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A 0.327-g sample of azulene (C10H8) is burned in a bomb calorimeter and the temperature increases from 25.20 °C to 27.60 °C. The calorimeter contains 1.17×103 g of water and the bomb has a heat capacity of 786 J/°C. Based on this experiment, calculate ΔE for the combustion reaction per mole of azulene burned (kJ/mol). C13H24O4(s) + 17 O2(g) 13 CO2(g) + 12 H2O(l) E =______ kJ/mol.

Respuesta :

Explanation:

The given data is as follows.

Molecular weight of azulene = 128 g/mol

Hence, calculate the number of moles as follows.

      No. of moles = [tex]\frac{mass}{\text{molecular weight}}[/tex]

                            = [tex]\frac{0.392 g}{128 g/mol}[/tex]

                            = 0.0030625 mol of azulene

Also,    [tex]-Q_{rxn} = Q_{solution} + Q_{cal}[/tex]

       [tex]Q_{rxn} = n \times dE[/tex]

         [tex]Q_{solution} = m \times C \times (T_{f} - T_{i})[/tex]

              [tex]Q_{cal} = C_{cal} \times (T_{f} - T_{i})[/tex]

Now, putting the given values as follows.    

     [tex]Q_{solution} = 1.17 \times 10^{3} g \times 10^{3} \times 4.184 J/g^{o}C \times (27.60 - 25.20)^{o}C[/tex]

                   = 11748.67  J

So,  [tex]Q_{cal} = 786 J/^{o}C \times (27.60 - 25.20)^{o}C[/tex]

                    = 1886.4 J

Therefore, heat of reaction will be calculated as follows.

        [tex]-Q_{rxn}[/tex] = (11748.67 + 1886.4) J

                      = 13635.07 J

As,  [tex]Q_{rxn} = n \times dE[/tex]

          13635.07 J = [tex]-n \times dE[/tex]

                dE = [tex]\frac{13635.07 J}{0.0030625 mol}[/tex]

                     = 4452267.75 J/mol

or,                 = 4452.26 kJ/mol       (as 1 kJ = 1000 J)

Thus, we can conclude that [tex]\Delta E[/tex] for the given combustion reaction per mole of azulene burned is 4452.26 kJ/mol.