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3 April, 08:43

A sample of the sugar d-ribose (C5H10O5) of mass 0.727 g was placed in a calorimeter and then ignited in the presence of excess oxygen. The temperature rose by 0.910 K. In a separate experiment in the same calorimeter, the combustion of 0.825 g of benzoic acid, for which the internal energy of combustion is - 3251 kJ mol-1, gave a temperature rise of 1.940 K. Calculate the enthalpy of formation of d-ribose.

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  1. 3 April, 12:29
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    The internal energy of combustion of d-ribose = - 2127 kJ/mol

    The enthalpy of formation of d-ribose = - 1269.65 kJ/mol

    Explanation:

    Step 1: Data given

    Mass of d-ribose = 0.727 grams

    The temperature rose by 0.910 K

    In a separate experiment in the same calorimeter, the combustion of 0.825 g of benzoic acid, for which the internal energy of combustion is - 3251 kJ mol-1, gave a temperature rise of 1.940 K.

    Molar mass of benzoic acid = 122.12 g/mol

    Step 2: Calculate ΔU for benzoic acid

    The calorimeter is a constant-volume instrument so:

    ΔU = q

    ΔU = (0.825 g / 122.12 g/mol) * (-3251 kJ / mol)

    ΔU = - 21.96 kJ

    Step 3: Calculate ΔU for d-ribose

    c = |q| / ΔT

    ⇒ with ΔT = 1.940 K

    c = 21.96 kJ / 1.940 K

    c = 11.32 kJ / K

    For d-ribose: ΔU = - cΔT

    ΔU = - 11.32 kJ/K * 0.910 K

    ΔU = - 10.3 kJ

    Step 4: Calculate moles of d-ribose

    moles ribose = 0.727 grams / 150.13 g/mol

    moles ribose = 0.00484 moles

    Step 5: Calculate the internal energy of combustion for d-ribose

    ΔrU = ΔU / n

    ΔrU = - 10.3 kJ / 0.004842 moles

    ΔrU = - 2127 kJ/mol

    Step 6: Calculate The enthalpy of formation of d-ribose

    The combustion of ribose is:

    C5H10O5 (s) + 5O2 (g) → 5CO2 (g) + 5H20 (l)

    Since there is no change in the number of moles of gas, ΔrH = ΔU

    For the combustion of ribose, we consider the following reactions:

    5CO2 (g) + 5H2O (l) → C5H10O5 (s) + 5O2 (g) ΔH = - 2127 kJ/mol

    C (s) + O2 (g) → CO2 (g) ΔH = - 393.5 kJ/mol

    H2 (g) + 1/2 O2 (g) → H2O (l) ΔH = - 285.83 kJ/mol

    ΔH = 2127 kJ/mol + 5 (-393.5 kJ/mol) + 5 (-285.83 kJ/mol)

    ΔH = 2127 kJ/mol - - 1967.5 kJ/mol - 1429.15 kJ/mol

    ΔH = - 1269.65 kJ/mol

    The internal energy of combustion of d-ribose = - 2127 kJ/mol

    The enthalpy of formation of d-ribose = - 1269.65 kJ/mol
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