Sulfur dioxide reacts with oxygen to form sulfur trioxide according to the equation
2SO2(g) + O2(g) <-> 2SO3(g)
Samples of sulfur dioxide, oxygen, and sulfur trioxide were added to a flask of volume 1.40 dm^3 and allowed to reach equilibrium at a given temperature. The flask contained 0.0550 mol of sulfur dioxide and 0.0720 mol of sulfur trioxide at equilibrium.
Kc has the numerical value of 27.9 under these conditions.
Calculate the amount, in moles, of oxygen gas in this equilibrium mixture.
Not really sure how to get the answer, as the only questions I've had up to this point made it easy to get the concentration, but here it's a gas
The only thing I've figured out is that
Kc = [SO3]^2 / [SO2]^2 x [O2]
So 27.9 = [SO3]^2 / [SO2]^2 x [O2]
Thank you so much!
You know from your Kc equation that you need concentrations. You have mols and volume; therefore, you know at equilibrium that [SO2] = 0.0550 mol/1.40 dm^3 and [SO3] = 0.0720 mol/1.40 dm^3. Therefore, [SO2] = approx 0.04 M and [SO3] = approx 0.05. You will need to recalculate both because I've only estimated. Now plug these into the equation.
........2SO2 + O2 ==> 2SO3
So plug in the [SO2] and [SO3] values into Kc expression and solve for [O2] and the answer will be in units of of mols/dm^3 = M. The problem asks for mols. You know the volume and you know M, solve for mols. Post your work if you get stuck.
This helped a lot, thanks for the easy to understand explanations ^_^
however many litters of sulphur trioxide are formed when 4800cm of sulphur dioxide is burned in air?
To solve this problem, we can use the given information and the equation for the equilibrium constant (Kc) to find the amount of oxygen gas in the equilibrium mixture.
First, let's assign variables to the amount of substances in the equilibrium mixture:
Let x be the amount (in moles) of oxygen gas (O2).
Since we have 0.0550 mol of sulfur dioxide (SO2) and 0.0720 mol of sulfur trioxide (SO3), the amounts of SO2 and SO3 are already known.
Next, we can set up the equation for Kc using the concentrations of the substances in the equilibrium mixture (remember that concentration is equal to moles divided by volume in dm^3):
Kc = [SO3]^2 / ([SO2]^2 * [O2])
Substituting the known values into the equation, we have:
27.9 = (0.0720 mol)^2 / ((0.0550 mol)^2 * x)
Now, we can rearrange the equation to solve for x:
27.9 = (0.0720)^2 / (0.0550)^2 * x
x = (0.0550)^2 * (0.0720)^2 / 27.9
Evaluating the expression on the right side of the equation, we get:
x ≈ 0.0820 mol
Therefore, the amount of oxygen gas in the equilibrium mixture is approximately 0.0820 mol.