Question

The combustion of 1.631 g of sucrose, C12H22O11(s) , in a bomb calorimeter with a heat capacity of 5.30 kJ/°C results in an increase in the temperature of the calorimeter and its contents from 22.68 °C to 27.75 °C. What is the internal energy change, ΔU , for the combustion of 1.631 g of sucrose?

Answer 1

Answer:

= - 26.31 kJ

Explanation:

we know that number of moles is calculated as

$Moles of C_{12}H_{22}O_{11} = \frac{mass}{molecular\ weight}$

               $= \frac{(1.631 g)}{(342.29 g/mol)}$

                      = 0.00476 mol

Heat absorbed by calorimeter $= heat\ capacity \times temperature\ rise$

$= 5.30 kJ/°C x (27.75 - 22.68)°C$

= 26.87 kJ

Enthalpy of combustion

$\Delta Hc = \frac{- 26.87}{0.00486}$

= - 55290.12 kJ/mol

Negative sign shows that the heat is released

The balanced reaction

$C_{12}H_{22} O_{11}(s) + 12 O_2(g) = 12 CO_2(g) + 11 H_2O(l)$

ΔHc = ΔU + Δng (RT)

-55290.12 = ΔU + (12 - 12) *(RT)

$\Delta U = - 55290.12 kJ/mol \times 0.00476 mol$

= - 26.31 kJ