Assuming enthalpy of combustion of hydrogen at 273 K, –286 kJ and enthalpy of fusion of ice at the same temperature to be + 6.0 kJ, calculate enthalpy change during formation of 100 g of ice
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Expert's answer
2017-05-04T06:12:34-0400
Answer: H2 + 1/2O2 = H2O(l) ΔH (H2) =-286 kJ/mol ΔH H2O (s) = H2O(l) =6.0 kJ/mol n = m/M M (H2O) = 18 g/mol n (H2O) = 100 / 18 = 5.56 mol Δ H (H2O) = 6.0 · 5.56 = 33.3 kJ
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