When 1.010 of sucrose (C12H22O11) undergoes combustions in a bomb calorimeter, the temperature rises from 24.92 ◦ C to 28.33 ◦ C. Find DErxn for the combustion of sucrose in kJ/mol sucrose. The heat capacity of the bomb calorimeter determined in a separate experiment is 4.90 kJ/◦ C. (Answer: - 5.66 x 103 kJ/mole).

Heat of combustion = 5.6 * 10³ kj/mol

Explanation:

Given dа ta:

Mass of sucrose = 1.010 g

Initial temperature = 24.92 °C

Final temperature = 28.33 °C

Heat capacity of calorimeter = 4.90 KJ/°C

Heat of combustion = ?

Solution:

ΔT = 28.33 °C - 24.92 °C = 3.41 °C

Q = - c. ΔT

Q = 4.90 KJ/°C. 3.41 °C

Q = - 16.7 kj

Number of moles of sucrose:

Number of moles of sucrose = mass / molar mass

Number of moles of sucrose = 1.010 g / 342.3 g/mol

Number of moles of sucrose = 0.003 mol

Heat of combustion:

Heat of combustion = Q/n

Heat of combustion = - 16.7 kj/0.003 mol

Heat of combustion = - 5.6 * 10³ kj/mol