Question

One of the cleanest-burning octanes is a compound called 2,3,4-trimethylpentane. The complete combustion of one mole of this compound to CO₂(g) and H₂O(g) leads to ΔH∘= -5064.9 kJ/mol. calculate ΔH∘f for 2,3,4-trimethylpentane. The Balanced Chemical equation is

C₈H₁₈(l) + (25/2)O₂(g) → 8CO₂(g) + 9H₂O(g).
ΔH∘f for CO₂ is -393.5, and the ΔH∘f for H₂O is -285.8

Answer 1

The standard enthalpy of formation (ΔH°f) for 2,3,4-trimethylpentane is calculated using the given enthalpy of combustion and the known ΔH°f for the combustion products, resulting in ΔH°f being -655.3 kJ/mol for 2,3,4-trimethylpentane.

Step-by-step explanation:

To calculate the standard enthalpy of formation (ΔH°f) for 2,3,4-trimethylpentane, we need to use the given enthalpy of combustion (ΔH° = -5064.9 kJ/mol) and the standard enthalpies of formation (ΔH°f) for the combustion products, which are CO2(g) and H2O(g).

The balanced combustion reaction for 2,3,4-trimethylpentane is:

C8H18(l) + (25/2)O2(g) → 8CO2(g) + 9H2O(g)

We can write the equation for the enthalpy changes involved in the combustion process:

ΔH° = ΣΔH°f(products) - ΣΔH°f(reactants) = -5064.9 kJ

For the products:

• 8 mol of CO2: 8 mol × (-393.5 kJ/mol) = -3148.0 kJ
• 9 mol of H2O: 9 mol × (-285.8 kJ/mol) = -2572.2 kJ

Sum of ΔH°f for products is: -3148.0 kJ + (-2572.2 kJ) = -5720.2 kJ

Now we substitute into the equation:
-5064.9 kJ = (-5720.2 kJ) - ΔH°f(2,3,4-trimethylpentane)

Solving for ΔH°f(2,3,4-trimethylpentane):

ΔH°f(2,3,4-trimethylpentane) = -5720.2 kJ + 5064.9 kJ = -655.3 kJ/mol

Therefore, the standard enthalpy of formation for 2,3,4-trimethylpentane is -655.3 kJ/mol.