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Question
The standard enthalpy of formation of water is - 286 kJ mol-1. Calculate the enthalpy change for the formation of 0.018 kg of water.
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Solution
Mass of H2O = 0.018 kg = 18 g
Number of moles of H2O = `("Mass of H"_2"O")/("Molar mass of H"_2"O") = (18 "g")/(18 "g mol"^-1)` = 1 mol
The thermochemical equation is,
\[\ce{H_{2(g)} + \frac{1}{2} O_{2(g)} -> H2O_{(l)}}\], ΔfH° = – 286 kJ mol–1
∴ Enthalpy change for formation of 1 mole H2O = - 286 kJ
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