Question

A 0.709 g sample of vanillin (C8H8O3, MM = 152.15) is combusted in a bomb calorimeter with a heat capacity of 5.97 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

Solution:

Heat of combustion (q) per mol of bomb calorimeter is calculated as:

q = - C ΔT / n

Where,

n = number of moles

C = heat capacity

ΔT = change in temperature

Number of moles of vanilin (n) = mass / molar mass

n = 0.709 g / 152.15 g mol-1 = 0.00466 mol

C = 5.97 kj / °C (given)

q = -3.83 x 10^3 KJ /mol = -3830 KJ/mol

Then,

q = - C ΔT / n

ΔT = - q x n / - C = - 3830 KJ mol-1 x 0.00466 mol / - 5.97 KJ/°C

= 2.99 °C

Hence, temperature change = 2.99 °C