Given:

• Reaction: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
• K_c = 4.0
• Current: [SO₂] = 0.40 M, [O₂] = 0.30 M, [SO₃] = 1.2 M

Calculate Q:

K/Q expression:

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

Substitute values:

$Q = \frac{(1.2)^2}{(0.40)^2(0.30)}$

$Q = \frac{1.44}{(0.16)(0.30)}$

$Q = \frac{1.44}{0.048}$

$Q = 30$

Answer: Q = 30

Compare Q and K:

Q = 30, K = 4.0

Q > K

Meaning:

• Too many products (SO₃) relative to equilibrium
• System will shift LEFT (reverse)
• Consume SO₃, make more SO₂ and O₂

Direction: ← Reverse (toward reactants)

Summary:

| Value | Amount | |-------|--------| | Q | 30 | | K | 4.0 | | Comparison | Q > K | | Direction | Reverse (←) |

Until Q = K, reaction proceeds in reverse direction.

Solution:

(a) Adding N₂ (reactant):

• Equilibrium shifts right (toward products)
• System consumes added N₂ to re-establish equilibrium
• More NH₃ is produced

(b) Removing NH₃ (product):

• Equilibrium shifts right (toward products)
• System produces more NH₃ to replace what was removed
• More N₂ and H₂ are consumed

(c) Increasing pressure:

• Count gas molecules: Reactants = 4 moles (1 N₂ + 3 H₂), Products = 2 moles NH₃
• Equilibrium shifts toward fewer gas molecules (right, toward products)
• This reduces pressure by decreasing total moles of gas

(d) Increasing temperature:

• Reaction is exothermic (heat is a product)
• Adding heat shifts equilibrium left (toward reactants)
• Think of heat as a product: adding a product shifts left
• K decreases as temperature increases for exothermic reactions

Reaction: N₂O₄(g) ⇌ 2NO₂(g) ΔH° = +57.2 kJ (endothermic)

(a) Add N₂O₄

Type: Concentration change

Effect:

• Adding reactant
• Shifts RIGHT (toward products)
• System consumes excess N₂O₄
• Makes more NO₂

Answer: Shifts right, increases [NO₂]

(b) Decrease volume

Type: Pressure/volume change

Count moles gas:

• Left: 1 mole N₂O₄
• Right: 2 moles NO₂
• Δn = 2 - 1 = +1

Decrease volume = increase pressure

Effect:

• Shifts toward fewer moles
• Shifts LEFT (toward N₂O₄)
• Relieves pressure by reducing total moles

Answer: Shifts left, decreases [NO₂]

(c) Increase temperature

Type: Temperature change (changes K!)

Reaction is endothermic: Heat is reactant

N₂O₄(g) + heat ⇌ 2NO₂(g)

Increase T:

• More heat available
• Shifts RIGHT (toward products)
• Uses added heat
• K increases

Answer: Shifts right, increases [NO₂], K increases

(d) Add catalyst

Type: Catalyst

Effect:

• Speeds up both forward and reverse equally
• NO shift in position
• NO change in K
• Reaches equilibrium faster only

Answer: No shift, no change in concentrations or K

Summary:

| Change | Direction | [NO₂] | K Changes? | |--------|-----------|-------|------------| | (a) Add N₂O₄ | Right → | Increases | No | | (b) Decrease volume | Left ← | Decreases | No | | (c) Increase T | Right → | Increases | Yes (↑) | | (d) Add catalyst | No shift | No change | No |

Solution:

(a) Calculate Q and compare to K: Q = [PCl₃][Cl₂] / [PCl₅] Q = (0.10)(0.10) / 0.20 = 0.010 / 0.20 = 0.050

Compare: Q = 0.050 vs K_c = 0.042 Q > K, so reaction shifts left (toward reactants)

(b) Calculate equilibrium concentrations:

Let x = change in concentration

| | PCl₅ | PCl₃ | Cl₂ | |----------|-------|-------|-------| | Initial | 0.20 | 0.10 | 0.10 | | Change | +x | -x | -x | | Equil. | 0.20+x| 0.10-x| 0.10-x|

K_c = (0.10-x)(0.10-x) / (0.20+x) = 0.042

(0.10-x)² = 0.042(0.20+x) 0.010 - 0.20x + x² = 0.0084 + 0.042x x² - 0.242x + 0.0016 = 0

Using quadratic formula: x = 0.0068 M

Equilibrium concentrations: [PCl₅] = 0.20 + 0.0068 = 0.21 M [PCl₃] = 0.10 - 0.0068 = 0.093 M [Cl₂] = 0.10 - 0.0068 = 0.093 M

Reaction: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) ΔH° = -198 kJ (exothermic)

Goal: Maximize SO₃ production

(a) Temperature choice

Reaction is exothermic: Heat is product

2SO₂(g) + O₂(g) ⇌ 2SO₃(g) + heat

Low temperature:

• Shifts RIGHT (toward products)
• Produces heat (relieves cold stress)
• Favors SO₃ formation
• K is larger at low T

High temperature:

• Shifts LEFT (toward reactants)
• Consumes heat
• Favors SO₂ and O₂
• K is smaller at high T

Answer (a): Low temperature favors SO₃

(b) Pressure choice

Count moles gas:

• Left: 2 + 1 = 3 moles
• Right: 2 moles
• Δn = 2 - 3 = -1

High pressure:

• Shifts toward fewer moles
• Shifts RIGHT (toward SO₃)
• Relieves pressure
• Favors products

Low pressure:

• Shifts toward more moles
• Shifts LEFT
• Favors reactants

Answer (b): High pressure favors SO₃

(c) Trade-offs

Temperature trade-off:

Low T (thermodynamics):

• ✓ Better yield (more SO₃)
• ✓ Higher K
• ✗ Reaction VERY slow
• ✗ Takes too long industrially

High T (kinetics):

• ✗ Lower yield (less SO₃)
• ✗ Lower K
• ✓ Much faster reaction
• ✓ Reaches equilibrium quickly

Industrial compromise:

• Use moderate temperature (~450°C)
• Acceptable yield
• Reasonable rate
• Use catalyst (V₂O₅) to speed up
• Multiple passes through reactor

Pressure:

• High pressure favored (1-2 atm)
• No major trade-off
• Equipment cost for very high pressure

Summary - Contact Process for SO₃:

| Factor | Ideal (equilibrium) | Industrial choice | Reason | |--------|---------------------|-------------------|---------| | Temperature | Low | Moderate (450°C) | Balance yield vs rate | | Pressure | High | Moderate (1-2 atm) | Good yield, low cost | | Catalyst | Yes | V₂O₅ | Speeds reaction at moderate T |

Key lesson: Industrial processes balance thermodynamics (K) and kinetics (rate)!