A 0.016076-mol sample of an organic compound was burned in oxygen in a bomb calorimeter. The temperature of the calorimeter increased from 23.7oC to 24.5oC. If the heat capacity of the calorimeter is 4.98 kJ (oC)-1, then what is the constant volume heat of combustion of this compound, in kilojoules per mole? (Remember to include a "+" or "-" sign in your answer, as appropriate.)
To calculate the constant volume heat of combustion, we need to use the formula:
q = ∆T * C
Where:
q = heat absorbed or released during the reaction
∆T = change in temperature
C = heat capacity of the calorimeter
Given:
∆T = (24.5oC - 23.7oC) = 0.8oC
C = 4.98 kJ (oC)-1
Substituting the values into the formula:
q = 0.8oC * 4.98 kJ (oC)-1
Next, we need to convert moles of the compound (0.016076 mol) to kilojoules per mole. To do this, we need to divide the total heat by the number of moles:
Heat per mole = q / moles
Calculating the value:
q = 0.8oC * 4.98 kJ (oC)-1 = 3.984 kJ
Heat per mole = 3.984 kJ / 0.016076 mol
Dividing these values, we find:
Heat per mole ≈ 248.086 kJ/mol
Hence, the constant volume heat of combustion of this compound is approximately 248.086 kJ/mol.