The combustion of 2.173 g of propanol (C3H7OH) increases the temperature bomb calorimeter from 298.00 K to 302.74 K.The heat capacity of the bomb calorimeter is 15.39 kJ/K

Question

. Determine Δ𝐻 for the combustion of propanol to carbon dioxide gas and liquid water.

Answer 1

Write the correctly balanced equation: 2C₃H₇OH + 9O₂ ==> 6CO₂ + 8H₂O ... balanced equation

q = Ccal∆T = (15.39 kJ/degree)(302.74 - 298.00) = (15.39 kJ/degree)(4.74 degrees) = 72.95 kJ

This is the heat released by combustion of 2.173 g of propanol. We want to find heat released by 1 mole of propanol.

molar mass propanol = 60.10 g/mol

moles propanol = 2.173 g x 1 mol/60.10 g = 0.03616 moles

∆H = 72.95 kJ/0.03616 moles = 2017 kJ/mole