To solve this problem, we need to analyze the equilibrium reaction and the conditions given. The reaction is represented as:
Understanding the Reaction
The balanced chemical equation is:
2 SO2 (g) + O2 (g) ⇌ 2 SO3 (g)
This indicates that two moles of sulfur dioxide (SO2) react with one mole of oxygen (O2) to produce two moles of sulfur trioxide (SO3).
Initial Conditions in the Second Vessel
In the second vessel, we have 32 grams of SO2. To find the number of moles of SO2, we can use the molar mass of SO2, which is approximately 64 g/mol:
- Moles of SO2 = Mass / Molar Mass = 32 g / 64 g/mol = 0.5 moles
Determining the Change in Moles
We want to oxidize 20% of the SO2 present in the vessel. Therefore, the amount of SO2 that will be oxidized is:
- 20% of 0.5 moles = 0.1 moles
This means that at equilibrium, we will have:
- Remaining SO2 = 0.5 moles - 0.1 moles = 0.4 moles
- Produced SO3 = 0.1 moles (since 2 moles of SO2 produce 2 moles of SO3)
Calculating the Required Moles of O2
According to the stoichiometry of the reaction, for every 2 moles of SO2 that react, 1 mole of O2 is consumed. Therefore, the moles of O2 required to oxidize 0.1 moles of SO2 can be calculated as follows:
- Moles of O2 needed = (0.1 moles SO2) × (1 mole O2 / 2 moles SO2) = 0.05 moles O2
Converting Moles of O2 to Mass
Now, we need to convert the moles of O2 into grams. The molar mass of O2 is approximately 32 g/mol:
- Mass of O2 = Moles × Molar Mass = 0.05 moles × 32 g/mol = 1.6 grams
Final Answer
To achieve the desired equilibrium condition where 20% of SO2 is oxidized to SO3, you must add 1.6 grams of O2 to the vessel containing 32 grams of SO2.
This approach combines stoichiometry with the principles of chemical equilibrium, allowing us to determine the necessary conditions for the reaction to proceed as desired. If you have any further questions or need clarification on any of the steps, feel free to ask!
