Chemical Equilibrium in Ammonia Synthesis
The Haber process for ammonia synthesis is given by: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = –92 kJ/mol. A chemical engineering student is studying the effect of conditions on the equilibrium yield of ammonia. The student has a 2.0 L reaction vessel initially containing 1.0 mol N₂ and 3.0 mol H₂ at 450°C. The equilibrium constant Kc at 450°C is 0.50 (for the reaction as written, with concentrations in mol/L). The student wants to determine the equilibrium concentrations of all species. Additionally, the student considers changing the pressure and temperature to improve the yield. The student recalls Le Chatelier's principle but needs to apply it quantitatively. Assume ideal gas behavior and that the volume remains constant. The student also has a catalyst available (iron with promoters) that speeds up the reaction but does not affect the equilibrium position.
1. Calculate the equilibrium concentrations of N₂, H₂, and NH₃ in the vessel at 450°C. Show your ICE table and the algebra used to solve for the equilibrium concentrations.
Read the worked answer
Step 1: Set up ICE table.
Initial concentrations:
[N₂]₀ = 1.0 mol / 2.0 L = 0.50 M.
[H₂]₀ = 3.0 mol / 2.0 L = 1.50 M.
[NH₃]₀ = 0 M.
Let x = change in [N₂] (mol/L). Then:
N₂: 0.50 – x
H₂: 1.50 – 3x
NH₃: 0 + 2x
Step 2: Write equilibrium expression.
Kc = [NH₃]² / ([N₂][H₂]³) = 0.50.
Substitute: (2x)² / ((0.50 – x)(1.50 – 3x)³) = 0.50.
Step 3: Simplify.
Numerator: 4x².
Denominator: (0.50 – x)(1.50 – 3x)³.
Note that 1.50 – 3x = 3(0.50 – x). So denominator = (0.50 – x) * 27(0.50 – x)³ = 27(0.50 – x)⁴.
Thus: 4x² / [27(0.50 – x)⁴] = 0.50.
Multiply both sides: 4x² = 0.50 * 27 * (0.50 – x)⁴ = 13.5 (0.50 – x)⁴.
Divide by 2: 2x² = 6.75 (0.50 – x)⁴.
Or x² = 3.375 (0.50 – x)⁴.
Step 4: Solve for x.
Take square root: x = sqrt(3.375) * (0.50 – x)² ≈ 1.837 * (0.50 – x)².
This is a nonlinear equation. Let y = 0.50 – x. Then x = 0.50 – y.
Equation: 0.50 – y = 1.837 y².
Rearrange: 1.837 y² + y – 0.50 = 0.
Solve quadratic: y = [-1 + sqrt(1 + 41.8370.50)] / (2*1.837) = [-1 + sqrt(1 + 3.674)] / 3.674 = [-1 + sqrt(4.674)] / 3.674 ≈ [-1 + 2.162] / 3.674 = 1.162 / 3.674 ≈ 0.316.
Thus y ≈ 0.316 M. Then x = 0.50 – 0.316 = 0.184 M.
Step 5: Calculate equilibrium concentrations.
[N₂] = 0.50 – x = 0.316 M.
[H₂] = 1.50 – 3x = 1.50 – 0.552 = 0.948 M.
[NH₃] = 2x = 0.368 M.
Step 6: Check.
Kc = (0.368)² / (0.316 * 0.948³) ≈ 0.1354 / (0.316 * 0.852) ≈ 0.1354 / 0.269 ≈ 0.503, close to 0.50 (rounding).
Answer: [N₂] ≈ 0.32 M, [H₂] ≈ 0.95 M, [NH₃] ≈ 0.37 M.
2. The student wants to increase the equilibrium yield of NH₃. Using Le Chatelier's principle, predict the effect of (a) increasing the total pressure by compressing the mixture, and (b) increasing the temperature. Explain your reasoning in terms of the reaction quotient and equilibrium shift. No calculations are required, but refer to the reaction stoichiometry and ΔH.
Read the worked answer
(a) Increasing total pressure:
The reaction has 4 moles of gas on the left (1 N₂ + 3 H₂) and 2 moles on the right (2 NH₃). Increasing pressure favors the side with fewer gas molecules to reduce the pressure. Therefore, the equilibrium shifts to the right, increasing NH₃ yield. Quantitatively, the reaction quotient Qc = [NH₃]²/([N₂][H₂]³). When volume decreases, all concentrations increase, but the denominator increases more because it has a higher total power (4 vs 2). Thus Qc becomes smaller than Kc, so the reaction proceeds forward to re-establish equilibrium.
(b) Increasing temperature:
The reaction is exothermic (ΔH = –92 kJ/mol). According to Le Chatelier's principle, increasing temperature favors the endothermic direction, which is the reverse reaction (decomposition of NH₃). Therefore, the equilibrium shifts left, decreasing NH₃ yield. The equilibrium constant Kc decreases with increasing temperature for an exothermic reaction. Thus, higher temperature reduces the yield, although it may increase the reaction rate. In practice, a compromise temperature (around 450°C) is used to balance yield and rate.
Answer: Increasing pressure shifts equilibrium right, increasing NH₃ yield; increasing temperature shifts equilibrium left, decreasing NH₃ yield.
3. The student then adds the iron catalyst. Explain how the catalyst affects the rate of reaction and the equilibrium position. Also, explain why the catalyst is necessary in the industrial process even though it does not change the equilibrium yield.
Read the worked answer
Effect on rate:
The iron catalyst provides an alternative reaction pathway with a lower activation energy. This increases the rate of both the forward and reverse reactions equally, so equilibrium is reached faster. The catalyst does not alter the equilibrium constant or the position of equilibrium; it only speeds up the attainment of equilibrium.
Effect on equilibrium position:
No effect. The equilibrium concentrations remain the same as calculated without the catalyst. The catalyst lowers the activation energy for both forward and reverse reactions by the same amount, so the thermodynamics (ΔG, K) are unchanged.
Why necessary industrially:
Without a catalyst, the reaction is extremely slow at the moderate temperatures used (around 450°C). The uncatalyzed reaction would take an impractically long time to reach equilibrium, making the process uneconomical. The catalyst allows the reaction to proceed at a commercially viable rate, so that a significant amount of ammonia can be produced in a reasonable time. Even though the equilibrium yield is not increased, the faster approach to equilibrium means more product is obtained per unit time, which is essential for industrial production.
Answer: The catalyst speeds up both forward and reverse reactions equally, reducing the time to reach equilibrium without changing the equilibrium position. It is necessary because without it the reaction is too slow to be practical industrially.