Question

(a) Find the no. of moles of ${\text{PC}}{{\text{l}}_{\text{5}}}$ to be heated in a closed container of 1 lit capacity so that $0.2$ mole chlorine is obtained at equilibrium. ${{\text{K}}_{\text{c}}}$ = $5 \times {10^{ - 2}}$ ${\text{mol/lit}}$
(b) Find the no. of moles of ${\text{PC}}{{\text{l}}_{\text{5}}}$ to be heated in a closed container of 2 lit capacity so that $1.6$ moles of chlorine is obtained at equilibrium. Find the number of moles of ${\text{PC}}{{\text{l}}_{\text{5}}}$ at equilibrium ${{\text{K}}_{\text{c}}}$ = $3.2$ ${\text{mol/lit}}$.

Answer 1

Hint: To solve this question, knowledge of chemical equilibrium is required which is defined as the state at which the rate of forward reaction is equal to the rate of the backward reaction. We shall write the equation and use the formula (below) to find the concentration of ${\text{PC}}{{\text{l}}_{\text{5}}}$

Formula Used: ${{\text{K}}_{\text{c}}}$=$\dfrac{{\left[ {{\text{PC}}{{\text{l}}_3}} \right]\left[ {{\text{C}}{{\text{l}}_2}} \right]}}{{\left[ {{\text{PC}}{{\text{l}}_{\text{5}}}} \right]}}$,
where ${{\text{K}}_{\text{c}}}$ is the equilibrium constant of the reaction.

Complete step by step solution:
a)The reaction for heating ${\text{PC}}{{\text{l}}_{\text{5}}}$ can be written as:

	${\text{PC}}{{\text{l}}_{\text{5}}}$	${\text{PC}}{{\text{l}}_3}$	${\text{C}}{{\text{l}}_2}$
Initial	1 mole	0 mole	0 mole
Final	$\left( {1 - x} \right)$ mole	x mole	x mole

Now according to the question, x = $0.2$ mol chlorine
Volume of the vessel being 1 litre, the conc. Of chlorine formed = $0.2$ ${\text{mol/lit}}$. The concentration of ${\text{PC}}{{\text{l}}_3}$ should also be the same as they are obtained from the same.
Therefore equilibrium constant, ${{\text{K}}_{\text{c}}}$ = $\dfrac{{\left[ {{\text{PC}}{{\text{l}}_3}} \right]\left[ {{\text{C}}{{\text{l}}_2}} \right]}}{{\left[ {{\text{PC}}{{\text{l}}_{\text{5}}}} \right]}}$
$ \Rightarrow \left[ {{\text{PC}}{{\text{l}}_{\text{5}}}} \right] = \dfrac{{\left[ {0.2} \right]\left[ {0.2} \right]}}{{5 \times {{10}^{ - 2}}}}$ = $0.8$ ${\text{mol/lit}}$.

b)For the second part:

	${\text{PC}}{{\text{l}}_{\text{5}}}$	${\text{PC}}{{\text{l}}_3}$	${\text{C}}{{\text{l}}_2}$
Initial	1 mole	0 mole	0 mole
Final	$\left( {1 - x} \right)$ mole	x mole	x mole

The value of x = $1.6$ moles of chlorine. So the number of moles for phosphorus pentachloride should also be the same. But here the volume of the container is 2 litre,
So the concentration of chlorine = $\dfrac{{1.6}}{2} = 0.8$ ${\text{mol/lit}}$
So the equilibrium constant, ${{\text{K}}_{\text{c}}}$ = $\dfrac{{\left[ {{\text{PC}}{{\text{l}}_3}} \right]\left[ {{\text{C}}{{\text{l}}_2}} \right]}}{{\left[ {{\text{PC}}{{\text{l}}_{\text{5}}}} \right]}}$
$ \Rightarrow \left[ {{\text{PC}}{{\text{l}}_{\text{5}}}} \right] = \dfrac{{\left[ {0.8} \right]\left[ {0.8} \right]}}{{3.2}}$ = $0.2$ ${\text{mol/lit}}$
Since ${\text{PC}}{{\text{l}}_3}$ molecules are formed along with the chlorine atoms from the same compound, the number of moles will be same as that of Chlorine, i.e. $0.8$${\text{mol/lit}}$.

Notes:
Any change in the value of the chemical equilibrium is guided by Le Chatelier’s Principle. According to it, a dynamic equilibrium tries to resist any change in the equilibrium of the system and acts in a direction to resist the change.