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Question
Calculate ΔsubH of the H2O from the given data:
\[\ce{H2O_{(s)}->H2O_{(l)},}\] ΔfusH = 6.01kJ mol−1
\[\ce{H2O_{(l)}-> H2O_{(g)},}\] ΔVapH = 45.07 kJ mol−1.
Numerical
Solution
Given:
\[\ce{H2O_{(s)}->H2O_{(l)}}\]
\[\ce{H2O_{(l)}-> H2O_{(g)}}\]
ΔsubH = ?
\[\ce{H2O_{(s)}->H2O_{(l)}}\]
+ \[\ce{H2O_{(l)}-> H2O_{(g)}}\]
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\[\ce{H2O_{(s)} + H2O_{(l)}-> H2O_{(l)} + H2O_{(g)}}\]
\[\ce{H2O_{(s)}->H2O_{(g)}}\]
ΔsubH = ΔfusH + ΔVapH
= 6.01kJ mol−1 + 45.07 kJ mol−1
∴ ΔsubH = 51.08 kJ mol−1
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Thermochemistry
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