A sample of n-octane (1.14 g) was completely burnt in excess of oxygen in a bomb calorimeter, whose heat capacity is 5 kJ K\(^{-1}\). As a result of combustion, the temperature of the calorimeter increased by 5 K. The magnitude of the heat of combustion at constant volume is ___
A sample of n-octane (1.14 g) was completely burnt in excess of oxygen in a bomb calorimeter, whose heat capacity is 5 kJ K\(^{-1}\). As a result of combustion, the temperature of the calorimeter increased by 5 K. The magnitude of the heat of combustion at constant volume is ___
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Correct Answer: 2500
Approach Solution - 1
Mass of n-octane = 1.14 g Heat capacity of the bomb calorimeter (C) = 5 kJ K\(^{-1}\)
Increase in temperature (\( \Delta T \)) = 5 K The heat evolved during the combustion of n-octane at constant volume (\( q_v \)) is absorbed by the calorimeter, causing the temperature increase.
Magnitude of heat evolved = \( q_v = C \times \Delta T \) \( q_v = 5 \, \text{kJ K}^{-1} \times 5 \, \text{K} = 25 \, \text{kJ} \)
This is the heat evolved from the combustion of 1.14 g of n-octane. We need to find the heat of combustion per mole of n-octane.
The molecular formula of n-octane is C\( _8 \)H\( _{18} \).
The molar mass of n-octane is: \( (8 \times 12) + (18 \times 1) = 96 + 18 = 114 \, \text{g mol}^{-1} \) Number of moles of n-octane burnt = \( \frac{\text{mass of n-octane}}{\text{molar mass of n-octane}} \) \[ \text{Moles of n-octane} = \frac{1.14 \, \text{g}}{114 \, \text{g mol}^{-1}} = 0.01 \, \text{mol} \] The heat evolved from the combustion of 0.01 mol of n-octane is 25 kJ. The heat of combustion per mole of n-octane (\( \Delta U \)) at constant volume is: \[ \Delta U = \frac{\text{Heat evolved}}{\text{Moles of n-octane}} = \frac{25 \, \text{kJ}}{0.01 \, \text{mol}} = 2500 \, \text{kJ mol}^{-1} \] The magnitude of the heat of combustion at constant volume is 2500 kJ mol\(^{-1}\).
The nearest integer is 2500.
Approach Solution -2
The heat of combustion of n-octane can be determined using the formula for heat transfer in a bomb calorimeter, \( q = C \Delta T \), where \( q \) is the heat evolved, \( C \) is the heat capacity, and \( \Delta T \) is the temperature change.
Given:
Heat capacity, \( C = 5 \) kJ K\(^{-1}\)
Temperature change, \( \Delta T = 5 \) K
Substituting these values into the formula, we get:
\( q = 5 \times 5 = 25 \) kJ.
This is the heat evolved for 1.14 g of n-octane. To find the heat of combustion per mole, we need the molar mass of n-octane (\( C_8H_{18} \)), which is 114 g/mol.
Amount of n-octane burnt = \(\frac{1.14 \text{ g}}{114 \text{ g/mol}} = 0.01 \text{ mol}\).
The heat of combustion per mole is then calculated as:
\(\frac{25 \text{ kJ}}{0.01 \text{ mol}} = 2500 \text{ kJ/mol}\).
Thus, the magnitude of the heat of combustion at constant volume is 2500 kJ/mol, which falls within the range 2500,2500.
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