Name the oil refining process shown in the diagram - Leaving Cert Chemistry - Question 6 - 2016

The physical property that permits these hydrocarbons to be isolated as naphtha is the boiling (condensing) point or temperature.

A major use for kerosene is as aviation (jet) fuel.

A major use for residue is in roofing, road making, or waterproofing.

The oil refining process that converts octane into 2,2,4-trimethylpentane is known as isomerisation.

These conversions are desirable because they improve the octane rating (number) which enhances engine performance and reduce the tendency of fuel to cause knocking. This is achieved by producing branched compounds that have higher octane ratings, thus resulting in smoother engine operation.

To calculate the heat of reaction, we use the following equation:

$\Delta H_{reaction} = \Sigma\Delta H_{f (products)} - \Sigma\Delta H_{f (reactants)}$

Given:

• Heat of formation of octane, (\Delta H_f (C_8H_{18}) = -250.1 , \text{kJ mol}^{-1})
• Heat of formation of ethylbenzene, (\Delta H_f (C_8H_{10}) = -12.5 , \text{kJ mol}^{-1})

Thus:

$\Delta H_{reaction} = -12.5 - (-250.1) = 237.6 \, \text{kJ}$

Given the two smaller hydrocarbons are octane (C₈H₁₈) and propene (C₃H₆), the formula of the third molecule can be deduced as follows:

$C_{18}H_{38} = C_8H_{18} + C_3H_6 + C_xH_y$

Solving for the third molecule:

$C_xH_y = C_{18}H_{38} - (C_8H_{18} + C_3H_6)$

Calculating:

• Carbonds: 18 - (8 + 3) = 7
• Hydrogens: 38 - (18 + 6) = 14

Thus, the third molecule is C₇H₁₄.

The molecular structure of octane is:

   H   H   H   H
   |   |   |   |
H–C–C–C–C–C–C–C–C–H
   |   |   |   |
   H   H   H   H

The molecular structure of 2,2,4-trimethylpentane is:

        H   H
        |   |
    H–C–C–C–H
        |   |   |
      H–C   H
        |        |
        H    H

The molecular structure of ethylbenzene is:

        H   H
        |   |
    H–C–C–C–C–C
        |   |   |
      H–C   H
        |   H
        H