For the given reaction $$C{l_2}(g)
+ 2{F_2}(g) \rightleftharpoons 2Cl{F_3}(g)$$ , the enthalpy change of the
reaction is given as$$\vartriangle {H_r} = - 329kJ$$ . We know that if the
enthalpy change is negative then the reaction is an exothermic reaction.
Therefore, according to Le Chatlier’s principle increasing the temperature will
shift the equilibrium towards the side of the reactants and hinder the
formation of the product.
Removing $$C{l_2}$$
will result in the decrease in the quantity of the product, since the equilibrium
of the reaction will move backward.
By increasing the volume of the container the value of $${Q_c}$$ will become greater than the value
of $${K_c}$$ hence favoring the backward
reaction and resulting in the decrease of the quantity of the product.
So, for the given reaction if we want to increase the
quantity of the product $$Cl{F_3}$$ then
the correct option will be to add $${F_2}$$
so that the to decrease the concentration of $${F_2}$$
and to bring the system into equilibrium the reaction will move forward and
hence the quantity of the product $$Cl{F_3}$$
will increase.
Hence, the correct
option is C.) Adding $${F_2}$$