O₂ reacts increasing the NO₂ to 0.718 moles. At the return to equilibrium,
the amount SO₃ in the container approximately <u>0.931 moles</u>.
<h3>How can the equilibrium values indicate the amount of SO₃?</h3>
The equilibrium constant for the reaction aA + bB ⇌ cC + dD is presented as follows;
The equilibrium constant for the reaction, whereby the volume is same for the contents is therefore;
Given that the NO reacts with the O₂ as follows;
2NO + O₂ 2NO₂
We have;
The number of moles of NO₂ added = 0.718 moles
The new number of moles are therefore;
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 ≈ 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 <u>0.931 moles</u>
Learn more about equilibrium constant here:
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