At a certain temperature, 0.760 mol of SOa is placed in a 5.00-L container. At equilibrium,...

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Answer 1

Answer #1

Answer:

Given moles of SO3=0.76 mol and volume=5 L

Therefore [SO3]=moles/L=0.76 mol/5L=0.152 M.

Moles of O2 at equilibrium=0.19 mol then

[O2]=0.19 mol/5L=0.038 M.

Now ICE table,

Reaction: 2SO3(g) <------> 2SO2(g) + O2(g)

Initial 0.152 0 0

Change -2x +2x +x

Equilibrium 0.152-2x 2x x

Given [O2] at equilbrium=x=0.038 M.

Therefore, [SO2]=2x=2x0.038=0.076 M

[SO3]=0.152-2x=0.152-0.076=0.076 M

Kc=[SO2]²[O2]/[SO3]²

Kc=(0.076)²(0.038)/(0.076)²

Kc=0.038 or Kc=3.8 x 10^-2.

Thanks

Answer 2

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