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.
