Reaction: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g)

Identify components:

• Products: SO₃ (coefficient: 2)
• Reactants: SO₂ (coefficient: 2), O₂ (coefficient: 1)
• All are gases → all included

K_c expression:

$K_c = \frac{[\text{products}]}{[\text{reactants}]}$

Apply coefficients as exponents:

$K_c = \frac{[SO3]^2}{[SO2]^2[O2]}$

Answer:

$K_c = \frac{[SO3]^2}{[SO2]^2[O2]}$

Key points:

• Products (SO₃) in numerator
• Reactants (SO₂, O₂) in denominator
• Coefficient 2 becomes exponent 2
• All concentrations at equilibrium
• Unitless (convention)

Solution:

Equilibrium expression: K_c = [NH₃]² / ([N₂][H₂]³)

Calculate K_c: K_c = (0.40)² / [(0.50)(0.20)³] K_c = 0.16 / [(0.50)(0.008)] K_c = 0.16 / 0.004 K_c = 40

Interpretation: K_c > 1 means the equilibrium favors products (ammonia formation).

Given:

• Reaction: N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
• K_c = 0.500
• T = 400°C = 673 K
• R = 0.08206 L·atm/(mol·K)

Relationship:

$K_p = K_c(RT)^{\Delta n}$

Calculate Δn:

Δn = moles gas products - moles gas reactants

Products: 2 mol NH₃ Reactants: 1 mol N₂ + 3 mol H₂ = 4 mol

Δn = 2 - 4 = -2

Calculate RT:

RT = (0.08206)(673) = 55.2 L·atm/mol

Calculate K_p:

$K_p = K_c(RT)^{\Delta n}$

$K_p = (0.500)(55.2)^{-2}$

$K_p = (0.500) \times \frac{1}{(55.2)^2}$

$K_p = (0.500) \times \frac{1}{3047}$

$K_p = 1.64 \times 10^{-4}$

Answer: K_p = 1.64 × 10⁻⁴

Interpretation:

K_p < K_c because Δn < 0

• Decreasing moles of gas
• Pressure-based K is smaller
• Both indicate reactants favored (K < 1)

Solution:

Relationship between K_p and K_c: K_p = K_c(RT)^Δn

where Δn = (moles of gaseous products) - (moles of gaseous reactants)

Calculate Δn: Products: 2 moles SO₃ Reactants: 2 moles SO₂ + 1 mole O₂ = 3 moles Δn = 2 - 3 = -1

Solve for K_c: K_c = K_p / (RT)^Δn K_c = K_p × (RT)^(-Δn) K_c = 3.0 × 10⁴ × [(0.08206)(700)]^(1) K_c = 3.0 × 10⁴ × 57.4 K_c = 1.7 × 10⁶

Note: K_c > K_p when Δn < 0

Given:

• Reaction: H₂(g) + I₂(g) ⇌ 2HI(g)
• At equilibrium:
  • [H₂] = 0.20 M
  • [I₂] = 0.20 M
  • [HI] = 1.60 M
• T = 500 K

(a) Calculate K_c

K_c expression:

$K_c = \frac{[HI]^2}{[H2][I2]}$

Substitute values:

$K_c = \frac{(1.60)^2}{(0.20)(0.20)}$

$K_c = \frac{2.56}{0.040}$

$K_c = 64$

Answer (a): K_c = 64

(b) Reversed reaction K_c

Reversed reaction: 2HI(g) ⇌ H₂(g) + I₂(g)

Rule: K_reverse = 1/K_forward

$K_{c,\text{reverse}} = \frac{1}{K_{c,\text{forward}}}$

$K_{c,\text{reverse}} = \frac{1}{64}$

$K_{c,\text{reverse}} = 0.0156$

Answer (b): K_c = 0.0156 or 1.56 × 10⁻²

Interpretation:

Forward reaction (K = 64):

• Products strongly favored
• Equilibrium lies far right
• HI formation favored

Reverse reaction (K = 0.0156):

• Reactants strongly favored
• Equilibrium lies far left
• HI decomposition unfavored

Note: K_forward × K_reverse = 64 × 0.0156 = 1 ✓