The given chemical reaction is:
[tex]2H_{2}O_{2}(aq)---> 2H_{2}O(l)+O_{2}(g)[/tex]
The standard heats of formation of [tex]H_{2}O and H_{2}O_{2}[/tex] are:
Δ[tex]H_{f}^{0}(H_{2}O[/tex]) = -285.8kJ/mol[/tex]
Δ[tex]H_{f}^{0}(H_{2}O_{2}[/tex]) = -187.6 kJ/mol[/tex]
Calculating the change in heat:
Δ[tex]H^{0}_{reaction)[/tex]=∑ΔH[tex]_{f}^{0}(products)[/tex]-∑ΔH[tex]_{f}^{0}(reactants)[/tex]
= [{2 * (-285.8 kJ/mol)} -{2*(-187.6 kJ/mol)}]
= -196.4 kJ/mol
Therefore, the change in enthalpy for the given reaction is -196.4 kJ/mol
