Consider the reaction equilibrium,2SO_{2(g)} | vedclass

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(C) The given reaction is $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$ with $\Delta H^\circ = -198 \ kJ$.Since $\Delta H^\circ < 0$,the react

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Lowering the temperature and increasing the pressure

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(C) The given reaction is $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$ with $\Delta H^\circ = -198 \ kJ$.Since $\Delta H^\circ According to Le Chatelier's principle,for an exothermic reaction,a decrease in temperature favors the forward reaction.Regarding pressure,the number of moles of gaseous products is $2$,while the number of moles of gaseous reactants is $2 + 1 = 3$.Since the number of moles decreases in the forward direction $(3 \rightarrow 2)$,an increase in pressure favors the forward reaction.Therefore,lowering the temperature and increasing the pressure are the favorable conditions.

Consider the reaction equilibrium,$2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$; $\Delta H^\circ = -198 \ kJ$. On the basis of Le Chatelier's principle,the condition favourable for the forward reaction is

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