At a certain temperature, 0.720 mol of SO3 is placed
in a 2.00-L container.
2SO3(g) <=> 2SO2(g) + O2(g)
At equilibrium, 0.100 mol of O2 is present.
Calculate Kc
....2SO3.....2SO2....O2
....0.720........0........0
......-2x.........+2x....+x
..0.7204-2x......2x.....0.1
therefore, x = 0.1M
to determine the [eq] of each , plug in 0.1 for x
[SO3] = 0.720 - (2 x 0.1) = 0.7M
[SO2] = 0.2M
[O2] = 0.1M
Kc = [SO2]^2[O2] / [SO3]^2
Kc = 0.2^2 x 0.1 / 0.7^2
Kc =8.1632*10^-3
2SO3(g) <=> 2SO2(g) + O2(g)
Initial: 0.72/2 0 0
Eq: 0.72-2x/2 = 0.26 2x/2= 0.100 x = 0.100, thus, x/2 = 0.05
Kc = 0.05*0.12 / 0.262
Kc = 7.396*10-3
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