Question

# State Le Chatelier’s principle and apply it to the following equilibrium.     $2S{{O}_{2}}_{_{(g)}}+{{O}_{2}}_{_{(g)}}\rightleftarrows 2S{{O}_{3}}_{_{(g)}};\Delta H=-189kJ$

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Hint: Any system at equilibrium when subjected to a disturbance like change in concentration, pressure or temperature responses in a way that tends to cancel or minimize the effect of that disturbance.

$2S{{O}_{2}}_{_{(g)}}+{{O}_{2}}_{_{(g)}}\rightleftarrows 2S{{O}_{3}}_{_{(g)}};\Delta H=-189kJ$
Effect of Temperature: Since the enthalpy of formation ($\Delta H$) of $S{{O}_{3}}$ is negative, the given equilibrium reaction is exothermic. If the temperature of the reaction is increased, then the reaction will proceed in the direction where the heat is absorbed, i.e. in the backward direction. Whereas, if the temperature is lowered, the equilibrium shifts in the forward direction where the heat is evolved. Therefore, being exothermic in nature, this reaction is favoured by low temperature. An optimum temperature of 673-723 K is used for the formation of $S{{O}_{3}}$ in the Contact process.
Effect of pressure: The given reaction proceeds forward with decrease in number of moles. If the pressure is increased, then as per Le Chatelier’s, the equilibrium is shifted in the direction in which the pressure decreases and hence, the number of moles also decreases, i.e. in the forward direction. Therefore, increase in pressure favours the formation of $S{{O}_{3}}$.
Effect of Concentration: If the concentrations of reactants $S{{O}_{2}}$ and ${{O}_{2}}$ are increased, then according to Le Chatelier’s principle, the reaction equilibrium shifts in the direction where the reactants are consumed, i.e. in the forward direction. Therefore, increase in concentration of $S{{O}_{2}}$ and ${{O}_{2}}$ leads to higher concentration of $S{{O}_{3}}$.