Question

A 0.623 g sample of vanillin (C8H8O3, MM = 152.15) is combusted in a bomb calorimeter with a heat capacity of 5.89 kJ/ºC. Given that the heat of combustion of vanillin is -3.83x103 kJ/mol, what must the temperature change have been in the bomb calorimeter?

Answer 1

In bomb calorimeter, the heat of combustion of the reaction can be found experimentally at constant volume

The heat of combustion can be calculated using the equation

Q = m*C*dT

Q = heat of combustion, m=mass of sample, C=specific heat capacity, dT=temp. Change

Given that the heat of combustion of vanillin is -3.83x10^3 kJ/mol,i.e., heat of combustion per mole = -3.83x10^3 kJ

For 0.623 g,no. of moles = m/Mw = 0.623/152.15 = 0.0041 mole

heat of combustion per mole = -3.83x10^3 kJ

heat of combustion per 0.0041mole = -3.83x10^3 kJ*0.0041/1mole = 15.68 KJ

Now for the combusiton of 0.0041 mole of Vanallin temp. change is

Q=m*C*dT

-15.68 KJ = 0.623 g*5.89 KJ/o C *dT

dT = 15.68/0.623*5.89 = 4.27 K

the temperature change in the bomb calorimeter is 4.27 K