Chemistry · General chemistry II · Worked example
Predict the shift when the volume is halved
For N₂O₄(g) ⇌ 2NO₂(g), Kc = 0.25 at a certain temperature, and an equilibrium mixture has [N₂O₄] = 1.00 M and [NO₂] = 0.500 M. The volume is suddenly halved. Which way does the reaction shift, and what are the new equilibrium concentrations?
Halve the volume
The amounts do not change, so every concentration doubles: [N₂O₄] = 2.00 M and [NO₂] = 1.00 M.
Compare Q with K
Before the change Q = K = 0.25; after it Q = 0.50, above K, so the reaction runs in reverse, toward N₂O₄. The left side has 1 mol of gas and the right side 2, so the shift is toward fewer moles of gas, as the principle predicts.
Set up the ICE table
If 2x mol/L of NO₂ combines, [NO₂] = 1.00 − 2x and [N₂O₄] = 2.00 + x. Substituting into K and expanding gives a quadratic.
Solve for x
The smaller root keeps [NO₂] positive: x = 0.135.
Read the new equilibrium
[N₂O₄] = 2.13 M and [NO₂] = 0.731 M, and Q is back at K. Both concentrations are higher than before the compression, but NO₂ rose by less than the factor of two the compression gave it.
Result
The reaction shifts toward N₂O₄. At the new equilibrium [N₂O₄] = 2.13 M and [NO₂] = 0.731 M.
Your turn
Which way does compressing the mixture shift 2SO₂(g) + O₂(g) ⇌ 2SO₃(g)? And H₂(g) + I₂(g) ⇌ 2HI(g)?
Show the answer and explanation
Toward SO₃; no shift for the HI reaction.
The SO₃ side has 2 mol of gas against 3 on the left, so compression favors it. H₂ + I₂ ⇌ 2HI has 2 mol of gas on each side: doubling every concentration leaves Q unchanged, so there is no shift.
Keep exploring
Equilibrium & ICE tables opens with the compressed mixture. It reports Q = 0.50 and a reverse shift, and its ICE table gives the new concentrations.
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