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

= - 26.31 kJ

Step-by-step explanation:

we know that number of moles is calculated as

`Moles of C_(12)H_(22)O_(11) = (mass)/(molecular\ weight)`

`= ((1.631 g))/((342.29 g/mol))`

= 0.00476 mol

Heat absorbed by calorimeter
`= heat\ capacity * temperature\ rise`

`= 5.30 kJ/°C x (27.75 - 22.68)°C`

= 26.87 kJ

Enthalpy of combustion

`\Delta Hc = (- 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 * 0.00476 mol`

= - 26.31 kJ