For 2 SO2(g) + O2(g) equilibrium reaction arrow 2 SO3(g), Kp = 3.0 104 at 700 K. In a 2.00-L vessel the equilibrium mixture contains 1.21 g of SO3 and 0.183 g of O2. How many grams of SO2 are in the vessel?
mols SO3 = grams/molar mass = ?
Then M SO3 = mols/L. Let's call this about 0.0075 M but that's just an estimate. You need to do it much more accurately.
mols O2 = grams/mola mass = ?
Then M O2 = mols/L. Let's call this about 0.003 M. Again, you should do it more accurately.
.......2SO2 + O2 ==> 2SO3
E........x..0.003.....0.0075
Substitute the E line into Kp expression and solve for x = (SO2)
Convert that from M to mols in 2 L to grams in 2 L.
To solve this problem, we need to use the given equilibrium constant (Kp) and the partial pressures of the gases to find the number of moles of each gas present in the mixture. From the number of moles, we can calculate the masses of the gases.
Here is how you can solve this problem step by step:
Step 1: Convert the given masses of SO3 and O2 to the number of moles.
To do this, we need to know the molar masses (also called molecular weights) of SO3 and O2. The molar mass of SO3 is 80.06 g/mol, and the molar mass of O2 is 32.00 g/mol.
Number of moles of SO3:
moles of SO3 = mass of SO3 / molar mass of SO3
moles of SO3 = 1.21 g / 80.06 g/mol
Number of moles of O2:
moles of O2 = mass of O2 / molar mass of O2
moles of O2 = 0.183 g / 32.00 g/mol
Step 2: Calculate the partial pressure of O2 using the ideal gas law.
To do this, we need to know the volume of the vessel and the temperature. The volume of the vessel is given as 2.00 L, and the temperature is given as 700 K.
PV = nRT
P = nRT/V
Partial pressure of O2:
P(O2) = moles of O2 * R * T / V
where R is the ideal gas constant (0.0821 L·atm/(mol·K)).
Step 3: Use the ideal gas law again to calculate the partial pressure of SO2.
The equation shows that 2 moles of SO2 are used to produce 2 moles of SO3. Since the stoichiometry is 1:1, the partial pressure of SO2 will be the same as the partial pressure of O2.
Partial pressure of SO2:
P(SO2) = P(O2)
Step 4: Calculate the number of moles of SO2 using the partial pressure and the ideal gas law.
P(SO2) = n(SO2) * R * T / V
Simplifying the equation:
n(SO2) = P(SO2) * V / (R * T)
Step 5: Convert the number of moles of SO2 to grams.
mass of SO2 = moles of SO2 * molar mass of SO2
Using the molar mass of SO2 (64.06 g/mol), you can calculate the mass of SO2.
Now that you have all the steps, you can calculate the answer:
1. Calculate the number of moles of SO3:
moles of SO3 = 1.21 g / 80.06 g/mol = 0.015 g
2. Calculate the number of moles of O2:
moles of O2 = 0.183 g / 32.00 g/mol = 0.006 mol
3. Calculate the partial pressure of O2:
P(O2) = (0.006 mol * 0.0821 L·atm/(mol·K) * 700 K) / 2.00 L = 0.172 atm
4. Calculate the number of moles of SO2:
n(SO2) = 0.172 atm * 2.00 L / (0.0821 L·atm/(mol·K) * 700 K) = 0.00576 mol
5. Calculate the mass of SO2:
mass of SO2 = 0.00576 mol * 64.06 g/mol = 0.369 g
Therefore, there are 0.369 grams of SO2 in the vessel.