Question

Two moles of $ PC{l_5} $ are heated in a $ 2L $ flask. At equilibrium $ 40\% PC{l_5} $ dissociates. What is the equilibrium constant for this reaction?

Answer 1

Hint: Given that $ PC{l_5} $ dissociates at equilibrium. From the balanced chemical equation and the number of moles of compounds in chemical reaction at equilibrium gives the number of moles of each species. From the obtained number of moles and volume mentioned the concentrations will be obtained. Substituting the concentrations in the equilibrium constant equation gives the equilibrium constant.

Complete Step By Step Answer:
Phosphorus pentachloride has a molecular formula $ PC{l_5} $ dissociates as follows:
 $ PC{l_5} \rightleftharpoons PC{l_3} + C{l_2} $
The ICE table for the above reaction will be

initial	$ 2mol $	$ 0mol $	$ 0mol $
change	$ - 2\alpha mol $	$ + 2\alpha mol $	$ + 2\alpha mol $
equilibrium	$ 2(1 - \alpha )mol $	$ + 2\alpha mol $	$ + 2\alpha mol $

Given that degree of dissociation, $ \alpha = 40\% $ which is equal to $ \alpha = 0.4 $
To obtain the number of moles of each species in the chemical reaction at equilibrium. Substitute the value of degree of dissociation in the obtained number of moles.
Given that the volume of flask is $ 2L $
The concentration will be obtained by dividing the number of moles with the volume which is $ 2L $
 $ \left[ {PC{l_5}} \right] = \dfrac{{2(1 - \alpha )}}{2} = \dfrac{{2(1 - 0.4)}}{2}0.6mol{L^{ - 1}} $
 $ \left[ {PC{l_3}} \right] = \dfrac{{2\alpha }}{2} = \dfrac{{2(0.4)}}{2}0.4mol{L^{ - 1}} $
 $ \left[ {C{l_2}} \right] = \dfrac{{2\alpha }}{2} = \dfrac{{2(0.4)}}{2}0.4mol{L^{ - 1}} $
The equilibrium constant which was represented by $ {K_c} $ for the above reaction can be written as
 $ {K_c} = \dfrac{{\left[ {PC{l_3}} \right]\left[ {C{l_2}} \right]}}{{\left[ {PC{l_5}} \right]}} $
Substitute the values of concentrations of each species in the above equation
 $ {K_c} = \dfrac{{0.4 \times 0.4}}{{0.6}} = 0.27mol.{L^{ - 1}} $
The equilibrium constant for the given reaction is $ 0.27mol.{L^{ - 1}} $

Note:
Molar concentration is also known as molarity. Here by using the number of moles and volume the molarity can be determined. While calculating the molarity the volume must be in litres if it is in millilitres it should multiply with a value of $ 1000 $ as one litre is equal to $ 1000 $ milliliters.