A 0.009298-mol sample of an organic compound was burned in oxygen in a bomb calorimeter. The temperature of the calorimeter increased from 23.7oC to 27.1oC. If the heat capacity of the calorimeter is 4.84 kJ (oC)-1, then what is the constant volume heat of combustion of this compound, in kilojoules per mole?

q = Ccal*dT = 4.84 kJ*(27.1-23.7) = ? kJ

heat combustion is -?kJ/0.009298 = -?kJ/mol

To find the constant volume heat of combustion of the compound, we need to use the formula:

ΔH = q / n

Where:
ΔH is the heat of combustion per mole of the compound,
q is the heat absorbed by the calorimeter,
n is the number of moles of the compound.

First, we need to calculate the heat absorbed by the calorimeter (q). We can use the formula:

q = CΔT

Where:
q is the heat absorbed by the calorimeter,
C is the heat capacity of the calorimeter,
ΔT is the change in temperature.

ΔT is calculated by subtracting the initial temperature from the final temperature:

ΔT = TF - TI

Substituting the given values:
ΔT = 27.1oC - 23.7oC
ΔT = 3.4oC

Now we can calculate the heat absorbed by the calorimeter:

q = 4.84 kJ (oC)-1 × 3.4oC

q = 16.456 kJ

Next, we need to determine the number of moles of the compound (n). In this case, we have 0.009298 moles of the compound.

Finally, we can find the constant volume heat of combustion (ΔH) using the formula:

ΔH = q / n

Substituting the known values:
ΔH = 16.456 kJ / 0.009298 mol

ΔH ≈ 1768.58 kJ/mol

Therefore, the constant volume heat of combustion of the compound is approximately 1768.58 kJ/mol.