Temperature, pressure, Kp and Le Chatelier’s principle
Answers are for parents and teachers.
1
A reaction takes 12 min at 60 °C. If its rate doubles for every 10 °C, how long would it take at 30 °C?
From 60 °C down to 30 °C is three steps, so the rate is 2³ = 8 times smaller and the time is 12 × 8 = 96 min.
2
For 2NH₃ ⇌ N₂ + 3H₂ the partial pressures are P(NH₃) = 0.20, P(N₂) = 0.10 and P(H₂) = 0.20 atm. Find Kp and the total pressure.
Kp = P(N₂) × P³(H₂) ÷ P²(NH₃) = 0.10 × 0.0080 ÷ 0.040 = 0.00080 ÷ 0.040 = 0.020. Total = 0.20 + 0.10 + 0.20 = 0.50 atm.
3
2SO₃ ⇌ 2SO₂ + O₂ is endothermic as written. Predict the shift when (a) the mixture is heated, (b) the pressure is raised.
(a) forward, because the forward reaction absorbs heat; (b) backward, toward the side with fewer moles of gas (2 mol against 3 mol).
4
A reaction has a forward activation energy of 70 kJ/mol and a backward one of 95 kJ/mol. Find ΔH. A catalyst lowers the forward value to 45 kJ/mol: what is the backward value now?
ΔH = 70 − 95 = −25 kJ/mol. The catalyst does not change ΔH, so the backward value is 45 + 25 = 70 kJ/mol.
5
The reaction 2SO₂ + O₂ ⇌ 2SO₃ is exothermic. A student wants the largest yield and suggests cooling the vessel as far as possible. Why is this not done in practice? Answer in two sentences.
At a very low temperature the reaction is far too slow to be useful. So a middle temperature is used, with a V₂O₅ catalyst, as a compromise between speed and yield.