Calculate the standard enthalpy change, ΔH°rxn, in kJ for the following chemical equation, using only the thermochemical equations below: 4KO2(s) + 2H2O(l) → 4KOH(aq) + 3O2(g) Report your answer to three significant figures in scientific notation.
Equations: ΔH°rxn (kJ)
4K(s) + O2(g) → 2K2O(s) -726.4
K(s) + O2(g) → KO2(s) -284.5
K2O(s) + H2O(l) → 2KOH(aq) -318
in order to get 4KO2, it is in the first reaction so we need to reverse the first reaction and multiply it by 4 times
multiply the third reaction by 2 to get 2H2O
now add the reaction 1 simply, hence the final
Delta H = 1138 - 636 - 726.4 = -224.4 KJ
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