calorimetry....

Question

When 1.231 grams of sucrose (Molar mass 342.3 g/mol) is burned in a bomb calorimeter, the temperature of the calorimeter increases from 22.41°C to 26.63°C. If the heat capacity of the calorimeter is 4.900 kJ/°C, what is the heat of combustion of sucrose?

Answer 1

∆H = Ccal x ∆T

∆H = Enthalpy of combustion = ?

Ccal = heat capacity of the calorimeter = 4.900 kJ/ºC

∆T = change in temperature = 26.63º - 22.41º = 4.22ºC

∆H = (4.900 kJ/deg)(4.22 deg) = -20.68 kJ (value is negative because temperature increased making this an exothermic reaction)

Since we have 1.231 g sucrose, we have 1.231 g / 342.2 g/mol = 0.00360 moles sucrose

Thus, ... ∆H/mole = -20.68 kJ/0.00360 moles = -5744 kJ/mole