# Chemistry

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At a certain temperature, 0.700 mol of SO3 is placed in a 3.50-L container.
At equilibrium, 0.100 mol of O2 is present. Calculate Kc.
2SO3=2SO2 +)2

• Chemistry -

(SO3) = 0.700/3.50 = 0.2 mol
.........2SO3 ==> 2SO2 + O2
I........0.200.....0......0
C........-2x......2x......x
E......0.200-2x....2x.....x
At equil, x = 0.1 mol.

Calculate the concns of each and substitute into the Kc expression; calculate Kc.

• Chemistry -

how do i find the concentrations?

• Chemistry -

The problem tells you that at equilibrium the amount of O2 = 0.1 mol.
That is in 3.50L; therefore, (O2) = 0.1mol/3.50L = x = 0.0286M
Then (SO3) = 0.200-2x
(SO2) = 2x
(O2) = x = 0.0286

• Chemistry -

Where do you substitute Kc into the expression?

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