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21 April, 07:09

Calculate the enthalpy change, ΔH, for the process in which 31.6 g of water is converted from liquid at 3.2 ∘C to vapor at 25.0°C. For water, ΔHvap = 44.0 kJ/mol at 25.0°C and Cs = 4.18 J / (g*°C) for H2O (l).

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  1. 21 April, 07:38
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    Enthalpy change = 74.36 kJ

    Explanation:

    Enthalpy change is defines as the heat absorbed or evolved during a chemical reaction at constant pressure and volume

    Reaction:

    H2O (l) - -> H2O (g)

    DHr = mCsDT

    Where m is the mass of water

    Cs is the specific heat capacity

    DT is the temperature difference

    = 31.6 * 4.18 * (25 - 3.2)

    = 2879.518 J

    DHvap = 44kJ/mol

    Moles of water = mass/molecular weight

    Molecular weight = (1*2) + 16

    = 18g/mol

    Moles of water = 31.6/18

    = 1.756 moles

    DHvap = 44 * 1.756

    = 77.244kJ

    DH = DHproduct - DHreactant

    DHproduct = DHvap = 77.24kJ

    DHreactant = DHr = 2879.518J = 2.880kJ

    = 77.24 - 2.880

    DH = 74.36kJ

    Enthalpy change = 74.36 kJ
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