Refinery gas, naphtha, kerosene and gas oil are products of the fractionation of crude oil - Leaving Cert Chemistry - Question 6 - 2018

The two major components of liquefied petroleum gas (LPG) are propane (C₃H₈) and butane (C₄H₁₀).

Methanethiol is added to natural gas and LPG to provide a distinct odor, allowing for the detection of gas leaks. Natural gas is odorless, and the addition of methanethiol adds a sulfur-like smell, enhancing safety by helping to prevent accidental inhalation or explosions.

To calculate the heat of combustion of methanethiol (CH₃SH), use the formula:

$ext{ΔH}_{ ext{reaction}} = ext{ΣΔH}_{ ext{f(products)}} - ext{ΣΔH}_{ ext{f(reactants)}}$

The heats of formation:

• Methanethiol: $ext{ΔH}_f = -228.3$ kJ/mol
• Carbon dioxide: $ext{ΔH}_f = -393.5$ kJ/mol (×1)
• Water: $ext{ΔH}_f = -285.8$ kJ/mol (×2)
• Sulfur dioxide: $ext{ΔH}_f = -296.8$ kJ/mol

Substituting these values:

ext{ΔH}_{ ext{reaction}} = ig( -393.5 + 2 imes -285.8 + -296.8 ig) - ig( -228.3 ig)

Calculating:

$ext{ΔH}_{ ext{reaction}} = -393.5 - 571.6 - 296.8 + 228.3 = -1239.1 ext{ kJ mol}^{-1}$

Thus, the heat of combustion of methanethiol is -1239.1 kJ/mol.

The IUPAC names for the four hydrocarbon products are:

1. 2,2,4-trimethylpentane
2. Butane
3. 1-butene
4. Ethylbenzene

Process B is catalytic cracking, and Process C is dehydrocyclization or reforming.

Process A is typically carried out in a distillation column, Process B (catalytic cracking) occurs in a catalytic cracking unit, and Process C (dehydrocyclization/reforming) takes place in a reformer.

Diesel has a higher boiling point compared to petrol. This is because diesel consists of larger hydrocarbons with longer carbon chains (14 to 19 carbon atoms), leading to greater van der Waals forces and thus requiring more energy (higher temperature) to overcome these intermolecular forces during phase transition compared to the smaller hydrocarbons in petrol, which typically contains 5 to 10 carbon atoms.