
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 100°C is 2257 J/g.
Part a: Calculate the change in enthalpy, ΔH, for these two processes.
H_{2}O(s) → H_{2}O(l); ΔH = ?
H_{2}O(l) → H_{2}O(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 (H_{2}O) is 18.02 g. Therefore:
fusion ΔH = 18.02 g x 333 J / 1 g
fusion ΔH = 6.00 x 10^{3} J
fusion ΔH = 6.00 kJ
vaporization ΔH = 18.02 g x 2257 J / 1 g
vaporization ΔH = 4.07 x 10^{4} J
vaporization ΔH = 40.7 kJ
So, the completed thermochemical reactions are:
H_{2}O(s) → H_{2}O(l); ΔH = +6.00 kJ
H_{2}O(l) → H_{2}O(g); ΔH = +40.7 kJ
b.) Now we know that:
1 mol H_{2}O(s) = 18.02 g H_{2}O(s) ~ 6.00 kJ
Using this conversion factor:
0.800 kJ x 18.02 g ice / 6.00 kJ = 2.40 g ice melted

Answer
a.)
H_{2}O(s) → H_{2}O(l); ΔH = +6.00 kJ
H_{2}O(l) → H_{2}O(g); ΔH = +40.7 kJ
b.) 2.40 g ice melted