Question

Given the equilibrium, $PC{l}_3\left(g\right)+C{l}_2\left(g\right)\rightleftharpoons PC{l}_5\left(g\right);\Delta H^o=-92kJ$. Which of the following would shift the equilibrium to the right?

• A. Increasing pressure
• B. Adding $PC{l}_3$ to the system
• C. Decreasing the temperature
• D. Addition of $He$

Answer 1

Answer: (A), (B), (C)

The correct options are
A Increasing pressure
B Adding $PC{l}_3$ to the system
C Decreasing the temperature

Increasing the pressure shifts the equilibrium to the right side as the system would be under stress. To relieve the stress the reactants collide and form the product.
By adding $PC{l}_3$, it reacts with more number of molecules of $C{l}_2$ and forms more amount of the product.
By decreasing the temperature more number of reactants collide and form the products and heat to maintain the temperature resulting in an exothermic reaction.