A 0.003913 mol sample of an organic compound was burned in oxygen in a bomb calorimeter. The temperature of the calorimeter increased from 23.1 °C to 25.7 °C. If the heat capacity of the calorimeter is 4.24 kJ (°C)−1, then what is the constant volume heat of combustion of this compound, in kilojoules per mole?
4.24(25.7-23.1) = ?
? / 0.0039 = ? should be ~ 2817 kJ/mol, but negative bc heat is going into surroundings
To find the constant volume heat of combustion of the organic compound, we need to use the equation:
q = C * ΔT
where:
q is the heat transferred,
C is the heat capacity of the calorimeter, and
ΔT is the change in temperature.
Given:
C = 4.24 kJ/°C,
ΔT = 25.7 °C - 23.1 °C = 2.6 °C.
Plugging in these values, we can calculate the heat transferred:
q = 4.24 kJ/°C * 2.6 °C
= 11.024 kJ
Now, the heat transferred during the combustion reaction of 0.003913 mol of the compound is 11.024 kJ. To find the constant volume heat of combustion per mole of the compound, we divide this value by the number of moles:
Heat of combustion = q / n
where:
q is the heat transferred (11.024 kJ), and
n is the number of moles (0.003913 mol).
Dividing the heat transferred by the number of moles, we get:
Heat of combustion = 11.024 kJ / 0.003913 mol
≈ 2816.968 kJ/mol
Therefore, the constant volume heat of combustion of the compound is approximately 2816.968 kJ/mol.