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Using standard heats of formation, calculate the standard enthalpy change for the following reaction. 2h2o(l)2h2(g) + o2(g).

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The standard enthalpy change for the reaction 2H2O (l) → 2H2 (g) + O2 (g) is -873.6 kJ/mol.

The standard enthalpy change (ΔH°) for a reaction is the enthalpy change when all reactants and products are in their standard states (at 1 atm pressure and a specified temperature, usually 25°C).

The standard enthalpy change for the reaction 2H2O (l) → 2H2 (g) + O2 (g) can be calculated using the standard heats of formation of the reactants and products as follows:

ΔH°rxn = [ΔH°f (H2O, l) x 2] + [ΔH°f (H2, g) x 2] + ΔH°f (O2, g)

ΔH°rxn = [-285.8 kJ/mol x 2] + [-241.8 kJ/mol x 2] + 0 kJ/mol

ΔH°rxn = -873.6 kJ/mol

Therefore, the standard enthalpy change for the reaction 2H2O (l) → 2H2 (g) + O2 (g) is -873.6 kJ/mol.

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