Question

If 6.28 g of glucose is combusted in a bomb calorimeter at constant pressure, the temperature of the calorimeter rises by 5.25°C. If the specific heat of the calorimeter is 18.6 kJ/°C, what is the molar heat of combustion for glucose?

Answer 1

mass of glucose = 6.28 g

moles of glucose = 6.28 / 180 = 0.0349 mol

specific heat = 18.6 kJ / oC

temperature rise = 5.25 oC

Q = Cp dT

= 18.6 x 5.25

Q = 97.65 kJ

molar heat of combustion = 97.65 / 0.0349

                                         = 2798 kJ/mol

molar heat of combustion = 2.80 x 10^3 kJ/mol