Respuesta :
Answer : The internal energy change for 1 mole of sucrose is [tex]-5.65\times 10^3kJ/mole[/tex]
Explanation :
First we have to calculate the heat released by calorimeter.
[tex]q_{calorimeter}=-c\times \Delta T[/tex]
where,
[tex]q_{calorimeter}[/tex] = heat released by calorimeter = ?
c = heat capacity of bomb calorimeter = [tex]7.50kJ/^oC[/tex]
[tex]\Delta T[/tex] = change in temperature of calorimeter = [tex]22.0^oC[/tex]
Now put all the given values in the above formula, we get
[tex]q_{calorimeter}=-7.50kJ/^oC\times 22.0^oC[/tex]
[tex]q_{calorimeter}=-165kJ[/tex]
Now we have to calculate the moles of sucrose.
[tex]\text{Moles of sucrose}=\frac{\text{Mass of sucrose}}{\text{Molar mass of sucrose}}=\frac{10.0g}{342.3g/mol}=0.0292mol[/tex]
Now we have to calculate the internal energy change.
As we know that, bomb calorimeter carried out at constant volume condition.
[tex]\Delta q=\Delta E+P\Delta V[/tex]
here, [tex]\Delta V=0[/tex]
So,
[tex]\Delta q=\Delta E=-165kJ[/tex]
Now we have to calculate the internal energy change for 1 mole of sucrose.
[tex]\Delta E=\frac{\Delta q}{n}=\frac{-165kJ}{0.0292mole}=-5650.7kJ/mole=-5.65\times 10^3kJ/mole[/tex]
Therefore, the internal energy change for 1 mole of sucrose is [tex]-5.65\times 10^3kJ/mole[/tex]
The combustion of sucrose, leading to an increase in the temperature of a bomb calorimeter of 22.°C, represents a change in the internal energy of -5.65 × 10³ kJ/mol.
Let's consider the combustion of sucrose.
C₁₂H₂₂O₁₁(s) + 12 O₂(g) → 12 CO₂(g) + 11 H₂O(l)
First, we will convert 10.0 g of sucrose to moles (n) using its molar mass (342.30 g/mol).
[tex]10.0 g \times \frac{1mol}{342.30 g} = 0.0292 mol[/tex]
Then, we will calculate the heat absorbed by the bomb calorimeter (Qb) using the following expression.
[tex]Qb = C \times \Delta T = \frac{7.50 kJ}{\° C} \times 22.0 \° C = 165 kJ[/tex]
where,
- C: heat capacity of the calorimeter
- ΔT: change in the temperature
According to the law of conservation of energy, the sum of the heat absorbed by the bomb calorimeter and the heat released by the combustion (Qc) is zero.
[tex]Qb + Qc = 0\\\\Qc = -Qb = -165 kJ[/tex]
Finally, we can calculate the change in the internal energy (ΔE) using the following expression.
[tex]\Delta E = \frac{Qc}{n} = \frac{-165 kJ}{0.0292mol} = -5.65 \times 10^{3} kJ/mol[/tex]
The combustion of sucrose, leading to an increase in the temperature of a bomb calorimeter of 22.°C, represents a change in the internal energy of -5.65 × 10³ kJ/mol.
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