Вопрос школьника
20 seconds after the start of the reaction 2SО2 + О2 → 2SО3 3,2 g remain unoxidised of 4,8 g of sulfur oxide (IV). Find the rate of oxidation of sulfur dioxide .
Ответ от учителя
The given chemical reaction is 2SO2 + O2 → 2SO3.
The balanced equation shows that for every 2 moles of SO2, 1 mole of O2 is required to produce 2 moles of SO3.
Given that 4.8 g of SO2 was present at the start of the reaction, we can calculate the number of moles of SO2 as follows:
Molar mass of SO2 = 32.06 g/mol
Number of moles of SO2 = 4.8 g / 32.06 g/mol = 0.15 mol
Since the reaction is 2SO2 + O2 → 2SO3, the number of moles of O2 required for the complete reaction is half the number of moles of SO2, which is 0.075 mol.
Now, after 20 seconds, only 3.2 g of SO2 remains unoxidized. We can calculate the number of moles of unoxidized SO2 as follows:
Molar mass of SO2 = 32.06 g/mol
Number of moles of unoxidized SO2 = 3.2 g / 32.06 g/mol = 0.1 mol
The difference between the initial number of moles of SO2 and the number of moles of unoxidized SO2 gives us the number of moles of SO2 that have reacted:
Number of moles of SO2 reacted = 0.15 mol — 0.1 mol = 0.05 mol
Now, we can calculate the rate of oxidation of SO2 as follows:
Rate of oxidation of SO2 = (Number of moles of SO2 reacted) / (Time taken for reaction)
Time taken for reaction = 20 seconds
Rate of oxidation of SO2 = 0.05 mol / 20 s = 0.0025 mol/s
Therefore, the rate of oxidation of SO2 is 0.0025 mol/s.
