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3 November, 06:52

A 35.6 g sample of ethanol (C2H5OH) is burned in a bomb calorimeter, according to the following reaction. If the temperature rose from 35.0 to 76.0 C and the heat capacity of the calorimeter is 23.3 kJ / C, what is the value of DH rxn? The molar mass of ethanol is 46.07 g/mol.

C2H5OH (l) + O2 (g) - -> CO2 (g) + H2O (g) ? H_ rxn = ?

A) - 1.24 * 103 kJ/mol

B) + 1.24 * 103 kJ/mol

C) - 8.09 * 103 kJ/mol

D) - 9.55 * 103 kJ/mol

E) + 9.55 * 103 kJ/mol

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Answers (1)
  1. 3 November, 07:30
    0
    A) - 1.24 * 10³ kJ/mol

    Explanation:

    To solve this problem we can use two formulas:

    Q = C*ΔT ΔH = Q/mol

    So first let's calculate Q, using the heat capacity C and the difference in temperature ΔT:

    Q = - 23.3 kJ/°C * (76.0-35-0) °C

    Q = 955.4 kJ

    Then let's calculate the moles of ethanol, using its molar mass:

    mol ethanol = 35.6 g : 46.07 g/mol = 0.773 mol

    Finally we calculate ΔH:

    ΔH = - 955.4kJ/0.773mol

    ΔH = - 1235.96 kJ/mol ≅ - 1.24 * 10³ kJ/mol

    It has a negative value because it is an exothermic reaction.
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