Temperature, pressure, Kp and Le Chatelier’s principle
الأسبوع 12 · الباب الثالث (٣): تأثير درجة الحرارة ونظرية التصادم وطاقة التنشيط والضغط وثابت الاتزان Kp وقاعدة لوشاتيليه والعوامل الحفازة والضوء · 3 دروس
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Pressure, Kp and Le Chatelier’s principle
Chemistry Sec 3 Term 1, Chapter 3: pp. 61–65
GoalWrite and use Kp for gases, and predict the shift of an equilibrium with Le Chatelier’s principle.
New words
partial pressure · the pressure one gas of a mixture would give if it filled the vessel alone; the total pressure is the sum of the partial pressuresKp · the equilibrium constant of a gas reaction written with partial pressures, in the same way as KcLe Chatelier’s principle · when a system at equilibrium is disturbed, it shifts in the direction that partly reduces the change
Explanation
In a mixture of gases each gas has a partial pressure, and the total pressure is the sum of them, here in atm. For gas equilibria the concentrations can be replaced by partial pressures, and the constant is Kp, written exactly like Kc: for N₂ + 3H₂ ⇌ 2NH₃, Kp = P²(NH₃) ÷ (P(N₂) × P³(H₂)). At one temperature Kp does not change when the pressures change. Le Chatelier, a French scientist, summed up the shifts of an equilibrium: if a system at equilibrium is disturbed by a change in concentration, pressure or temperature, it shifts in the direction that partly reduces the change. Adding a reactant, or removing a product, shifts it forward. Raising the pressure shifts it toward the side with fewer moles of gas, and nothing happens if both sides have the same number. Raising the temperature shifts it toward the endothermic direction, and lowering it toward the exothermic one. The word is “partly”: the change is reduced, never cancelled. For ammonia, 4 mol of gas (1 + 3) give 2 mol, so a higher pressure raises the yield of NH₃; it is the yield, not the speed, that the pressure argument is about.
N₂ + 3H₂ ⇌ 2NH₃, ΔH = −92 kJ
Position shifts
Reason
add N₂
forward
uses up some of the added N₂
remove NH₃
forward
replaces some of the NH₃ removed
raise the pressure
forward
4 mol of gas become 2 mol
raise the temperature
backward
the backward direction absorbs heat
add a catalyst
no shift
both rates rise equally
Worked examples
ƒKp of N₂O₄ ⇌ 2NO₂
Kp = P²(NO₂) ÷ P(N₂O₄) = 0.60² ÷ 0.90 = 0.40
Kp from partial pressures. For N₂O₄ ⇌ 2NO₂ the partial pressures are P(NO₂) = 0.60 atm and P(N₂O₄) = 0.90 atm. Kp = P²(NO₂) ÷ P(N₂O₄) = 0.60² ÷ 0.90 = 0.36 ÷ 0.90 = 0.40. The total pressure is 0.60 + 0.90 = 1.50 atm.
Three gases. For 2SO₂ + O₂ ⇌ 2SO₃ the partial pressures are P(SO₂) = 0.20, P(O₂) = 0.10 and P(SO₃) = 0.40 atm. Kp = P²(SO₃) ÷ (P²(SO₂) × P(O₂)) = 0.16 ÷ (0.04 × 0.10) = 0.16 ÷ 0.004 = 40. The total pressure is 0.20 + 0.10 + 0.40 = 0.70 atm.
Equilibrium
Compress
Heat
H₂ + I₂ ⇌ 2HI
no shift
depends on ΔH
N₂O₄ ⇌ 2NO₂ (endothermic)
backward
forward
2SO₂ + O₂ ⇌ 2SO₃ (exothermic)
forward
backward
Predict the shift. (a) H₂ + I₂ ⇌ 2HI has 2 mol of gas on each side, so a change of pressure causes no shift. (b) N₂O₄ ⇌ 2NO₂, endothermic as written: compressing shifts it backward, to the side with 1 mol, and heating shifts it forward. (c) 2SO₂ + O₂ ⇌ 2SO₃ is exothermic: lowering the volume, which raises the pressure, shifts it forward (3 mol to 2 mol), and heating shifts it backward. Each answer is one sentence of cause and effect, not a calculation.
Class activity
Two-room crowd model, paper tokens and two boxes, with a grown-up. Rule for each round: a quarter of the tokens in room A move to room B, and half of the tokens in room B move to room A. Start with 12 tokens in A and 6 in B: 3 move each way, so nothing changes (equilibrium). Now add 6 tokens to B. Play several rounds, rounding to whole tokens, and count A and B after each. See what happens: the system settles near 16 in A and 8 in B. B went from 6 to 12 and came back only to 8: the change was partly reduced, not cancelled. That is Le Chatelier’s principle.
Practice
1
For 2NO + O₂ ⇌ 2NO₂ the partial pressures are P(NO) = 0.20, P(O₂) = 0.050 and P(NO₂) = 0.30 atm. Find Kp and the total pressure.
N₂ + 3H₂ ⇌ 2NH₃ (ΔH = −92 kJ) is at equilibrium. Predict the shift (forward, backward or none) when (a) NH₃ is removed, (b) the mixture is heated, (c) the pressure is lowered by enlarging the vessel.
(a) forward, to replace NH₃; (b) backward, because the forward reaction gives out heat; (c) backward, toward the side with more moles of gas (4 mol).
4
Does a change of pressure shift the position of H₂(g) + Cl₂(g) ⇌ 2HCl(g)? Give the reason.
No. There are 2 mol of gas on each side, so a change of pressure affects both sides equally.