Case of Study

Question 1

These are the thermodynamics of the Haber-Bosh process
What does ΔH and ΔG tell us?

Answer 1

• ΔH is negative = exothermic process
• ΔG is negative = reaction is spontaneous

Question 2

If the thermodynamic for the Haber-Bosh process are favourable, why is this reaction hard to undertake?

Answer 2

• Due to kinetics
• Activation energy for the uncatalysed reaction is really high

Question 3

How can we increase the rate of the Haber-Bosh Process?

Answer 3

• Using a catalyst - to lower the activation energy
• Cannot increase temperature as it is an exothermic reaction (favour LHS of reaction)
• Increase pressure, as it favours the side of the reaction with the fewst molecules

Question 4

What type of Catalyst is used for the Haber-Bosh Process?

Answer 4

• Fe acts as a catalyst, potash and alumina as promoters
• This is one of the very few examples in heterogeneous catalyst where metal catalyst is not supported

Question 5

N₂ activation/bond dissociation: 945 kJ/mol
What are the different modes of adsorption for nitrogen?

Answer 5

• End-on OR Side-on (depending on the geometry of the catalyst)
• Sigma-dontion from the lone pairs on the nitrogen to the empty orbitals of the metals
• Absorption of the molecule
• pi-back donation from the metal to the anti-bonding orbital of the nitrogen, weakening the triple N-N bond so it breaks

Question 6

H₂ activation/bond dissociation: 436 kJ/mol
How does hydrogen absorb onto the surface of the catalyst?

Answer 6

• Only side on (as low probability of finding e- at end) where sigma donation occurs to the metal catalyst

Question 7

Why is Fe + promoters (alumina + potash) used as the catalyst

Answer 7

• They have empty orbitals allowing for sigma donation from nitrogen and hydrogen (balance between being electron rich but also accepting electrons)

Question 8

Why are the promoters potash (K₂O) and Alumina used on Fe catalysed in the Haber-Bosh process

Answer 8

• Alumina: structural promoter - allowing for high thermal stability for the catalyst and prevents sintering
• Potash: Electronic promoter - donates electron density to the Fe catalyst, allowing for more π back-donation

Question 9

Sketch a relative Leonard-Jones potential diagram for the catalyst (K/Fe) and without K promoter.

Answer 9

Question 10

Why is Fe(III) eaiser to activate over Fe(100), Fe(11) and Fe(211)

Answer 10

• The way the nitrogen bind on the top down atom distorts the symmetry of the molecule, it is easier to activate the molecule - called C7 sites
• Harder than Fe(100) say for example because there is not distortion in the symmetry of the molecule

Question 11

Assuming that N₂ activation is the rate determining step, there are 3 different mechanisms for ammonia synthesis on iron: Dissociative, Associative Alternating, and Associative Distal Pathway
Describe the graph which can be used to tell which mechanism is occuring?

Answer 11

• Graph shows a measure of the conc of N₂ on the surface of the Fe catalyst
• Chaning pressure of H₂ only, with constant pressure of N₂ and temperature (in a ratio of 450: 150 Torr or 3:1)

Question 12

Answer 12

• Not consuming nitrogen as part of the associative pathway

Question 13

Answer 13

• Have little hydrogen on the reactor that you don’t have much hydrogen to form ammonia
• When you get the right stiochiometry then you have a big consumption of this atomic nitrogen, resulting in high rates of ammonia production

Question 14

The experimental data related to which type of pathway?

Answer 14

• A dissociative pathway
• Nad concentration becomes very small, which indicates that the dissociation of N₂ is the rate determining step: N₂ → 2N