Question

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

Answer 1

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