Using the following information calculate the heat of formation of NaOH. 2Na(s)+2H2O(l)→2NaOH(s)+H2(g);ΔH∘=−281.9kJ [Given that ΔH∘fH2O(l)=−285.8kJ/mol].
A
-141.6
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B
-712.6
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C
-426.6
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D
-650.4
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Solution
The correct option is C -426.6 The enthalpy change for the reaction is the difference between the total enthalpy of products and the total enthalpy of reactants.
ΔHr=[2(ΔHf)NaOH−2(ΔHf)H2O]
−281.9=(2ΔHf)NaOH+2×285.8
(ΔHf)NaOH=−426.6kJ
Hence, the heat of formation of NaOH is -426.6 kJ/mol.
Note: The enthalpies of formation of elements in their standard states are considered to be zero.
Hence, the enthalpies of Na(s) and H2(g) do not appear in the above expression.