Question

At 1000K, the equilibrium constant, ${ K }_{ c }$ for the reaction,
\[CO_{ \left( g \right) }\quad +\quad Cl_{ 2\left( g \right) }\quad ⇌\quad COCl_{ 2\left( g \right) }\]
is equal to 3.04. If 1 mole of CO and 1 mole of $Cl_{ 2 }$ are introduced into a 1 liter box at 1000K, what will be the final concentration of $COCl_{ 2 }$ at equilibrium?

Answer 1

Hint: We know that equilibrium constant can be found out by knowing the concentration of reactants and products at equilibrium. Assume some degree of dissociation of the reactants at equilibrium taking a concentration of every chemical species to find out equilibrium constant and equate it to the given value.

Complete step by step solution:
We know that equilibrium constant is the product of products concentration raised to the power of their respective stoichiometric coefficients divided by-product of reactants concentration raised to the power of their respective stoichiometric coefficients. Assuming some degree of dissociation for reactants we can calculate the equilibrium constant in terms of the degree of dissociation. When we equate it to a given equilibrium constant we get the value of the degree of dissociation.
As volume is 1L concentration will be numerically the same as that of the number of moles.
The concentration of CO, $Cl 2 $, $COCl_{ 2 }$ initially are 1, 1, 0 respectively.

Let the degree of dissociation be x at equilibrium.
The concentration of CO,$Cl_{ 2 }$, $COCl_{ 2 }$ finally are 1-x, 1-x, x respectively.
Equilibrium constant = $\dfrac { x }{ \left( 1-x \right) ^{ 2 } } \quad =\quad K_{ c }$
\[\dfrac { x }{ \left( 1-x \right) ^{ 2 } } \quad =\quad 3.04\]
\[x\quad =\quad 3.04(1-x)^{ 2 }\]
\[x\quad =\quad 3.04(1-2x+{ x }^{ 2 })\]
\[3.04x^{ 2 }\quad -\quad 7.08x\quad +\quad 3.04\quad =0\]
\[x\quad =\quad 0.568\quad or\quad 1.761\]
But the degree of dissociation cannot be greater than 1 so x = 0.568.
Therefore, the final concentration of $COCl_{ 2 }$ at equilibrium will be 0.568$mol{ L }^{ -1 }$.

Note: We need to be careful with the value of the degree of dissociation after calculating the roots of the quadratic equation. We need to be careful with concentration and moles while calculating equilibrium constant. Here volume was 1L but it may not be the same everywhere.