For the reaction 2SO2(g) + 02(g) = 2S03(g) + Heat, the yield of S03 will be maximum if

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Q: 18 (CDS-I/2002)
For the reaction 2SO2(g) + 02(g) = 2S03(g) + Heat, the yield of S03 will be maximum if

question_subject: 

Science

question_exam: 

CDS-I

stats: 

0,12,37,14,12,22,1

keywords: 

{'s03': [0, 0, 1, 0], 'reaction': [1, 0, 5, 18], 'temperature': [0, 1, 1, 7], 'pressure': [0, 0, 0, 1], 'heat': [10, 3, 13, 46], 'yield': [0, 1, 0, 1]}

In the given reaction 2SO2(g) + O2(g) = 2SO3(g) + Heat, the yield of SO3 can be maximized by manipulating the temperature and pressure conditions.

Option 1 suggests increasing the temperature while keeping the pressure constant. When the temperature is increased, the reaction rate is accelerated, promoting the formation of more product. However, by maintaining constant pressure, the system can become limited by the reactant concentration, resulting in a lower yield of SO3. Therefore, option 1 is not the best choice.

Option 2 proposes decreasing the temperature and increasing the pressure. Lowering the temperature will slow down the reaction rate, allowing the reaction to reach equilibrium and favor the formation of the product. Increasing the pressure also favors the production of more SO3 as it helps to shift the equilibrium towards the product side. This option is correct as it provides the ideal conditions for maximizing the yield of SO3.

Option 3 suggests increasing both the temperature and pressure. This can increase the reaction rate, but if the pressure is too high, it may decrease the yield due to the limited availability of reactants. Therefore, option 3 is not the best choice.

Option 4 suggests decreasing both the temperature and pressure. This would slow

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