified
404.7k+ views
Hint: In order to solve this question we have to use the law of mass action which states that the equilibrium constant for a reaction can be written as the ratio of the product of the concentrations of the products to the concentration of the reactants such that each reactant and the product is raised to the power that is equal to their stoichiometric coefficients in the balanced chemical equation.

Complete step by step answer:
${ PCl }_{ 5 }$ on dissociation gives ${ PCl }_{ 3}$ and ${ Cl }_{ 2 }$. It is an endothermic reaction. The reaction is given below:
${ PCl }_{ 5 }(g)\xrightarrow { \Delta } { PCl }_{ 3 }(g)+{ Cl }_{ 2 }(g)$
If C is the initial concentration of ${ PCl }_{ 5 }$ and x is the concentration of ${ Cl }_{ 2 }$ at equilibrium, then:
$\begin{matrix} Reaction: \\ Initial: \\ At\quad equilibrium: \end{matrix}\begin{matrix} { PCl }_{ 5 }(g) \\ C \\ C-x \end{matrix}\rightleftharpoons \begin{matrix} { PCl }_{ 3 }(g) \\ 0 \\ x \end{matrix}+\begin{matrix} { Cl }_{ 2 }(g) \\ 0 \\ x \end{matrix}$

Now, it is given that the concentration of $\left[ { Cl }_{ 2 } \right] =0.1 M$ and the value of the equilibrium constant ${ K }_{ c }=0.04$.
The equilibrium constant ${ K }_{ c }$ can be written as the ratio of the product of the concentrations of ${ PCl }_{ 3}$ and ${ Cl }_{ 2 }$ to the concentration of ${ PCl }_{ 5 }$ at equilibrium ach raised to the power equal to their stoichiometric coefficients:
${ K }_{ c }=\cfrac { \left[ { PCl }_{ 3 } \right] \left[ { Cl }_{ 2 } \right] }{ \left[ { PCl }_{ 5 } \right] } $....(1)

Substituting the value of the concentrations in equation (1) we get,
${ K }_{ c }=\cfrac { { x }^{ 2 } }{ C-x } $
Substituting the value of ${ K }_{ c }$ and x in the above equation, we get:
$0.04=\cfrac { { 0.1 }^{ 2 } }{ C-0.1 } $
$\Rightarrow 0.04(C-0.1)={ 0.1 }^{ 2 }$
$\Rightarrow C=0.35$
So, the correct answer is “Option C”.

Note: In this question we cannot neglect the value of x with respect to C since the value to x is 0.1 which is greater than the 5% of C i.e. it is greater than 0.0175. x can only be neglected with respect to C if the value of x is less than the value of 5% of C.