Answer:
The equilibrium partial pressure of O2 is 0.545 atm
Explanation:
Step 1: Data given
Partial pressure of SO2 = 0.409 atm
Partial pressure of O2 = 0.601 atm
At equilibrium, the partial pressure of SO2 was 0.297 atm.
Step 2: The balanced equation
2SO2 + O2 ⇆ 2SO3
Step 3: The initial pressure
pSO2 = 0.409 atm
pO2 = 0.601 atm
pSO3 = 0 atm
Step 4: Calculate the pressure at the equilibrium
pSO2 = 0.409 - 2X atm
pO2 = 0.601 - X atm
pSO3 = 2X
pSO2 = 0.409 - 2X atm = 0.297
X = 0.056 atm
pO2 = 0.601 - 0.056 = 0.545 atm
pSO3 = 2*0.056 = 0.112 atm
Step 5: Calculate Kp
Kp = (pSO3)²/((pO2)*(pSO2)²)
Kp = (0.112²) / (0.545 * 0.297²)
Kp = 0.261
The equilibrium partial pressure of O2 is 0.545 atm
To find the empirical formula you would first need to find the moles of each element:
58.8g/ 12.0g = 4.9 mol C
9.9g/ 1.0g = 9.9 mol H
31.4g/ 16.0g = 1.96 O
Then you divide by the smallest number of moles of each:
4.9/1.96 = 2.5
9.9/1.96 = 6
1.96/1.96 = 1
Since there is 2.5, you find the least number that makes each moles a whole number which is 2.
So the empirical formula is C5H12O2.
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