Question

1.576 g of an unknown hydrocarbon (66.5 g/mol) burns in bomb calorimeter in excess oxygen. The heat capacity of the calorimeter,Cv, = 5.820 kJ/ºC and ΔT =6.626 ºC. Find ΔE for this hydrocarbon in kJ/mol.

Answer 1

Qcal = Ccal * delta T

= 5.820 KJ/oC * 6.626 oC

= 38.563 KJ

This is heat absorbed by calorimeter which must be released by experiment

mass of hydrocarbon = 1.576 g

we have below equation to be used:

number of mol of hydrocarbon,

n = mass of hydrocarbon/molar mass of hydrocarbon

=(1.576 g)/(66.5 g/mol)

= 2.37*10^-2 mol

delta E = -Qcal/number of mol

= 38.563 KJ / (2.37*10^-2 mol)

= 1627 KJ/mol

Answer: 1627 KJ/mol