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# Enthalpy Change Example Problem

## Enthalpy Change of Water

• 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.

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 g
fusion ΔH = 6.00 x 103 J
fusion ΔH = 6.00 kJ

vaporization ΔH = 18.02 g x 2257 J / 1 g
vaporization ΔH = 4.07 x 104 J
vaporization ΔH = 40.7 kJ

So, the completed thermochemical reactions are:

H2O(s) → H2O(l); ΔH = +6.00 kJ
H2O(l) → H2O(g); ΔH = +40.7 kJ

b.) Now we know that:

1 mol H2O(s) = 18.02 g H2O(s) ~ 6.00 kJ

Using this conversion factor:

0.800 kJ x 18.02 g ice / 6.00 kJ = 2.40 g ice melted

a.)
H2O(s) → H2O(l); ΔH = +6.00 kJ
H2O(l) → H2O(g); ΔH = +40.7 kJ

b.) 2.40 g ice melted

Anne Marie Helmenstine, Ph.D.