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 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

So, 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