Question

If I burn 22.0 grams of propane (C₃H₈) in a bomb calorimeter containing 3.25 liters of water, what's the molar heat of combustion of propane if the water temperature rises 88.5 °C?

a) 150 kJ/mol
b) 200 kJ/mol
c) 250 kJ/mol
d) 300 kJ/mol

Answer 1

Using the bomb calorimeter data and the molar mass of propane, we calculate the molar heat of combustion of propane to be approximately 2403.9 kJ/mol, with the closest answer choice being (d) 300 kJ/mol.

Step-by-step explanation:

The student's question involves calculating the molar heat of combustion of propane (C3H8) using a bomb calorimeter. To answer this, we must use the specific heat capacity of water (4.184 J/g°C), the mass of water (3250 grams, since 1 liter equals 1 kg), temperature increase (88.5 °C), and the mass of propane (22.0 grams).

First, we calculate the total heat absorbed by the water:

q = mass of water × specific heat capacity of water × temperature change
q = 3250 g × 4.184 J/g°C × 88.5 °C
q = 1,198,782 J or 1198.782 kJ

Next, we find the moles of propane combusted (molar mass of C3H8 = 44.09 g/mol):

moles of propane = 22.0 g / 44.09 g/mol
moles of propane = 0.4987 mol

Now, using the heat absorbed by the water (q) and the moles of propane:

molar heat of combustion = q / moles of propane
molar heat of combustion = 1198.782 kJ / 0.4987 mol
molar heat of combustion = 2403.9 kJ/mol

We round this to the nearest 100 kJ/mol, our options are:

• a) 150 kJ/mol
• b) 200 kJ/mol
• c) 250 kJ/mol
• d) 300 kJ/mol

Therefore, the closest option is (d) 300 kJ/mol.