Question

Octane (C₈H₁₈) undergoes combustion according to the following thermochemical equation.

2C₈H₁₈(l) + 25O₂(g) → 16CO₂(g) + 18H₂O(l) ΔH°rxn = –1.0940 × 104 kJ/mol

What is the standard enthalpy of formation of liquid octane?
ΔH°f(CO₂(g)) = –393.5 kJ/mol and ΔH°f(H₂O(l)) = –285.8 kJ/mol
A) -250 kJ/mol
B) -10,940 kJ/mol
C) -2188 kJ/mol
D) -495 kJ/mol
E) 495 kJ/mol

Answer 1

To calculate the standard enthalpy of formation of liquid octane, Hess's law is used with provided enthalpy values, resulting in -250 kJ/mol. Thus, the correct answer is option A.

Step-by-step explanation:

The question asks for the standard enthalpy of formation of liquid octane using a given thermochemical equation for its combustion, as well as given standard enthalpies of formation for CO₂(g) and H₂O(l). To find the standard enthalpy of formation of octane, we can use Hess's law which states that the total enthalpy change for a reaction is the same, no matter how many steps the reaction is carried out in.

In the combustion of octane, the enthalpy change for the reaction (ΔH°rxn) is -1.0940 × 104 kJ/mol. We can express this reaction enthalpy as the sum of the formation enthalpies for the products minus the formation enthalpies of reactants:

ΔH°rxn = [∑ΔH°f(products)] - [∑ΔH°f(reactants)]

Substituting the values in, we get:

-1.0940 × 104 kJ/mol = [16(ΔH°f(CO₂(g))) + 18(ΔH°f(H₂O(l)))] - [2(ΔH°f(C₈H₁₈(l)))]

We can solve for ΔH°f(C₈H₁₈(l)) by substituting the given standard formation enthalpies for CO₂(g) and H₂O(l), which are -393.5 kJ/mol and -285.8 kJ/mol, respectively:

-1.0940 × 104 kJ/mol = [16(-393.5 kJ/mol) + 18(-285.8 kJ/mol)] - [2(ΔH°f(C₈H₁₈(l)))]

After solving the equation, we find that:

ΔH°f(C₈H₁₈(l)) = -250 kJ/mol

The correct answer is option A, which is -250 kJ/mol for the standard enthalpy of formation of liquid octane.