Sulfur trioxide decomposes into sulfur dioxide and oxygen in an equilibrium. You have a 3.00 L vessel that is charged with 0.755 mol of SO3. At equilibrium, the amount of SO3 is 0.250 mol.

2 SO3 (g) ----> 2 SO2 (g) +O2 (g)

A) calculate the equilibrium concentrations of SO2 and O2.
B) calculate the value of Kc for the equilibrium.

Please help me. I don't know where to start.

3 answers

Calculate concentrations.
Set up an ICE chart.
Solve. I like to do the concentrations first but it may be easier to do it in mols first, THEN make the conversion. The only problem is that we ofte forget to mak the conversion.
.........2SO3 ==> 2SO2 + O2
I......0.755........0.....0
C.......-2x........2x.....x
E....0.755-2x.......2x....x

The problem tells you that 0.755-2x = 0.250. Solve for x, then evaluate 2x. Then convert all of the mols into concentrations in mols/L. You have mols and L = 3.00 L. Finally, substitute concentrations into the Kc expression and solve for Kc. Post your work if you get stuck.

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