Question

Which of the following factor shifts the reaction,
$PC{l}_3\left(g\right)+C{l}_2\left(g\right)\rightleftharpoons PC{l}_5\left(g\right);\Delta H=-ve$ towards the reactant side?

• A. Addition of $PC{l}_5$
• B. Increasing the pressure
• C. Decreasing the temperature
• D. Addition of a catalyst

Answer 1

The correct option is A Addition of $PC{l}_5$

Option A: According to Le-Chatelier's principle, increasing concentrations of the products results in shifting the equilibrium in favour of the reactants. So, by adding $PC{l}_5$ will shift the reaction to the reactant side.

Option B: If there is an increase in the pressure. The position of equilibrium will move in such a way as to decrease the pressure again. It is done by favouring the direction which produces fewer gaseous moles. So, the equilibrium shifts to the right with increasing pressure as reactant side has more number of gaseous moles.

Option C: When we decrease the temperature, the reaction goes towards the side where temerature increases. Since the reaction is exothermic, the temperature increases towards the products side. So decrease in temperature favours product formation.

Option D: Using catalyst doesn't affect the state of equilibrium.