Question

At a certain temperature, 0.800 mol of SO3 is placed in a 3.00 -L container. 2SO3(g)↽−−⇀2SO2(g)+O2(g)...

At a certain temperature, 0.800 mol of SO3 is placed in a 3.00 -L container. 2SO3(g)↽−−⇀2SO2(g)+O2(g) At equilibrium, 0.130 mol of O2 is present. Calculate ?c.

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Answer #1

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

Equillibrium expression(Kc) for this reaction is as follows

Kc = [SO2]2[O2]/[SO3]

Initial concentration

[SO3] = (0.800mol/3L)×1L = 0.2667M

[SO2] = 0

[O2] = 0

Change in concentration

[SO3] = -2x

[SO2] = +2x

[O2] = +x

Equillibrium concentration

[SO3] = 0.2667 - 2x

[SO2] = 2x

[O2] = x

Therefore,

Kc = x(2x)2/(0.2667 - 2x)2

at equillibrium ,

[O2] = (0.130mol/3L) × 1L = 0.0433M

Therefore, at equillibrium

[SO2] = 2× (0.0433M) = 0.0866M

[SO3] = 0.2667 - (2×0.0433M) = 0.1801M

Kc = 0.0433M(0.0866M)2/(0.1801M)2

Kc = 0.010

  

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