Question

3.00 g of a fuel (MW = 650.0 g/mol) is combusted in a bomb calorimeter. The temperature of the calorimeter increases by 25.0 degrees Celsius. The heat capacity for the calorimeter has been determined to be 9.001kJ/C. What is the heat of reaction (in kJ/mol)?​

Answer 1

-4.88 × 10⁴ kJ/mol

Step-by-step explanation:

Step 1: Calculate the heat gained by the calorimeter (Qcal)

We will use the following expression.

Qcal = C × ΔT

where,

• C: heat capacity of the calorimeter

• ΔT: temperature change

Qcal = 9.001 kJ/°C × 25.0°C = 225 kJ

Step 2: Calculate the heat released by the fuel

According to the law of conservation of energy, the sum of the heat released by the fuel (Qf) and the heat absorbed by the calorimeter is zero.

Qf + Qcal = 0

Qf = -Qcal = -225 kJ

Step 3: Calculate the heat of reaction in kJ/mol

3.00 g of a fuel (MW = 650.0 g/mol) released 225 kJ of heat. The heat of reaction is:

-225 kJ/3.00 g × 650.0 g/mol = -4.88 × 10⁴ kJ/mol