To find the change in energy, [tex]\(\Delta E_{\text{comb}}\)[/tex], for the combustion of naphthalene using the described bomb calorimeter experiment, we can follow these steps:
### Step 1: Calculate the change in temperature
The change in temperature ([tex]\(\Delta T\)[/tex]) is given by the difference between the final temperature and the initial temperature of the calorimeter:
[tex]\[
\Delta T = T_{\text{final}} - T_{\text{initial}}
\][/tex]
Given:
- [tex]\(T_{\text{initial}} = 25.10^\circ C\)[/tex]
- [tex]\(T_{\text{final}} = 31.56^\circ C\)[/tex]
Thus,
[tex]\[
\Delta T = 31.56^\circ C - 25.10^\circ C = 6.46^\circ C
\][/tex]
### Step 2: Calculate the heat absorbed by the calorimeter
The heat absorbed by the calorimeter ([tex]\(q_{\text{calorimeter}}\)[/tex]) can be calculated using the given heat capacity of the calorimeter and the temperature change:
[tex]\[
q_{\text{calorimeter}} = C_{\text{calorimeter}} \times \Delta T
\][/tex]
Given:
- [tex]\(C_{\text{calorimeter}} = 5.11 \, \text{kJ} / ^\circ \text{C}\)[/tex]
- [tex]\(\Delta T = 6.46^\circ C\)[/tex]
Thus,
[tex]\[
q_{\text{calorimeter}} = 5.11 \, \text{kJ}/^\circ \text{C} \times 6.46^\circ \text{C} = 33.0 \, \text{kJ}
\][/tex]
### Step 3: Convert the mass of naphthalene to moles
The number of moles of naphthalene ([tex]\(n\)[/tex]) can be calculated using the given mass and the molar mass of naphthalene:
[tex]\[
n = \frac{\text{mass of naphthalene}}{\text{molar mass of naphthalene}}
\][/tex]
Given:
- [tex]\(\text{mass of naphthalene} = 0.820 \, \text{g}\)[/tex]
- [tex]\(\text{molar mass of naphthalene} = 128.17 \, \text{g/mol}\)[/tex]
Thus,
[tex]\[
n = \frac{0.820 \, \text{g}}{128.17 \, \text{g/mol}} = 0.00640 \, \text{mol}
\][/tex]
### Step 4: Calculate the change in energy per mole
The change in energy per mole ([tex]\(\Delta E_{\text{comb}}\)[/tex]) can be calculated using the heat absorbed by the calorimeter and the number of moles of naphthalene:
[tex]\[
\Delta E_{\text{comb}} = \frac{q_{\text{calorimeter}}}{n}
\][/tex]
Given:
- [tex]\(q_{\text{calorimeter}} = 33.0 \, \text{kJ}\)[/tex]
- [tex]\(n = 0.00640 \, \text{mol}\)[/tex]
Thus,
[tex]\[
\Delta E_{\text{comb}} = \frac{33.0 \, \text{kJ}}{0.00640 \, \text{mol}} = 5160 \, \text{kJ/mol}
\][/tex]
Thus, the change in energy for the combustion of naphthalene is [tex]\( \Delta E_{\text{comb}} = 5160 \, \text{kJ/mol} \)[/tex].
