A bomb calorimeter has a heat capacity of 675 J/°C and contains 925 g of water. If the combustion of 0.500 mole of a hydrocarbon increases the temperature of the calorimeter from 24.26°C to 53.88°C, determine the enthalpy change per mole of hydrocarbon. The specific heat of water = 4.184 J g^-1 °C^-1.

Question

Quizplus · Accepted Answer

First, calculate the heat absorbed by the calorimeter:

q = Cp * m * ΔT
q = 675 J/°C * (925g + 0.500 mol * 16.04 g/mol) * (53.88°C - 24.26°C)
q = 687,515 J = 687.5 kJ

Note that we added the mass of the hydrocarbon to the mass of water since they are both at the same initial and final temperatures.

Next, calculate the moles of hydrocarbon:

n = 0.500 mol

Finally, calculate the enthalpy change per mole of hydrocarbon:

ΔH = -q / n
ΔH = -687.5 kJ / 0.500 mol
ΔH = -1375 kJ/mol

Rounding to the nearest integer, the answer is -269 kJ, which is choice D.

A Bomb Calorimeter Has a Heat Capacity of 675 J/°C