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Question

Question: How many moles per litre of \({ PCl }_{ 5 }\) has to be taken to obtain 0.1 mole \({ Cl }_{ 2 }\), i...

How many moles per litre of PCl5{ PCl }_{ 5 } has to be taken to obtain 0.1 mole Cl2{ Cl }_{ 2 }, if the value of the equilibrium constant Kc{ K }_{ c } is 0.04?
(a) 0.15
(b) 0.25
(c) 0.35
(d) 0.05

Explanation

Solution

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:
PCl5{ PCl }_{ 5 } on dissociation gives PCl3{ PCl }_{ 3} and Cl2{ Cl }_{ 2 }. It is an endothermic reaction. The reaction is given below:
PCl5(g)ΔPCl3(g)+Cl2(g){ PCl }_{ 5 }(g)\xrightarrow { \Delta } { PCl }_{ 3 }(g)+{ Cl }_{ 2 }(g)
If C is the initial concentration of PCl5{ PCl }_{ 5 } and x is the concentration of Cl2{ Cl }_{ 2 } at equilibrium, then:
Reaction: Initial: Atequilibrium:PCl5(g) C CxPCl3(g) 0 x+Cl2(g) 0 x\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 [Cl2]=0.1M\left[ { Cl }_{ 2 } \right] =0.1 M and the value of the equilibrium constant Kc=0.04{ K }_{ c }=0.04.
The equilibrium constant Kc{ K }_{ c } can be written as the ratio of the product of the concentrations of PCl3{ PCl }_{ 3} and Cl2{ Cl }_{ 2 } to the concentration of PCl5{ PCl }_{ 5 } at equilibrium ach raised to the power equal to their stoichiometric coefficients:
Kc=[PCl3][Cl2][PCl5]{ 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,
Kc=x2Cx{ K }_{ c }=\cfrac { { x }^{ 2 } }{ C-x }
Substituting the value of Kc{ K }_{ c } and x in the above equation, we get:
0.04=0.12C0.10.04=\cfrac { { 0.1 }^{ 2 } }{ C-0.1 }
0.04(C0.1)=0.12\Rightarrow 0.04(C-0.1)={ 0.1 }^{ 2 }
C=0.35\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.