An intermediate step in the production of battery acid, sulfuric acid, invoves the information of sulfur trioxide gas from sulfur dioxide and oxygen gases. (this reaction is also the usual first step in the production of acid rain).
The Kp for this reaction at 830degreeC is 0.13. In one experiment, 96.9g of sufur dioxide and 34.6g of oxygen were added to a rxn flask. What must the total pressure in the flask be at eqilibrium if the reaction has an 80% yield of sulfur trioxide. Assume the volume is 3.5L.
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As I see the the first problem.
2SO2 + O2 ==> 2SO3.
Convert g SO2 to mols, convert that to mols SO3, take 80%. That is mols SO3 we should have at the end of the reaction for an 80% yield. Then calculate SO2 remaining, O2 used up (at 80%) and O2 remaining. Add mols of each and use PV = nRT to determine total P.
Check my thinking.
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Taking away 80% means what you have at 100% - 80% ?
Since the molar ratios are the same for S03 and S02, are the mols going to be the same? So that it would make 0.3+0.3?
Will the total mols be: mols of O2 + 0.3+ 0.3?
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Thanks for your help much appreciated!
I don't think so.
Mols SO2 = about 0.3 but mols SO3 will be what formed at 80% yield which is not 0.3. I have that as about 1.2 or so. Then O2 used in the reaction is 1/2 SO3 formed so that is about 0.6 (1.2 * 1.2) and that subtracted from the O2 initially is 1.08 - 0.6 = about 0.48 O2 remaining unreacted. You need to refine all of the numbers and don't round too much in doing so. I think the total mols is more like 0.3 something for SO2, 0.48 or so for O2, and about 1.2 or so for SO3. Check my thinking.
You're on the right track! To calculate the amount of sulfur trioxide (SO3) you should have at the end of the reaction for an 80% yield, you first need to convert the grams of sulfur dioxide (SO2) given to moles.
To convert grams of SO2 to moles, you need to use the molar mass of SO2. From the periodic table, you can find that sulfur has a molar mass of 32.06 g/mol and oxygen has a molar mass of 16.00 g/mol. Since sulfur dioxide (SO2) has one sulfur atom and two oxygen atoms, its molar mass is 32.06 g/mol + (2 x 16.00 g/mol) = 64.06 g/mol.
So, for the 96.9 g of SO2, you divide the mass by the molar mass to get the number of moles:
Moles of SO2 = 96.9 g / 64.06 g/mol ≈ 1.515 mol
Since the molar ratio for the reaction is 2:1 for SO2 to SO3, you should have half as many moles of SO3:
Moles of SO3 = 1.515 mol / 2 = 0.7575 mol
Now, since the yield is given as 80%, you multiply the moles of SO3 by 0.80 to get the actual moles of SO3 you will have at the end:
Moles of SO3 (actual) = 0.7575 mol x 0.80 = 0.606 mol
To determine the moles of SO2 remaining, you subtract the moles of SO3 (actual) from the moles of SO2 initially:
Moles of SO2 remaining = 1.515 mol - 0.606 mol = 0.909 mol
For the remaining oxygen (O2) gas, you divide the given mass by its molar mass. From the periodic table, the molar mass of oxygen is 16.00 g/mol. So, for the 34.6 g of O2, you have:
Moles of O2 = 34.6 g / 16.00 g/mol ≈ 2.1625 mol
Since the molar ratio for the reaction is 1:1 for O2 and SO3, you can use the moles of SO3 (actual) as the moles of O2 (used up), as they will be equal in this case:
Moles of O2 remaining = Moles of O2 - Moles of O2 (used up) = 2.1625 mol - 0.606 mol = 1.5565 mol
Now, you can calculate the total moles of gas in the flask by adding the moles of SO2 remaining, SO3 (actual), and O2 remaining:
Total moles of gas = Moles of SO2 remaining + Moles of SO3 (actual) + Moles of O2 remaining
Total moles of gas = 0.909 mol + 0.606 mol + 1.5565 mol = 3.0715 mol
Finally, you can use the ideal gas law equation PV = nRT, where P is the total pressure, V is the volume (given as 3.5 L), n is the total moles of gas, R is the ideal gas constant (0.0821 L·atm/mol·K), and T is the temperature in Kelvin (830 degrees Celsius = 1103 K):
P x 3.5 L = 3.0715 mol x 0.0821 L·atm/mol·K x 1103 K
Rearranging the equation to solve for P, we have:
P = (3.0715 mol x 0.0821 L·atm/mol·K x 1103 K) / 3.5 L
P ≈ 27.342 atm
Therefore, the total pressure in the flask at equilibrium should be approximately 27.342 atm.