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Sagot :
Given the problem, we need to calculate the change in internal energy, [tex]\(\Delta E\)[/tex], for the combustion reaction of sucrose in a bomb calorimeter. We are provided with the following data:
- Mass of sucrose: 10.0 grams
- Heat capacity of the calorimeter: 7.50 kJ/°C
- Temperature increase: 22.0 °C
- Molar mass of sucrose ([tex]\(\text{C}_{12}\text{H}_{22}\text{O}_{11}\)[/tex]): 342.30 g/mol
We proceed in a step-by-step manner as follows:
1. Calculate the heat absorbed by the calorimeter:
Given the heat capacity of the calorimeter ([tex]\(C\)[/tex]) and the temperature increase ([tex]\(\Delta T\)[/tex]), we use the formula:
[tex]\[ q = C \times \Delta T \][/tex]
Substituting the given values:
[tex]\[ q = 7.50 \, \text{kJ/°C} \times 22.0 \, \text{°C} = 165.0 \, \text{kJ} \][/tex]
Hence, the heat absorbed by the calorimeter is 165.0 kJ.
2. Calculate the number of moles of sucrose burned:
To find the number of moles, we use the molar mass of sucrose and the given mass:
[tex]\[ \text{moles of sucrose} = \frac{\text{mass of sucrose}}{\text{molar mass of sucrose}} \][/tex]
Substituting the values:
[tex]\[ \text{moles of sucrose} = \frac{10.0 \, \text{g}}{342.30 \, \text{g/mol}} \approx 0.0292 \, \text{mol} \][/tex]
So, 10.0 grams of sucrose corresponds to approximately 0.0292 moles.
3. Calculate the change in internal energy per mole of sucrose ([tex]\(\Delta E\)[/tex]):
The heat absorbed by the calorimeter represents the total heat released by the combustion of the given amount of sucrose. To find [tex]\(\Delta E\)[/tex] per mole, we divide the total heat absorbed by the number of moles of sucrose:
[tex]\[ \Delta E \, (\text{per mole}) = \frac{q}{\text{moles of sucrose}} \][/tex]
Substituting in the values:
[tex]\[ \Delta E \, (\text{per mole}) = \frac{165.0 \, \text{kJ}}{0.029214139643587496 \, \text{mol}} \approx 5647.95 \, \text{kJ/mol} \][/tex]
Thus, the change in internal energy [tex]\(\Delta E\)[/tex] for the reaction per mole of sucrose is [tex]\(5647.95\)[/tex] kJ/mol.
Expressing this to three significant figures, we get:
[tex]\[ \Delta E \approx 5650 \, \text{kJ/mol} \][/tex]
Therefore, the change in internal energy for this reaction per mole of sucrose is approximately [tex]\(5650 \, \text{kJ/mol}\)[/tex].
- Mass of sucrose: 10.0 grams
- Heat capacity of the calorimeter: 7.50 kJ/°C
- Temperature increase: 22.0 °C
- Molar mass of sucrose ([tex]\(\text{C}_{12}\text{H}_{22}\text{O}_{11}\)[/tex]): 342.30 g/mol
We proceed in a step-by-step manner as follows:
1. Calculate the heat absorbed by the calorimeter:
Given the heat capacity of the calorimeter ([tex]\(C\)[/tex]) and the temperature increase ([tex]\(\Delta T\)[/tex]), we use the formula:
[tex]\[ q = C \times \Delta T \][/tex]
Substituting the given values:
[tex]\[ q = 7.50 \, \text{kJ/°C} \times 22.0 \, \text{°C} = 165.0 \, \text{kJ} \][/tex]
Hence, the heat absorbed by the calorimeter is 165.0 kJ.
2. Calculate the number of moles of sucrose burned:
To find the number of moles, we use the molar mass of sucrose and the given mass:
[tex]\[ \text{moles of sucrose} = \frac{\text{mass of sucrose}}{\text{molar mass of sucrose}} \][/tex]
Substituting the values:
[tex]\[ \text{moles of sucrose} = \frac{10.0 \, \text{g}}{342.30 \, \text{g/mol}} \approx 0.0292 \, \text{mol} \][/tex]
So, 10.0 grams of sucrose corresponds to approximately 0.0292 moles.
3. Calculate the change in internal energy per mole of sucrose ([tex]\(\Delta E\)[/tex]):
The heat absorbed by the calorimeter represents the total heat released by the combustion of the given amount of sucrose. To find [tex]\(\Delta E\)[/tex] per mole, we divide the total heat absorbed by the number of moles of sucrose:
[tex]\[ \Delta E \, (\text{per mole}) = \frac{q}{\text{moles of sucrose}} \][/tex]
Substituting in the values:
[tex]\[ \Delta E \, (\text{per mole}) = \frac{165.0 \, \text{kJ}}{0.029214139643587496 \, \text{mol}} \approx 5647.95 \, \text{kJ/mol} \][/tex]
Thus, the change in internal energy [tex]\(\Delta E\)[/tex] for the reaction per mole of sucrose is [tex]\(5647.95\)[/tex] kJ/mol.
Expressing this to three significant figures, we get:
[tex]\[ \Delta E \approx 5650 \, \text{kJ/mol} \][/tex]
Therefore, the change in internal energy for this reaction per mole of sucrose is approximately [tex]\(5650 \, \text{kJ/mol}\)[/tex].
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