Q. Consider the reaction equilibrium, $2SO_{2}(g) + O_{2}(g) \rightleftharpoons 2SO_{3}(g);\ \ \ \ \$ $\ \mathrm{\Delta}H{^\circ} = - 198\ kJ$ On the basis of Le-Chatelier’s principle, the condition favourable for the forward reaction is
A) Lowering of temperature as well as pressure
B) Increasing temperature as well as pressure
C) Lowering the temperature and increasing the pressure
D) Any value of temperature and pressure
Explanation
(c)
[table] $\mathbf{\mathrm{\Delta}n = -}\mathbf{ve}$ | $\mathbf{\mathrm{\Delta}H{^\circ} = -}\mathbf{ve}$ ; Takes place with decrease in number of mole or pressure, hence increase in pressure shifts equilibrium in forward side. | Takes place with evolution of heat or increase in temperature, hence decrease in temperature shifts this equilibrium in forward side. [/table]
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.