determine the enthalpy change for the combustion of 1 mole of C2H6g with all reactants and product in their standard state.

Answer 1

The balanced chemical equation for the combustion of C2H6 is:

C2H6(g) + 7/2 O2(g) -> 2 CO2(g) + 3 H2O(l).

From the standard enthalpies of formation data, we have:

ΔHf°[C2H6(g)] = -84.7 kJ/mol
ΔHf°[CO2(g)] = -393.5 kJ/mol
ΔHf°[H2O(l)] = -285.8 kJ/mol
ΔHf°[O2(g)] = 0 kJ/mol

The enthalpy change for the reaction is given by the equation:

ΔH° = ΣΔHf°(products) - ΣΔHf°(reactants)

Substituting the values, we get:

ΔH° = [2(-393.5 kJ/mol) + 3(-285.8 kJ/mol)] - [-84.7 kJ/mol - 7/2(0 kJ/mol)]
ΔH° = [-787.0 kJ/mol - 857.4 kJ/mol] + 42.4 kJ/mol
ΔH° = -1602 kJ/mol + 42.4 kJ/mol
ΔH° = -1559.6 kJ/mol

Therefore, the enthalpy change for the combustion of 1 mole of C2H6 in its standard state is approximately -1559.6 kJ/mol.