Question

Part 1: A 1.07 g sample of caffeine (C8H10N4O2) burns in a constant-volume calorimeter that has a heat capacity of 7.28 kJ/K. The temperature increases from 297.65 K to 301.38 K. Determine the heat (qv) associated with this reaction.

Part 2:Now use the data above to find ΔE for the combustion of one mole of caffeine

Answer 1

Part 1:

Given that mass of caffeine , m = 1.07 g

molar mass of caffeine = 194.2 g/mol

Moles of caffeine = mass / molar mass = 1.07 g/ 194.2 g/mol = 0.0055 mol

heat capacity of calorimeter at constant volume, Cv = 7.28 kJ/K

Temperature increases from 297.65 K to 301.38 K.

Hence, temperature difference, dT = 301.38 K - 297.65 K = 3.73 K

Therefore,

Heat capacity of the calorimeter , Cv = q/dT

q = Cv x dT

= 7.28 kJ/K x 3.73 K

= 27.15 kJ

qv = 27.15 kJ

Heat associated with burning of  0.0055 mol (1.07 g) of caffeine = qv/moles of caffeine

= 27.15 kJ x 0.0055

= 0.149 kJ/mol

Therefore, heat associated with burning of  0.0055 mol (1.07 g) of caffeine = 0.149 kJ/mol

Part 2:

Given that moles of Caffeine , n =1

  heat capacity at constant volume, Cv = 7.28 kJ/K

temperature difference, dT = 3.73 K

We know that

ΔE = nCvdT

= 1 x 7.28 kJ/K x 3.73 K

= 27.15 kJ

ΔE = 27.15 kJ

Therefore, ΔE for the combustion of one mole of caffeine = 27.15 kJ