The conditions favourable for the reaction: $2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g); △ H^{\circ} = - 198 K J$ are

low temperature,high pressure

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any value of T and P

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low temperature and low pressure

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high temperature and high pressure

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Solution

The correct option is A low temperature,high pressure
The negative value of enthalpy change indicates that the forward reaction is exothermic.
When the temperature is lowered, the reaction will proceed in the forward direction so that heat is liberated which will nullify the effect of lowered temperature.
The forward reaction proceeds with decrease in the number of moles, the increased pressure will shift the equilibrium towards right which will decrease the number of moles and nullify the effect of increased pressure. This will increase the yield of the forward reaction.
Thus the formation of $S O_{3} (g)$ will be favored at low temperature and high pressure.

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