Nickel, iron and V₂O₅ as catalysts

Which catalyst for which process

1. Magnetic properties and catalytic activity

2. One catalyst for all?

One factory makes ammonia, one makes sulphuric acid, one turns vegetable oil into solid fat. Each uses a transition-element catalyst.

Pick the right answer

Predict: can all three use the same catalyst?

A. Yes: any transition element speeds up any reaction
B. No: each process has its own catalyst
C. No catalyst is needed if the factory heats enough

Explanation

Each process has its own catalyst: finely divided nickel for the hydrogenation of oils, iron for ammonia (Haber-Bosch method) and

V2O5 V X_{2} O X_{5}

for the first step towards sulphuric acid (contact method). Heating alone does not replace the catalyst the book names for each process.

3. Nickel: hydrogenation

Hydrogenation

Adding hydrogen to a substance; here, to oils.

Finely divided nickel is used as a catalyst in the hydrogenation of oils. It turns liquid oils into solid fats.

liquid oil + $H2 H X_{2}$ → solid fat (finely divided Ni (catalyst))

Hydrogenation of oils

Finely divided means a very fine powder. A powder has a huge surface, so many surface atoms meet the molecules.

4. Iron: Haber-Bosch

Divided iron is used as a catalyst in the preparation of ammonia gas by the Haber-Bosch method.

N2 + 3H2 →[500\,^C / 200\,atm\\ Fe] 2NH3

Book p.15

Iron is also the catalyst in the Fischer-Tropsch method, which turns water gas into a fuel (from iron's uses).

5. V₂O₅: contact method

Contact method

The industrial preparation of sulphuric acid.

Vanadium pentoxide,

V2O5 V X_{2} O X_{5}

, is used as a catalyst in the first step:

2SO2 + O2 →[V₂O₅\\ 450\,^C] 2SO3

SO3 + H2O \to H2SO4 SO X_{3} + H X_{2} O H X_{2} SO X_{4}

Then sulphur trioxide and water give sulphuric acid.

6. Match the catalyst

Add one you know from fig 1-3:

MnO2 MnO X_{2}

in the decomposition of

H2O2 H X_{2} O X_{2}

Sort into groups

Which catalyst does each reaction or process use?

Vegetable oil + hydrogen → solid fat $N2 + 3H2 \to 2NH3 N X_{2} + 3 H X_{2} 2 NH X_{3}$ $2SO2 + O2 \to 2SO3 2 SO X_{2} + O X_{2} 2 SO X_{3}$ $2H2O2 \to 2H2O + O2 2 H X_{2} O X_{2} 2 H X_{2} O + O X_{2}$ Water gas → fuel (Fischer-Tropsch)

7. Spot the error

Pick the right answer

Which is the correct way to write the first step of the contact method?

A. $2SO2 + O2 + V2O5 \to 2SO3 2 SO X_{2} + O X_{2} + V X_{2} O X_{5} 2 SO X_{3}$
B. $2SO2 + O2 →[V₂O₅\\ 450\,^C] 2SO3$
C. $2SO2 + O2 →[Fe\\ 450\,^C] 2SO3$

Explanation

A catalyst is not used up, so it is written over the arrow with the conditions, not as a reactant. The contact method uses

V2O5 V X_{2} O X_{5}

at 450 °C; iron is the catalyst for ammonia.

Pick the right answer

A student writes

N2 + 3H2 \to[V2O5] 2NH3 N X_{2} + 3 H X_{2} V X_{2} O X_{5} 2 NH X_{3}

. What is wrong?

A. The catalyst: the Haber-Bosch method uses iron
B. Nothing: any transition-element catalyst can go over the arrow
C. $V2O5 V X_{2} O X_{5}$ should be written as a reactant

Explanation

The catalyst over the arrow does not belong to this process:

V2O5 V X_{2} O X_{5}

is the contact method's catalyst, and the Haber-Bosch method uses iron. Over the arrow is where a catalyst goes, because it is not used up.

8. Conditions from memory

Fill in the blanks

In the Haber-Bosch method the catalyst is _____, at 500 °C and _____ atm.

Explanation

The Haber-Bosch method uses iron (Fe) as the catalyst, at 500 °C and 200 atm, all written over the arrow.

Fill in the blanks

In the contact method the catalyst is _____, at _____ °C.

Explanation

The contact method uses

V2O5 V X_{2} O X_{5}

as the catalyst, at 450 °C, written over the arrow.

Need a hint?

Each process has its own catalyst. Picture each equation: what is written over its arrow?
Ammonia's catalyst is a metal; the contact method's catalyst is an oxide.
The two temperatures in this lesson are 500 °C and 450 °C.

9. Summary

Finely divided Ni: hydrogenation of oils.
Fe: Haber-Bosch (500 °C / 200 atm); also Fischer-Tropsch.
$V2O5 V X_{2} O X_{5}$ : contact method (450 °C). $MnO2 MnO X_{2}$ : decomposition of $H2O2 H X_{2} O X_{2}$ .

Catalyst	Process	Over the arrow / product
Finely divided Ni	Hydrogenation of oils	oil → solid fat
Fe	Haber-Bosch method	500 °C / 200 atm, Fe → $NH3 NH X_{3}$
Fe	Fischer-Tropsch method	water gas → fuel
$V2O5 V X_{2} O X_{5}$	Contact method	$V2O5 V X_{2} O X_{5}$ / 450 °C → $SO3 SO X_{3}$ , then $H2SO4 H X_{2} SO X_{4}$
$MnO2 MnO X_{2}$	Decomposition of $H2O2 H X_{2} O X_{2}$ (fig 1-3)	$H2O2 H X_{2} O X_{2}$ → $H2O + O2 H X_{2} O + O X_{2}$

Each process has its own catalyst.

Back to the hook: each process has its own catalyst. Can you explain why the nickel is used as a fine powder?

Concepts in this lesson