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Enthalpy Change Example Problem

Enthalpy Change of Water

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  • 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

  • Answer

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

    b.) 2.40 g ice melted

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