Question

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

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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