Question

At a certain temperature, 0.860 mol of SO3 is placed in a 1.50-L container. At equilibrium, 0.100 mol of O2 is present. Calculate Kc.

Answer 1

Answer – Given, moles of SO₃ = 0.860 , volume = 1.50 L , at equilibrium moles of O₂ = 0.100 moles

We know reaction –

2 SO₃ ---> 2 SO₂ + O₂

First we need to calculate the molarity

[SO₃] = 0.860 mole / 1.50 L = 0.573 M

[O₂] = 0.10 mole / 1.5 L = 0.0667 M

Now we need to put ICE chart

   2 SO₃ ---> 2 SO₂ + O₂

I   0.573           0         0

C    -2x            +2x     +x

E 0.573-2x      +2x     0.0667

So, x = 0.0667 M

So, at equilibrium,

[SO₃] = 0.573-2x

           = 0.573 -2*0.0667

           = 0.440 M

[SO₂] = 2x

          = 0.133 M

[O₂] = x = 0.0667 M

So, Kc = [SO₂]²[O₂] / [SO₃]²

           = (0.133)²*90.0667) / (0.44)²

           = 0.00612