There are a couple of ways to approach this. The simpler approach may be to change Kp to Kc, convert grams SO3 and grams O2 to M, set up an ICE(you need only the E line) chart and substitute into the Kc expression, solve for (SO2) then change to grams.
(SO3) = mols/L = 1.58/2*80 = ?
(O2) = mols/L = 0.136/2*32
........2SO2 + O2 ==> 2SO3
E........x.0.00213....0.00988
The second approach is to use Kp as is. Convert grams SO3 and grams O2 to mols and plug into PV = nRT; solve for pO2 and pSO3 separately. Then construct an ICE chart (you need only the E line) and substitute into Kp expression and solve for pSO2. Plug that back into PV = nRT to solve for n and convert to grams.
For 2 SO2(g) + O2(g) 2 SO3(g), Kp = 3.0 104 at 700 K. In a 2.00-L vessel the equilibrium mixture contains 1.58 g of SO3 and 0.136 g of O2. How many grams of SO2 are in the vessel?
1 answer