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

The standard enthalpy of formation of liquid octane is -250.2 kJ/mol

Explanation:

The given balanced chemical reaction is,

First we have to calculate the enthalpy of reaction .

where,

We are given:

Putting values in above equation, we get:

Thus the standard enthalpy of formation of liquid octane is -250.2 kJ/mol

we can conclude that the melting point of table salt is above room temperature (it's actually 801 deg c). the melting point of mercury must be below room temperature (- 39 deg c).