SOLUTION: Sch4u1 equilibrium ice calculations worksheet - Studypool - Free Printable

Let’s solve Problem 1 step by step.

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We are given the reaction:

2 SO₂ (g) + O₂ (g) ⇌ 2 SO₃ (g)

Initial amounts:
- 4.0 mol SO₂
- 2.2 mol O₂
- Volume = 10 L
- At equilibrium, there are 2.6 mol SO₂

We need to find the equilibrium constant K.

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Step 1: Set up an ICE table (in moles first, then convert to concentration)

ICE stands for Initial, Change, Equilibrium.

Reaction:
2 SO₂ + O₂ ⇌ 2 SO₃

| Species | Initial (mol) | Change (mol) | Equilibrium (mol) |
|---------|---------------|--------------|-------------------|
| SO₂ | 4.0 | -2x | 4.0 - 2x |
| O₂ | 2.2 | -x | 2.2 - x |
| SO₃ | 0 | +2x | 2x |

We’re told that at equilibrium, SO₂ = 2.6 mol

So:

> 4.0 - 2x = 2.6
→ 2x = 4.0 - 2.6 = 1.4
→ x = 0.7

Now plug x back in to find all equilibrium moles:

- SO₂ = 2.6 mol (given)
- O₂ = 2.2 - x = 2.2 - 0.7 = 1.5 mol
- SO₃ = 2x = 2 × 0.7 = 1.4 mol

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Step 2: Convert moles to concentrations (since K uses Molarity = mol/L)

Volume = 10 L

- [SO₂] = 2.6 mol / 10 L = 0.26 M
- [O₂] = 1.5 mol / 10 L = 0.15 M
- [SO₃] = 1.4 mol / 10 L = 0.14 M

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Step 3: Write the expression for K

For the reaction:
2 SO₂ + O₂ ⇌ 2 SO₃

K = [SO₃]² / ([SO₂]² × [O₂])

Plug in the values:

K = (0.14)² / [(0.26)² × (0.15)]

Calculate numerator:
0.14² = 0.0196

Calculate denominator:
0.26² = 0.0676
0.0676 × 0.15 = 0.01014

Now divide:

K = 0.0196 / 0.01014 ≈ 1.933

Rounded to two decimal places → 1.93

✔ Matches the answer key!

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Final Answer:
[1.93]