Haber process: higher temperature and lower pressure proposal
A chemical plant produces ammonia by the Haber process: N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = −92 kJ mol⁻¹. The plant currently operates at 450°C and 200 atm with an iron catalyst and recycle of unreacted gases. Management proposes two changes to cut costs: raise the temperature to 550°C to increase the rate, and lower the pressure to 100 atm to reduce compression costs. The engineers provide equilibrium constants: Kp = 6.0 × 10⁻⁵ atm⁻² at 450°C and 4.0 × 10⁻⁶ atm⁻² at 550°C. At 200 atm and 450°C the equilibrium mixture contains about 35% NH₃ by mole. The catalyst is certified for operation up to 500°C; above that it sinters and loses surface area. The plant must decide whether to approve either change.
1. Use the Kp values to explain the effect of raising the temperature from 450°C to 550°C on the equilibrium yield of ammonia.
Read the worked answer
Kp falls from 6.0 × 10⁻⁵ atm⁻² at 450°C to 4.0 × 10⁻⁶ atm⁻² at 550°C, a decrease of about 15 times.
For this exothermic reaction, increasing temperature shifts the equilibrium in the endothermic direction, i.e. back to N₂ and H₂. The smaller Kp means that at equilibrium the partial pressure of NH₃ must be lower relative to the reactants.
The rate does increase at 550°C, but the yield per pass falls, so more unreacted gas must be recycled and more energy is used. In addition, 550°C exceeds the catalyst's 500°C limit and will sinter the iron catalyst.
2. Explain, using the equilibrium expression, why lowering the pressure from 200 atm to 100 atm reduces the equilibrium amount of ammonia.
Read the worked answer
For Kp = (pNH₃)² / [pN₂ (pH₂)³], write each partial pressure as p_i = x_i P. Then
Kp = P⁻² × (xNH₃)² / [xN₂ (xH₂)³].
If the pressure is halved while the mole fractions were unchanged, the calculated reaction quotient would become Q = (P/2)⁻² × ... = 4Kp. Since Q > Kp, the reaction shifts to the left, reducing NH₃.
This agrees with Le Chatelier's principle: the forward reaction reduces 4 mol of gas to 2 mol, so high pressure favours ammonia and low pressure favours the reactants.
3. Recommend whether the plant should approve either change. Justify your answer using equilibrium, rate and catalyst considerations.
Read the worked answer
Neither change should be approved as stated. Raising the temperature to 550°C lowers Kp and therefore the equilibrium yield, and it exceeds the catalyst's safe limit, causing sintering and loss of activity. Lowering the pressure to 100 atm shifts the equilibrium towards reactants because the forward reaction produces fewer gas moles; the ammonia fraction at equilibrium would fall.
The plant should keep the high pressure (about 200 atm) despite compression costs, because high pressure gives a much better equilibrium yield. It should keep the temperature near 450°C, or at most below 500°C, to balance a reasonable rate against yield and catalyst life. Rate can be improved by a more active catalyst, better heat exchange and recycle rather than by conditions that destroy the equilibrium yield.