Question

1) Consider the dissolution of CaCO3 compound in aqueous medium.

Answer 1

Explanation:

1a) CaCO₃(s) → Ca²⁺(aq) + CO₃²⁻(aq)

1b) Remember, solids are not included in the equilibrium equation.

K = [Ca²⁺] [CO₃²⁻]

1c) Adding CO₃²⁻ ions will shift the reaction to the left, producing CaCO₃(s) until equilibrium is restored.

2) 2 SO₂(g) + O₂(g) → 2 SO₃(g)

Kc = 2.5×10¹⁰ = [SO₃]² / ([SO₂]² [O₂])

2a) SO₂(g) + ½ O₂(g) → SO₃(g)

Kc = [SO₃] / ([SO₂] [O₂]^½)

Kc² = [SO₃]² / ([SO₂]² [O₂])

Kc² = 2.5×10¹⁰

Kc ≈ 1.58×10⁵

2b) SO₃(g) → SO₂(g) + ½ O₂(g)

Kc = [SO₂] [O₂]^½ / [SO₃]

Kc = 1 / (1.58×10⁵)

Kc ≈ 6.33×10⁻⁶

2c) 3 SO₂(g) + ³/₂ O₂(g) → 3 SO₃(g)

Kc = [SO₃]³ / ([SO₂]³ [O₂]^³/₂)

Kc = ([SO₃] / ([SO₂] [O₂]^½))³

Kc = (1.58×10⁵)³

Kc ≈ 3.95×10¹⁵

3) H₂(g) + I₂(g) → 2 HI(g)

K = [HI]² / ([H₂] [I₂])

Make an ICE table.

$\left[\begin{array}{cccc}&Initial&Change&Equilibrium\\H_{2}&0.400&-0.240&0.160\\I_{2}&1.60&-0.240&1.360\\HI&0&+0.480&0.480\end{array}\right]$

K = (0.480)² / (0.160 × 1.360)

K = 1.06