Answer:
[tex]\Delta H^0 _{reaction} = - 54.04 \ kJ/mol[/tex]
Explanation:
The given equation for the chemical reaction can be expressed as;
[tex]2H_2O_{(l)} \to 2H_2O_{(g)} + O_{2(g)}[/tex]
Using Hess Law to determine how much heat is produced by the decomposition of exactly 1 mole of H2O2 under standard conditions; we have the expression showing the Hess Law as follows:
[tex]\Delta H^0 _{reaction} = \sum n* \Delta H^0 _{products} - \sum n* \Delta H^0 _{reactants}[/tex]
At standard conditions;
the molar enthalpies of the given equation are as follows:
[tex]\Delta H_2O_{(g)} =-241.82\ kJ/mol[/tex]
[tex]\Delta H_ O_{2(g)} = 0 \ kJ/mol[/tex]
[tex]\Delta H _{H_2O_{(l)}}= -187.78 \ kJ/mol[/tex]
Replacing them into above formula; we have:
[tex]\Delta H^0 _{reaction} = (2*(-241.82\ kJ/mol) + 0 \ kJ/mol + (2 *(-187.78 \ kJ/mol))[/tex]
[tex]\Delta H^0 _{reaction} =-108.08 \ kJ/mol[/tex]
The above is the amount of heat of formation for two moles of hydrogen peroxide; thus for 1 mole hydrogen peroxide ; we have :
[tex]\Delta H^0 _{reaction} = \dfrac{-108.08 \ kJ/mol}{2}[/tex]
[tex]\Delta H^0 _{reaction} = - 54.04 \ kJ/mol[/tex]
Hence; the heat produced after the decomposition of 1 mole of hydrogen peroxide is -54.04 kJ/mol
