The standard enthalpy of formation of H2O (l) is -286 kJ/mol and the standard enthalpy of combustion of ethane is -1560 kJ/mol. Given that the enthalpy change when one mole of...

Question

The standard enthalpy of formation of H2O (l) is -286 kJ/mol and the standard enthalpy of combustion of ethane is -1560 kJ/mol. Given that the enthalpy change when one mole of ethene is redcued by hydrogen gas to ethane is - 138kJ/mol, calculate the standard enthalpy of combustion of ethene.

Answer 1

This is an example of using Hess' Law.

C₂H₄ + 3O₂ ==> 2CO₂ + 2H₂O TARGET EQUATION (combustion of ethene)

Given is the following:

eq.1 C₂H₄ + H₂ ==> C₂H₆ ∆H = -138 kJ/mol

eq.2 2C₂H₆ + 7O₂ ==> 4CO₂ + 6H₂O ∆H = -1560 kJ/mol

eq.3 2H₂ + O₂ ==> 2H₂O ∆H = -286 kJ/mol

Copy eq.1: C₂H₄ + H₂ ==> C₂H₆ ∆H = -138 kJ/mol

Copy eq.2 ÷ 2: C₂H₆ + 7/2O₂ ==> 2CO₂ + 3H₂O ∆H = -780 kJ/mol

Reverse eq.3 ÷ 2: H₂O ==> H₂ + 1/2O₂ ∆H = +143 kJ/mol

Adding up equations 1, 2 and 3 we have...

C₂H₄ + H₂ + C₂H₆ + 7/2O₂ + H₂O ==> C₂H₆ + 2CO₂ + 3H₂O + H₂ + 1/2O₂, which reduces to...

C₂H₄ + 3O₂ ==> 2CO₂ + 2H₂O which is the TARGET EQUATION

∆H = -138 + (-780) + 143 = -775 kJ/mole