The synthesis of ammonia is a prime example of how chemistry can be used to improve human life. Even — Analytical Chemistry Chemistry Question
Catalysts for the synthesis of ammonia
The synthesis of ammonia is a prime example of how chemistry can be used to improve human life. Even though primitive living systems had been “fixing” nitrogen to make compounds of nitrogen for hundreds of millions of years, human beings learned to prepare ammonia only about 100 years ago.
Ammonia is a source of nitrogen atom required for all amino acids and is essential in the production of fertilizer. Amino groups can be easily transformed into nitro groups found commonly in explosives such as TNT. More than 100 million tons of ammonia are produced annually worldwide, second only to sulfuric acid. However, Nature produces even more ammonia than the chemical industry. Ammonia is synthesized from nitrogen and hydrogen, however, the chemical bond of the nitrogen molecule is extremely stable, keeping ammonia from being synthesized without proper conditions or use of catalyst. In the early 20th century, Haber–Bosch method was developed for ammonia synthesis using high pressure and temperature, which is still employed in today’s chemical industry. Haber (1918) and Bosch (1931) were awarded the Nobel Prize in chemistry for these contributions.
First, let us see if the reaction is feasible from a thermodynamic standpoint. Calculate the standard entropy change of the system in the following reaction:
N2(g) + 3 H2(g) → 2 NH3(g)
The standard entropy is 191.6, 130.7, and 192.5 J K−1mol−1 for N2, H2, and NH3, respectively. Does the entropy of the system increase or decrease? If it decreases, what must be the case for the reaction to proceed spontaneously?
Model Answer
So = (2 × 192.5) – (191.6 + 3×130.7) = –198.7 J K−1 mol−1
The reaction must be exothermic and produce enough heat to increase the entropy of the surroundings and thereby offset the decrease in system entropy.
In order to see whether the reaction is likely to be exothermic, consider a similar reaction between oxygen and hydrogen to form water. Is that reaction exothermic? Match the compounds with the standard enthalpy of formation (∆fH o) in kJ mol–1 .
H2O(g) • • – 46.11
HF(g) • • –241.82
NH3(g) • • –271.1
Model Answer
Combination of hydrogen with a more electronegative element will be more exothermic.
H2O(g): – 241.82 kJ mol−1
HF(g): – 271.1 kJ mol−1
NH3(g): – 46.11 kJ mol−1
Using the value of ∆fH o you selected above, calculate the entropy change at 25 °C of the system and the surroundings combined.
Model Answer
∆Stot = ∆Ssys + ∆Ssur = ∆Ssys – ∆Hsys / T
= – 198.7 J mol−1 K−1 + 3 1 92.22 10 J mol / 298 K = + 111 J mol−1K−1
Consider now the rate of the reaction. The rate determining step of the reaction, N2(g) + 3 H2(g) → 2 NH3(g) is the atomization of the nitrogen molecule. Assuming that the activation energy of the atomization is the bond energy of the nitrogen molecule (940 kJ mol−1) and that the A factor of the rate determining step is 1×1013 sec−1, calculate the rate constant of atomization at 800 °C using the Arrhenius rate law. Calculate the rate constant at the same temperature when the activation energy is lowered by half with a catalyst.
Model Answer
NH3(g): – 46.11 kJ mol−1
k1 = A exp(– aE R T ) = 1×1013 × exp(– 3940 10 8.3145 1073 × × ) = 1.74×10–33 sec–1
k2 = A exp(– aE R T ) = 1×1013 × exp(– 3470 10 8.3145 1073 × × ) = 1.32×10–10 sec–1
k2/k1 = 7.6×1022
The amount of catalyst used by the chemical industry is enormous. More than 100 tons of catalyst are used in a factory where 1000 tons of ammonia can be produced daily. In addition to the Fe catalyst that has been used since Haber and Bosch, a Ru catalyst is used in ammonia synthesis. Metal complex binding with elemental nitrogen and hydrogen is also studied as homogeneous catalyst for ammonia synthesis in solution.
Reactions between reactants and undissolved metal catalyst can occur at the metal surface so that the catalyst surface area affects the catalysis rate. Calculate the amount of substance of nitrogen N2 adsorbed on 1.00 kg of Fe catalyst. Assume that the catalyst is composed of a 1 µm3 cube (very fine powder) and that all six faces of the cube are available for nitrogen adsorption. The density of Fe is 7.86 g cm−3 and the adsorption area for a nitrogen molecule is 0.16 nm2.
Model Answer
Mass of cube = 7.86 g cm−3 × (1.00×10–4 cm)3 = 7.86×10−15 kg
Number of cubes in 1 kg = 1.00 kg / (7.86 * 10^-15 kg) = 1.27×1014
Surface area of Fe powder = 6×10–12 m2 × 1.27×1014 = 763 m2
Area for N2 adsorption = 0.16×10–18 m2
n = area of Fe powder / area for N2 = 4.77×1021 = 7.92×10−3 mol
If a soluble, homogeneous catalyst with a molar mass of 500 g mol−1 is synthesized for nitrogen molecule binding, calculate how many nitrogen molecules bind to 1.00 kg of catalyst? Assume that one catalyst molecule binds one nitrogen molecule. Compare the result with the number of nitrogen molecules adsorbed on the Fe surface in problem 16.2.
Model Answer
1.00 kg / 0.500 kg mol−1 = 2.00 mol N(N2) = 1.20×1024
While ammonia is synthesized under high pressure and temperature in the chemical industry, natural ammonia is synthesized from atmospheric nitrogen, ~0.8 atm. Enzymes for ammonia synthesis in nature called nitrogenases are proteins with cofactors that contain Fe or Mo. The ammonia synthesis reaction by nitrogenases is an electron transfer reaction: N2(g) + 8 H+ + 8 e– → 2 NH3(g) + H2(g). 16 ATP molecules are consumed in the reaction. ATP molecule decomposes into ADP and inorganic phosphate, and releases an energy of 30.5 kJ mol–1. Calculate the energy required to synthesize 1 mol of ammonia using nitrogenase. At least 400 kJ of energy is used for the synthesis of 1 mol of ammonia in the chemical industry these days.
Model Answer
8 × 30.5 kg mol−1 = 244 kJ
E(nitrogenase) < E(chemical industry)