Question

For a particular isomer of C 8 H 18 , the combustion reaction produces 5113.3 kJ of heat per mole of C 8 H 18 ( g ) consumed, under standard conditions. C 8 H 18 ( g ) + 25 2 O 2 ( g ) ⟶ 8 CO 2 ( g ) + 9 H 2 O ( g ) Δ H ∘ rxn = − 5113.3 kJ / mol What is the standard enthalpy of formation of this isomer of C 8 H 18 ( g ) ?

Answer 1

Answer:

ΔH°f(C₈H₁₈(g)) = -210.9 kJ/mol

Explanation:

Let's consider the combustion of C₈H₁₈.

C₈H₁₈(g) + 25/2 O₂(g) ⟶ 8 CO₂(g) + 9 H₂O(g) ΔH°rxn = − 5113.3 kJ

We can calculate the standard enthalpy of formation of C₈H₁₈(g) using the following expression.

ΔH°rxn = 8 mol × ΔH°f(CO₂(g)) + 9 mol × ΔH°f(H₂O(g)) - 1 mol × ΔH°f(C₈H₁₈(g)) - 25/2 mol × ΔH°f(O₂(g))

1 mol × ΔH°f(C₈H₁₈(g)) = 8 mol × ΔH°f(CO₂(g)) + 9 mol × ΔH°f(H₂O(g)) - 25/2 mol × ΔH°f(O₂(g)) - ΔH°rxn

1 mol × ΔH°f(C₈H₁₈(g)) = 8 mol × (-393.5 kJ/mol) + 9 mol × (-241.8 kJ/mol) - 25/2 mol × 0 kJ/mol - (− 5113.3 kJ)

ΔH°f(C₈H₁₈(g)) = -210.9 kJ/mol