In the following reversible reaction 2SO_2(g) | vedclass

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(C) The given reaction is $2SO_2(g) + O_2(g) ightleftharpoons 2SO_3(g) + Q \ 	ext{cal}$.Since the reaction is exothermic (indicated by $+ Q \ 	ext

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Low temperature and high pressure

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(C) The given reaction is $2SO_2(g) + O_2(g) ightleftharpoons 2SO_3(g) + Q \ 	ext{cal}$.Since the reaction is exothermic (indicated by $+ Q \ 	ext{cal}$),according to Le Chatelier's principle,a low temperature will shift the equilibrium in the forward direction to produce more $SO_3$.Also,the number of moles of gaseous reactants is $3$ ($2$ moles of $SO_2$ and $1$ mole of $O_2$) and the number of moles of gaseous products is $2$ $(SO_3)$.Since the volume decreases from left to right,a high pressure will shift the equilibrium in the forward direction to produce more $SO_3$.Therefore,the most suitable conditions are low temperature and high pressure.

In the following reversible reaction $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) + Q \ \text{cal}$. The most suitable condition for the higher production of $SO_3$ is:

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