Question

The combustion of 1.961 g of sucrose, C12H22O11(s), in a bomb calorimeter with a heat capacity of 4.00 kJ/°C results in an increase in the temperature of the calorimeter and its contents from 22.92 °C to 31.00 °C. Calculate the enthalpy of combustion, Δ?c, for sucrose in kilojoules per mole.

Δ?c= kJ/mol

What is the internal energy change, Δ?, for the combustion of 1.961 g of sucrose in the bomb calorimeter?

Δ?= kJ

Answer 1

Moles of C12H22O11 = mass/molecular weight

= (1.961 g) / (342.29 g/mol)

= 0.005729 mol

Heat absorbed by calorimeter = heat capacity x temperature rise

= 4.00 kJ/°C x (31 - 22.92)°C

= 32.32 kJ

Enthalpy of combustion

ΔHc = - 32.32/0.005729

= - 5641.47 kJ/mol

Negative sign shows that the heat is released

The balanced reaction

C12H22O11(s) + 12 O2(g) = 12 CO2(g) + 11 H2O(l)

ΔHc = ΔU + Δng (RT)

-5641.47 = ΔU + (12 - 12) *(RT)

ΔU = - 5641.47 kJ/mol x 0.005729 mol

= - 32.32 kJ