equilibrium constant for;
so3 <----> so2 + 1/2o2
is found to be 0.471M^1/2 @ 600deg C
after a quantity of sulfur trioxide is allowed to come to equilibrium in a 1L flask @ 600deg C, 0.2 mol oxygen was found to be present. What mass of sulfur trioxide was initially placed in the flask?
answer is 62.3g but i get 20.9g...?
I get 62.38 g SO3 which rounds to 62.4 g.
............SO3 ==> SO2 + 1/2 O2.
initial......x.......0......0
at equil, O2 = 0.2 mol; therefore, SO2 must be 0.4 mol(2*0.2) and SO3 must be x-0.4. Then
Keq = 0.471 = (SO2)(O2)1/2/(SO3)
0.471 = (0.4)(0.2)1/2/(x-0.4)
Solve for x and multiply by 80 to convert to grams.
I thought so2 would be 0.4, I didn't have so3 correct on my ice table.
I get x as 0.485;
0.471(x-0.4)=(0.4)(0.2)^1/2
0.471x-0.1884=0.04
0.471x=0.2284
But x needs to be 1.18 for so3 to be 0.78M(1.18 - 0.4) to multiply with 80, am i missing something?
Yes, you are missing something. You need work on your algebra.
0.471(x-0.4) = (0.4)(0.2)^1/2 is ok to here. Your error is in the next step. 0.2^1/2 = 0.4472 and that times 0.4 = 0.17889.
0.471x - 0.1884 = 0.17889
0.471x = 0.17889 + 0.1884 = 0.36729
x = 0.36729/0.471 = 0.7798 moles which in 1L makes it 0.7798M. (And I don't agree with 1.18 for SO3 being needed, either).
0.7798 moles x 80g/mol = 62.38 g SO3. Technically, if the problem uses ONLY 0.2 and 0.4 then there is only one significant figure and we would round 62.38g to 6E1 grams.
Your last statement is not correct. You are SOLVING for x, not x-0.4. x is what we started with (which is what the problem asks for) and x-0.4 is what it is at equilibrium. So we start with 0.7798 moles and at equilibrium it is that - 0.4
Check my answer versus Keq.
(0.4)(0.2)^1/2/(0.7798-0.4) = 0.471, right on the money.
:D thankyou very much for explaining it all,
it seems i was treating the 0.2^1/2 as 0.2*0.5 (being half of one...so wrong!).
To get 0.4472 though, 0.2 needs *2.236, it dawned to me as 2.24, bringing so3 to 62.4g after rounding.
I promise to practice my algebra.
I appreciate your brilliance.