Answer:
-1235.3 kJ/mol
Explanation:
The balanced equation for the reaction can be represented by:
[tex]C_2H_5OH_{(l)} + 3 O_2_{(g)} -----> 2 CO_2_{(g)} + 3 H_2O_{(g)}[/tex]
One mole of ethanol reacts with 3 moles of oxygen to produce 2 moles of carbon dioxide and 3 moles of water.
Applying Hess's law to the reaction:
ΔH° = [tex][(2)CO_{2(g)} + (3)H_2O_{(g)}] - [C_2H_5OH_{(l)} + (3)O_{2(g)}][/tex]
ΔH°f [tex]C_2H_5OH_{(l)}[/tex] = –277.7 kJ/mol
ΔH°f [tex]CO_{2(g)}[/tex] = –393.5 kJ/mol
ΔH°f [tex]H_2O_{(g)}[/tex] = –241.8 kJ/mol
Hence;
ΔH° = [ (2)-393.5 + (3)-241.8 kJ/mol] - [ -277.1 + (3)0 kJ/mol]
= [-787 -725.4 kJ/mol ] - [ -277.1 kJ/mol ]
= -1512.4 + 277.1 kJ/mol
= -1235.3 kJ/mol
The ΔH° for the combustion of ethanol is -1235.3 kJ/mol
