Question

Sulfur dioxide and nitrogen dioxide react at 298 K via the reaction

SO2(g)+NO2(g)⇌SO3(g)+NO(g)

A system at equilibrium contains 0.369 mol SO2 , 0.369 mol NO2 , 0.654 mol SO3 , and 0.718 mol NO. O2 is added to the container and NO reacts completely with the O2. Calculate the amount of SO3 in the container after the system returns to equilibrium.

Answer 1

O₂ reacts increasing the NO₂ to 0.718 moles. At the return to equilibrium,

the amount SO₃ in the container approximately 0.931 moles.

How can the equilibrium values indicate the amount of SO₃?

The equilibrium constant for the reaction aA + bB ⇌ cC + dD is presented as follows;

$K_c = \mathbf{\dfrac{[C]^c_{eq} +[D]^d_{eq} }{[A]^a_{eq} +[B]^b_{eq} }}$

The equilibrium constant for the reaction, whereby the volume is same for the contents is therefore;

$K = \dfrac{0.654 \times 0.718}{0.369 \times 0.369} \approx 3.45$

Given that the NO reacts with the O₂ as follows;

2NO + O₂ $\longrightarrow$ 2NO₂

We have;

The number of moles of NO₂ added = 0.718 moles

The new number of moles are therefore;

• $K = \mathbf{ \dfrac{(0.654 + x) \times (0+ x)}{(0.369-x) \times (0.369 + 0.718-x)}} \approx 3.45$

Which gives;

0 = (0.654 + x) × x - 3.45 × (0.369 × (0.369 + 0.718)

2.45·x² - 5.6772·x + 1.384 = 0

$x = \dfrac{5.6772 \pm \sqrt{(-5.6772)^2 - 4 \times 2.45 \times (1.3834)} }{2\times 2.45}$

x ≈ 0.2768 or x ≈ 2.04

The possible number of moles of SO₃ in the container after equilibrium is reestablished is therefore;

n ≈ 0.654 + 0.2768 ≈ 0.931

• The number of moles of SO₃ in the container after returning to equilibrium is 0.931 moles

Learn more about equilibrium constant here:

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