Equilibrium Calculations

Question

 0.2000 mol each of SO₂, O₂ and SO₃ is placed in a 2.000 L flask and allowed to come to equilibrium. The equilibrium [SO₂] is found to be 0.0040 mol/L. What is the value of K_eqfor the reaction?

2 SO_2(g) + O_2(g)  2 SO_3(g)

Answer 1

I just wanted to add an alternate approach to that provided by @Anthony T.

Set up an ICE table using molar concentrations to visualize the changes taking place...

[SO₂] = 0.2000 mol / 2.000 L = 0.1000 M

[O₂] = 0.2000 mol / 2.000 L = 0.1000 M

[SO₃] = 0.2000 mol / 2.000 L = 0.1000 M

2SO₂ + O₂ <==> 2SO₃

0.1000....0.1000.....0.1000....Initial

-2x...........-x............+2x.........Change

0.0040......?..............?..........Equilibrium

Since we know that 0.1000 - 2x = 0.0040, we solve for x, which is x = 0.048. We can now fill in the equilibrium concentrations for all of the species present...

[SO₂] = 0.0040 M

[O₂] = 0.1000 - 0.048 = 0.052 M

[SO₃] = 0.100 + 0.096 = 0.196 M

The equilibrium expression is

Keq = [SO₃]² / [SO₂]²[O₂]

Plugging in the equilibrium concentrations and solving for Keq, we have...

Keq = [0.196]² / [0.0040]²[0.052]

Keq = 46173 = 4.617x10⁴

Answer 2

From the balanced equation, we see that for every mole of SO2 that reacts, 1/2 mole of O2 is used up, and 1 mole of SO3 is produced.

Set up and ICE table.

                               S02          O2       SO3

Initial                     0.2         0.2        0.2 

Change              - 0.192    -0.096   +0.192

Equilibrium          0.008   0.104    +0.392

The values in the table are in absolute moles, not concentrations

The equilibrium concentrations are as follows: SO2 (0.004) given,

O2 ( 0.104/2L = 0.052) , SO3 (0.392/2L  = 0.196 all in moles/L.

K_C = [SO3}^2 / [SO2]^2 x [O2]

= 0.196^2 / (0.004 ^2 x 0.052) = 46173

Check math