Question

At a certain temperature, 0.920 mol of SO3 is placed in a 2.50-L container.

2 SO₃ (g) ⇌ 2 SO₂ (g) + O₂ (g)

At equilibrium, 0.140 mol of O2 is present. Calculate Kc.

Answer 1

At a certain temperature, 0.920 mol of SO3 is placed in a 2.50-L container.

2 SO₃ (g) ⇌ 2 SO₂ (g) + O₂ (g)

The equilibrium constant will relate product and reactants distribution. It is similar to a ratio

The equilibrium is given by

rReactants -> pProducts

Keq = [products]^p / [reactants]^r

For a specific case:

aA + bB = cC + dD

Keq = [C]^c * [D]^d / ([A]^a * [B]^b)

now...

Keq = [SO2]^2 [ O2 ] /[SO3]^2

initial conditions

[SO2] = 0

[O2] = 0

[SO3] = mol/V = 0.92/2.5 = 0.368 M

in equilibrium

[SO2] = 0 + 2x

[O2] = 0 + x

[SO3] = 0.368 -2x

and we know

At equilibrium, 0.140 mol of O2 is present. Calculate Kc.

[O2] = 0 + x = mol/V

x = 0.14/2.5 = 0.056

substitute

[SO2] = 0 + 2*0.056 = 0.112 M

[O2] = 0 + 0.056 = 0.056 M

[SO3] = 0.368 -2*0.056 = 0.256 M

Keq = [SO2]^2 [ O2 ] /[SO3]^2

Keq = (0.112 ^2)* (0.056 ) /(0.256 ^2)

Keq = 0.01071875