Ammonia is one of the most important intermediates. It is used, for example, for the production of f — Physical Chemistry Chemistry Question
Haber-Bosch Process
Ammonia is one of the most important intermediates. It is used, for example, for the production of fertilizers. Usually, ammonia is produced from hydrogen and nitrogen in the Haber–Bosch process.
Table 1:
Chemical Substance ΔfHo ( kJ mol–1 K–1 ) –1 So (J mol–1 K–1 ) –1 Cpo (J mol–1 K–1 ) –1
N2 (g) 0.0 191.6 29.1
NH3 (g) – 45.9 192.8 35.1
H2 (g) 0.0 130.7 28.8
Table 2:
Chemical Substance a (Jmol–1 K–1 ) –1 b (Jmol–1 K–1 ) –1 c (Jmol–1 K–1 ) –1 d (Jmol–1 K–1 ) –1 e (Jmol–1 K–1 ) –1 f (Jmol–1 K–1 ) –1
N2 (g) 27.3 5.2·10–3 –1.7·10–9 170.5 8.1·10–2 –2.3·10–5
NH3 (g) 24.2 4.0·10–2 –8.2·10–6 163.8 1.1·10–1 –2.4·10–5
H2 (g) 28.9 –.5.8·10–4 1.9·10–6 109.8 8.1·10–2 –2.4·10–5
Write down the chemical equation for this reaction.
Model Answer
N2(g) + 3 H2(g) 2 NH3(g)
Calculate the thermodynamic properties (reaction enthalpy, entropy, and Gibbs energy) for this reaction under standard conditions. Use the values in Table 1. Is the reaction exothermic or endothermic? Is it exergonic or endergonic?
Model Answer
ΔHo = – 91.8 kJ mol–1
ΔSo = –198.1 J mol–1 K–1
ΔGo = ΔHo – T ΔSo = –32.7 kJ mol–1
The reaction is exothermic and exergonic under standard conditions.
What will happen if you mix nitrogen and hydrogen gas at room temperature? Explain your reasoning.
Model Answer
Ammonia will form instantaneously, but the activation energy for the reaction will be so high that the two gases won`t react. The reaction rate will be very low.
Calculate the thermodynamic properties (reaction enthalpy, entropy, and Gibbs energy) for this chemical reaction at 800 K and 1300 K at standard pressure. Is the reaction exothermic or endothermic? Is it exergonic or endergonic?
The temperature dependence of the heat capacity and the entropy are described by Cp(T) = a + b T + c T 2 and S(T) = d + e∙T + f∙T 2. The values of the constants a – f can be found in Table 2.
Model Answer
The enthalpy of formation is described by ΔfH(T) = ΔfHo + ∫ Cp(T) dT
For N2: ΔfH(800 K) = 15.1 kJ mol–1, ΔfH (1300 K) = 31.5 kJ mol–1.
For H2: ΔfH (800 K) = 14.7 kJ mol–1, ΔfH (1300 K) = 29.9 kJ mol–1.
For NH3: ΔfH (800 K) = – 24.1 kJ mol–1, ΔfH (1300 K) = 4.4 kJ mol–1.
This leads to a reaction enthalpy of:
ΔH(800 K) = – 107.4 kJ mol–1, ΔH(1300 K) = –112.4 kJ mol–1.
Entropy can be calculated directly with this equation..
For N2: S(800 K) = 220.6 J mol–1 K–1, S(1300 K) = 236.9 J mol–1 K–1.
For H2: S(800 K) = 159.2 J mol–1 K–1, S(1300 K) = 174.5 J mol–1 K–1.
For NH3: S(800 K) = 236.4 J mol–1 K–1, S(1300 K) = 266.2 J mol–1 K–1.
This leads to a reaction entropy of:
S(800K) = – 225.4 J mol–1 K–1, S(1300K)= – 228.0 J mol–1 K–1.
Gibbs energy is:
ΔG(800K) = 72.9 kJ mol–1, ΔG(1300K) = 184.0 kJ mol–1.
The reaction is still exothermic but now endergonic.
Calculate the mole fraction of NH3 that would form theoretically at 298.15 K, 800 K and 1300 K and standard pressure.
Assume that all the gases behave like ideal gases and that the reactants are added in the stochiometric ratio.
In an industrial process, the reaction has to be fast and result in high yields. Task 2.3 shows that the activation energy of the reaction is high and task 2.5 shows that the yield decreases with increasing temperatures. There are two ways of solving this contradiction.
Model Answer
The equilibrium constant can be calculated from Gibbs energy according to Kx(T) = exp(–ΔG(RT)–1).
This leads to the following equilibrium constants:
Kx(298.15 K) = 5.36×105,
Kx(800 K) = 1.74×10–5,
Kx(1300 K) = 4.04×10–8.
Using x_NH3^2 / (x_N2 * x_H2^3) = Kx, x_H2 = 3 * x_N2, and 1 = x_N2 + x_H2 + x_NH3 we obtain (1 - 4 * x_N2)^2 / (27 * x_N2^4) = Kx. This equation can be converted into 27 * x_N2^4 * Kx + 16 * x_N2^2 - 8 * x_N2 + 1 = 0 which has only one solution, since Kx and x_N2 are always positive.
We obtain the following table:
T K–1 | x_N2 | x_H2 | x_NH3
298.15 | 0.01570 | 0.04710 | 0.03720
800 | 0.24966 | 0.74898 | 0.00136
1300 | 0.24998 | 0.74994 | 0.00008
The reaction can proceed at lower temperatures by using a catalyst (for example iron oxide). How does the catalyst influence the thermodynamic and kinetic properties of the reaction?
Model Answer
The catalyst reduces the activation energy of the process and increases the reaction rate. The thermodynamic equilibrium is unchanged.
It is also possible to increase pressure. How does the pressure change influence the thermodynamic and kinetic properties of the reaction?
Model Answer
Higher pressures will result in a higher mol fraction of NH3, since Kx = Kp p2 increases. An increase in pressure shifts the equilibrium toward the products but does not change the reaction rate.
What are the best conditions for this reaction?
Model Answer
The best conditions are: high pressure, temperature as low as possible and the presence of a catalyst. The temperature has to be optimized such that the turnover is fast and the yield still acceptable.