Question

For the reaction
$2Cl{F}_3\left(g\right)\rightleftharpoons C{l}_2\left(g\right)+3F_2\left(g\right)$
$log{K}_{eq}v/s\frac{1}{T}$ (where temperature is in K) curve is obtained as following
Which of the following change will increase the concentration of $C{l}_2$ in an equilibrium mixture of $C{l}_2,F_{2}andCl{F}_3:$

• A. Addition of inert gas at constant pressure
  
• B. Increase in temperature at constant volume
  
• C. Addition of catalyst at equilibrium
  
• D. Removal of $F_2\left(g\right)$ at equilibrium

Answer 1

Answer: (B), (D)

Removal of $F_2\left(g\right)$ at equilibrium

$logK=\frac{-△H}{R}\left[\frac{1}{T}\right]$
$y=mx+cx=\frac{1}{T}m=\frac{-△H}{R}$

From graph we can see that slope is negative.
$\therefore △H$ must be positive
$\therefore$ reaction is endothermic
In product side we have 4 gaseous mole
In reactant side we have 2 gaseous moles
$\to$ Addition of inert gas does not affect concentration of any reactants and products since it will not react.
$\to$ The reaction is endothermic hence if we increase temperature concentration of product will increase.
$\to$ Catalyst is used to attain equilibrium fast, increases forward and backward rate equally.
$\to$ If the concentration of products is decreased then reaction will go in forward side to again reduce the change so if $F_2$ is removed then $C{l}_2$ concentration will increase. Hence, option B and D are correct.