combustionof 1.943g of glucose,C6H12O6(s),in Bomb calorimeter with a heat capacity of 4.10kJ/°C results in an increase in the temperature of the calorimeter and its contents from 23.00 °C to 30.39°C.

Question

What is the internal energy change, Δ𝑈 , for the combustion of 1.943 g of glucose? Calculate the enthalpy of combustion, Δ𝐻c, of glucose in kilojoules per mole.

Answer 1

q = Ccal∆T

q = heat = ?

Ccal = calorimeter constant = 4.10 kJ/º

∆T = change in temp =30.39º - 23.00º = 7.39º

q = (4.10 kJ/º)(7.39º) = 30.30 kJ = ∆U since it is constant volume and no work was performed

∆Hc = 30.30 kJ/moles glucose

molar mass glucose = 180.2 g/mol

moles glucose = 1.943 g glucose x 1 mol/180.2 g = 0.01078 moles

∆Hc = 30.30 kJ/0.01078 moles = 2811 kJ/mol