Question

When 0.478 g of biphenyl (C12H10) undergoes combustion in a bomb calorimeter, the temperature rises from 25.8 ∘C to 30.3 ∘C Find ΔErxn for the combustion of biphenyl. The heat capacity of the bomb calorimeter, determined in a separate experiment, is 5.86 kJ/∘C Express the energy in kilojoules per mole to three significant figures.

Answer 1

The enthalpy change
`(\(\Delta H_{\text{rxn}}\)) for the combustion of biphenyl is approximately \(8.50 \, \text{kJ/mol}\).`

To find the enthalpy change
`(\(\Delta H_{\text{rxn}}\))` for the combustion of biphenyl (
`C_12H_10`), you can use the equation:

`\[\Delta H_{\text{rxn}} = q_{\text{calorimeter}} = (q)/(n)\]`

where:

`\(q\)` is the heat absorbed by the calorimeter,

`\(n\)` is the number of moles of biphenyl.

First, let's calculate the heat absorbed by the calorimeter
`(\(q_{\text{calorimeter}}\))` using the formula:

`\[q_{\text{calorimeter}} = C_{\text{calorimeter}} \cdot \Delta T\]`

where:

`\(C_{\text{calorimeter}}\)` is the heat capacity of the bomb calorimeter,

`\(\Delta T\)`is the temperature change.

Given:

`\(C_{\text{calorimeter}} = 5.86 \, \text{kJ/\degree C}\)- \(\Delta T = 30.3 \, \degree C - 25.8 \, \degree C = 4.5 \, \degree C\)`

Substitute these values to find \(q_{\text{calorimeter}}\):

`\[q_{\text{calorimeter}} = 5.86 \, \text{kJ/\degree C} * 4.5 \, \degree C = 26.37 \, \text{kJ}\]`

Now, we need to find the number of moles
`(\(n\))`of biphenyl. To do this, we can use the molar mass of biphenyl
`(\(C_(12)H_(10)\))`.

`\[\text{Molar mass of } C_(12)H_(10) = (12 * \text{atomic mass of C}) + (10 * \text{atomic mass of H})\]`

Using the atomic masses:

- Atomic mass of C ≈ 12.01 g/mol

- Atomic mass of H ≈ 1.01 g/mol

`\[\text{Molar mass of } C_(12)H_(10) = (12 * 12.01) + (10 * 1.01) \, \text{g/mol} = 154.22 \, \text{g/mol}\]`

Now, calculate the number of moles
`(\(n\))` using the given mass of biphenyl
`(\(0.478 \, \text{g}\))`:

`\[n = \frac{\text{mass}}{\text{molar mass}} = \frac{0.478 \, \text{g}}{154.22 \, \text{g/mol}} \approx 0.0031 \, \text{mol}\]`

Finally, substitute
`\(q_{\text{calorimeter}}\) and \(n\) into the first equation to find \(\Delta H_{\text{rxn}}\)`:

`\[\Delta H_{\text{rxn}} = \frac{26.37 \, \text{kJ}}{0.0031 \, \text{mol}} \approx 8500 \, \text{kJ/mol}\]`

Expressed to three significant figures, the enthalpy change
`(\(\Delta H_{\text{rxn}}\)) for the combustion of biphenyl is approximately \(8.50 \, \text{kJ/mol}\).`