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0.5 mole of each of H2, SO2 and CH4 are kept in a container. A hole was made in the container. After 3 hours, the order of partial pressures in the container will be:
- a)pSO2 > pCH4 > pH2
- b)pH2 > pSO2 > pCH4
- c)pH2 > pCH4 > pSO2
- d)pSO2 > pH2 > pCH4
Correct answer is option 'A'. Can you explain this answer?
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0.5 mole of each of H2, SO2 and CH4 are kept in a container. A hole wa...
To determine the order of partial pressures in the container after 3 hours, we need to consider the stoichiometry of the reaction that occurs when the hole is made in the container.
The reaction that occurs when the hole is made is the combustion of methane (CH4) with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O):
CH4 + 2O2 -> CO2 + 2H2O
From the stoichiometry of this reaction, we can see that for every mole of CH4 that reacts, we need 2 moles of O2. Therefore, all the CH4 will react with the available O2, and we will have no CH4 left in the container after 3 hours.
The molar ratio of CH4 to SO2 is 1:1, so if all the CH4 reacts, the same amount of SO2 will also react. Therefore, after 3 hours, there will be no SO2 left in the container as well.
The molar ratio of CH4 to H2 is 1:2, so if all the CH4 reacts, we will have twice as much moles of H2 left in the container. Therefore, after 3 hours, the order of partial pressures in the container will be:
pH2 > 0.5 mole
pSO2 = 0 mole
pCH4 = 0 mole
Therefore, the order of partial pressures in the container after 3 hours will be:
pH2 > pSO2 = pCH4
The reaction that occurs when the hole is made is the combustion of methane (CH4) with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O):
CH4 + 2O2 -> CO2 + 2H2O
From the stoichiometry of this reaction, we can see that for every mole of CH4 that reacts, we need 2 moles of O2. Therefore, all the CH4 will react with the available O2, and we will have no CH4 left in the container after 3 hours.
The molar ratio of CH4 to SO2 is 1:1, so if all the CH4 reacts, the same amount of SO2 will also react. Therefore, after 3 hours, there will be no SO2 left in the container as well.
The molar ratio of CH4 to H2 is 1:2, so if all the CH4 reacts, we will have twice as much moles of H2 left in the container. Therefore, after 3 hours, the order of partial pressures in the container will be:
pH2 > 0.5 mole
pSO2 = 0 mole
pCH4 = 0 mole
Therefore, the order of partial pressures in the container after 3 hours will be:
pH2 > pSO2 = pCH4
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0.5 mole of each of H2, SO2 and CH4 are kept in a container. A hole was made in the container. After 3 hours, the order of partial pressures in the container will be:a)pSO2 > pCH4 > pH2b)pH2 > pSO2 > pCH4c)pH2 > pCH4 > pSO2d)pSO2 > pH2 > pCH4Correct answer is option 'A'. Can you explain this answer?
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