Question

The heat of combustion (∆H) for an unknown hydrocarbon is -8.21 kJ/mol. If 0.424 mol of the hydrocarbon is burned in a bomb calorimeter, what is the change in temperature (°C) of the calorimeter? The heat capacity of the bomb calorimeter is 1.12 kJ/°C.

Answer 1

When 0.424 moles of an unknown hydrocarbon (∆Hc = -8.21 kJ/mol) is burned in a bomb calorimeter (C = 1.12 kJ/°C), the change in the temperature is 3.10 °C.

The heat of combustion (∆Hc) for an unknown hydrocarbon is -8.21 kJ/mol. The heat released by the combustion of 0.424 moles of the hydrocarbon is:

$Qc = 0.424 mol \times \frac{(-8.21kJ)}{mol} = -3.48 kJ$

According to the law of conservation of energy, the sum of the heat released by the combustion (Qc) and the heat absorbed by the bomb calorimeter (Qb) is zero.

$Qc + Qb = 0\\\\Qb = -Qc = 3.48 kJ$

Given the heat absorbed by the bomb calorimeter (Qb) and the heat capacity of the bomb calorimeter (C), we can calculate the temperature change (ΔT) using the following expression.

$Qb = C \times \Delta T\\\\\Delta T = \frac{Qb}{C} = \frac{3.48 kJ}{1.12 kJ/\° C } = 3.10 \° C$

When 0.424 moles of an unknown hydrocarbon (∆Hc = -8.21 kJ/mol) is burned in a bomb calorimeter (C = 1.12 kJ/°C), the change in the temperature is 3.10 °C.

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