RAJASTHAN PMT Rajasthan - PMT Solved Paper-2008

Consider the reaction equilibrium: \[2S{{O}_{2}}(g)+{{O}_{2}}(g)2S{{O}_{3}}(g);\]                                                 \[\Delta {{H}^{o}}=-198\,kJ\] On the basis of Le-Chateliers 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

Correct Answer: C

Solution :

\[\Delta {{n}_{g}}=-vc\]	\[\Delta {{H}^{o}}=-vc\]
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.