Answer:
-250.40 kJ = ΔH°f(C8H18)(l)
Explanation:
The enthalpy change for a given chemical reaction is given by the sum of the standard heats of formation of products multiplied by their respective coefficients in the balanced equation minus the sum of the standard heat of formation of reactants again multiplied by their coefficients.
For this particular reaction,
2C8H18(l) + 25O2(g) → 16CO2(g) + 18H2O(l)
the ΔH°rxn will be given by
ΔH°rxn = 16ΔH°f(CO2(g)) + 18ΔH°f(H2O(l)) - (2ΔH°f(C8H18)(l) + 25 ΔH°fO2(g))
Notice we are given the information for the ΔH°rxn and all the ΔH°fs except for O2 which is zero and the ΔH°(C8H18)(l) which is what we are looking for.
So plugging our values and solving:
–1.0940 × 10⁴ kJ = 16mol x (–393.5 kJ/mol) + 18mol x (–285.8 kJ/mol) - 2(ΔH°f(C8H18)(l)
-1.094 x10⁴ kJ = -6296 kJ + ( -5144.40 kJ ) - 2(ΔH°f(C8H18)(l)
-500.40 kJ/2 = ΔH°f(C8H18)(l)
-250.40 kJ = ΔH°f(C8H18)(l)
The standard enthalpy of formation of liquid octane is -250.40 kJ
The enthalpy change for a given chemical reaction is given by the sum of the standard heats of formation of products multiplied by their respective coefficients in the balanced equation minus the sum of the standard heat of formation of reactants again multiplied by their coefficients.
Balanced chemical reaction:
2 C₈H₁₈ (l) + 25 O₂ (g) → 16 CO₂ (g) + 18 H₂O (l)
The ΔH°rxn will be given by:
ΔH°rxn = 16 ΔH°f (CO₂ (g)) + 18 ΔH°f (H₂O (l)) - (2 ΔH°f (C₈H₁₈) (l) + 25 ΔH°f O₂ (g))
ΔH°rxn and all the ΔH°fs except for O₂ which is zero and the ΔH°(C₈H₁₈) (l) is then calculated:
On substituting the values:
-1.0940 × 10⁴ kJ = 16mol x (-393.5 kJ/mol) + 18mol x (-285.8 kJ/mol) - 2(ΔH°f(C₈H₁₈)(l) -1.094 x10⁴ kJ
= -6296 kJ + ( -5144.40 kJ ) - 2(ΔH°f(C₈H₁₈)(l)
-500.40 kJ/2 = ΔH°f(C₈H₁₈)(l)
-250.40 kJ = ΔH°f(C₈H₁₈)(l)
Find more information about Enthalpy change here:
brainly.com/question/14047927
