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:
2SO2 + O2 <> 2SO3.
If the # of moles of SO2 and SO3 are =, how many moles of O2 are present?
I let SO2 and SO3 =x and O2=y. Since SO2 and SO3 will cancel in the equil. eqn, I got 1.81 X 10-2 mol O2 but book says 0.208???
Thanks for sharing your work and the answer. It helps tremendously. There are a number of problems and not all are yours.
2SO2 + O2 ==> 2SO3
(SO2) = (2x/11.5)M
(SO3) = (2x/11.5)M
O2 = (x/11.5)M
Kc = 55.2 = (SO3)^2/(SO2)^2(O2)
55.2 = (2x/11.5)^2/(2x/11.5)^2(x/11.5) = 55.2
The book answer is for moles = 0.208, which is correct. If you solve it as I indicated above, x = moles = 0.208.
If you solve it as you indicated, your x = molarity (since molarity is what is substituted into Kc), so your answer of 0.0181 is for molarity. To find moles, M x L = moles = 0.0181 x 11.5L = 0.208 moles.
To solve this problem, we can start by setting up an expression for the equilibrium constant, Kc:
Kc = [SO3]^2 / ([SO2]^2 * [O2])
Since the number of moles of SO2 and SO3 are equal, we can assume the concentrations of SO2 and SO3 to be x moles/L each. Therefore, [SO2] = x and [SO3] = x.
To find the concentration of O2, we need to express it in terms of x. Since the volume of the flask is given as 11.5 L, and the total moles of gas present is x + x + y = 2x + y, we can set up the ideal gas law equation to relate the concentration of O2 to the number of moles (y):
(number of moles of O2) / (volume of flask in L) = (concentration of O2)
y / 11.5 = [O2]
We can substitute these values into the expression for Kc:
Kc = (x)^2 / ((x)^2 * (y / 11.5))
Simplifying the equation, we get:
Kc = 11.5 / (x^2 * y)
Now, we can rearrange the equation to solve for y:
y = 11.5 / (Kc * (x^2))
Given that Kc = 55.2 and assuming x = 1 mol/L for simplicity, we can substitute these values into the equation to find y:
y = 11.5 / (55.2 * (1^2))
y = 11.5 / 55.2
y ≈ 0.208 moles
Therefore, the number of moles of O2 present is approximately 0.208, which is consistent with the answer given in the book.