Enthalpy Change Example Problem Ice to Water Vapor

This enthalpy change example problem is the enthalpy change as ice changes state from solid to liquid water and finally to water vapor. Enthalpy Review You may wish to review the Laws of Thermochemistry and Endothermic and Exothermic Reactions before you begin. Problem Given: The heat of fusion of ice is 333 J/g (meaning 333 J is absorbed when 1 gram of ice melts). The heat of vaporization of liquid water at 100Â °C is 2257 J/g. Part a: Calculate the change in enthalpy, ΔH, for these two processes. H2O(s) → H2O(l); ΔH ? H2O(l) → H2O(g); ΔH ? Part b: Using the values you just calculated, determine the number of grams of ice that can be melted by 0.800 kJ of heat. Solution a) Did you notice that the heats of fusion and vaporization were given in joules and not kilojoules? Using the periodic table, we know that 1 mole of water (H2O) is 18.02 g. Therefore: fusion ΔH 18.02 g x 333 J / 1 gfusion ΔH 6.00 x 103 Jfusion ΔH 6.00 kJ vaporization ΔH 18.02 g x 2257 J / 1 gvaporization ΔH 4.07 x 104 Jvaporization ΔH 40.7 kJ So, the completed thermochemical reactions are: H2O(s) → H2O(l); ΔH 6.00 kJH2O(l) → H2O(g); ΔH 40.7 kJ​ b) Now we know that: 1 mol H2O(s) 18.02 g H2O(s) ~ 6.00 kJ So, using this conversion factor: 0.800 kJ x 18.02 g ice / 6.00 kJ 2.40 g ice melted Answer a)Â  H2O(s) → H2O(l); ΔH 6.00 kJÂ   Â   H2O(l) → H2O(g); ΔH 40.7 kJ​ b) 2.40 g ice melted