Question

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.

Answer 1

Step-by-step 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 =
`\frac{mass}{\text{molecular weight}}`

=
`(0.392 g)/(128 g/mol)`

= 0.0030625 mol of azulene

Also,
`-Q_(rxn) = Q_(solution) + Q_(cal)`

`Q_(rxn) = n * dE`

`Q_(solution) = m * C * (T_(f) - T_(i))`

`Q_(cal) = C_(cal) * (T_(f) - T_(i))`

Now, putting the given values as follows.

`Q_(solution) = 1.17 * 10^(3) g * 10^(3) * 4.184 J/g^(o)C * (27.60 - 25.20)^(o)C`

= 11748.67 J

So,
`Q_(cal) = 786 J/^(o)C * (27.60 - 25.20)^(o)C`

= 1886.4 J

Therefore, heat of reaction will be calculated as follows.

`-Q_(rxn)` = (11748.67 + 1886.4) J

= 13635.07 J

As,
`Q_(rxn) = n * dE`

13635.07 J =
`-n * dE`

dE =
`(13635.07 J)/(0.0030625 mol)`

= 4452267.75 J/mol

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

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