Question

3.11 g of toluene (C7H8) is burned in a bomb calorimeter. The following reaction occurs. C7H8(l) + 9 O2(g) → 7 CO2(g) + 4 H2O(l) The temperature of the calorimeter changes from 24.2°C to 40.1°C. Given that the heat capacity of the calorimeter is 8.32 kJ/°C, find the amount of heat generated per mole of toluene.

___ kJ/mol toluene

Answer 1

Ans. Heat released by combustion of toluene is absorbed by the calorimeter system to rise its temperature.

Thus, amount of heat released during toluene combustion = heat gained by calorimeter

Now using, q = c x dT           

Where, q= heat change,                                                c= specific heat of calorimeter                                   dT = change in temperature

Note: the formula does show ‘mass’ because it is not mentioned in the question as well as the unit of heat capacity of the calorimeter also lacks the dimension for mass (g). It means that the calorimeter system works at pre-set mass of the system (water is injected into the system), as most of the modern day bomb calorimeters works on.

or, q = 8.32 kJ/ ⁰C x (40.1 – 24.2)⁰C = 8.32 kJ/ ⁰C x 15.9⁰C = 132.288 J

Therefore, amount of heat released during combustion of 3.11 g gram toluene = 132.288 J

Number of moles of toluene in 3.11 g sample = (Mass / molecular mass) of toluene

                                                            = 3.11 g / 92.140 g mol^-1

                                                            = 0.033753 moles

Now,

Amount of heat produced by 1 mole = 132.288 J / (0.033753 moles)

                                                = 3919.3 J / mol

                                                = 3.9 kJ / mol