An equilibrium mixture of SO2,SO3 and O2 gases is maintained in a 11.5 L flask at a temp at which Kc=55.2 for the rxn shown below. If the number of moles of SO3 in the flask at equilibrium is twice the num of moles of SO2, how much O2 is present in the flask at equilibrium?
2SO2 (g) + O2 --> 2SO3(g)
To solve this problem, we need to use the information given about the number of moles of SO3 in the flask at equilibrium. Let's assume that the number of moles of SO2 is x, which means the number of moles of SO3 is 2x.
We are given the volume of the flask (11.5 L), which allows us to find the initial concentrations of the gases. Since the total pressure of the gases is not given, we will assume it to be 1 atm.
Using the ideal gas law, we can find the initial concentration of each gas in moles per liter (M):
Initial concentration of SO2 = (moles of SO2) / (volume of flask in liters)
Initial concentration of SO2 = x / 11.5 M
Initial concentration of SO3 = (moles of SO3) / (volume of flask in liters)
Initial concentration of SO3 = (2x) / 11.5 M
Initial concentration of O2 = (moles of O2) / (volume of flask in liters)
Since the stoichiometric coefficient in front of O2 is 1, the initial concentration is the same as its moles per liter, which is what we want to find.
Now, let's look at the equation and the equilibrium expression:
2SO2 (g) + O2 (g) --> 2SO3 (g)
Kc = [SO3]^2 / ([SO2]^2 * [O2])
Since the number of moles of SO3 at equilibrium is twice the number of moles of SO2, the equilibrium concentration of SO3 is 2x / 11.5 M, and the equilibrium concentration of SO2 is x / 11.5 M.
Let's substitute these values into the equilibrium expression:
55.2 = (2x / 11.5 M)^2 / ((x / 11.5 M)^2 * [O2])
Simplifying the equation:
55.2 = 4x^2 / (11.5 M)^2 * [O2]
Now, let's solve for [O2]:
55.2 * [(11.5 M)^2 / 4] = x^2 * [O2]
Multiply both sides by 4 / ((11.5 M)^2 * 55.2):
[O2] = 4 * 55.2 / (11.5 M)^2
Now we can substitute the value of M (11.5 L) into the equation to find the value of [O2]:
[O2] = 4 * 55.2 / (11.5)^2
Calculating the value:
[O2] ≈ 0.25 M
Therefore, there is approximately 0.25 M of O2 present in the flask at equilibrium.
To find the amount of O2 present in the flask at equilibrium, we need to follow these steps:
Step 1: Write the balanced equation for the reaction:
2SO2 (g) + O2 → 2SO3(g)
Step 2: Set up the expression for the equilibrium constant (Kc):
Kc = [SO3]^2 / ([SO2]^2 * [O2])
Given that Kc = 55.2, we can substitute this value into our equation.
Step 3: Let's assume that the initial number of moles of SO2 is "x". This means that the number of moles of SO3 is twice that, so it is "2x".
Therefore, at equilibrium, the number of moles of SO2 would be (x - 2x) = -x.
Step 4: The number of moles of O2 will be the same as the number of moles of SO2 added to the number of moles of SO3:
Moles of O2 = -x + 2x = x
Step 5: Calculate the concentration of O2 (since the volume of the flask is given):
Concentration (O2) = (Moles of O2) / (Volume of flask)
Concentration (O2) = x / 11.5 L
Step 6: Substitute the values into the equilibrium expression:
55.2 = (2x)^2 / ((x)^2 * (x / 11.5))
Step 7: Solve for x:
Simplifying the equation,
55.2 = 4 / (x / 11.5)
220.8 = x / 11.5
x = 220.8 * 11.5
x = 2540.8
Step 8: Calculate the moles of O2 using x:
Moles of O2 = x = 2540.8
Therefore, the amount of O2 present in the flask at equilibrium is approximately 2540.8 moles.
.........2SO2 + O2 ==> 2SO3
equil.....x.............2x
At equilibrium, concns are
(SO2) = (x/11.5) M
(SO3) = (2x/11.5)M
(O2) = y M
Kc = 55.2 = (2x/11.5)^2/(x/11.5)^2*y
Solve for y which gives (O2) in moles/L, then multiply by 11.5L to obtain moles.
You can do it another way by assuming any convenient number for moles SO2 (e.g., 11.5 moles) for SO2 so (SO3) = 11.5/11.5 = 1M
Then moles SO3 = 23.0 moles or (SO3) = 23.0/11.5 = 2.0M
O2 = xM and solve for x, then multiply by 11.5 to find moles O2.