Question

When a 0.235-g sample of benzoic acid is combusted in a bomb calorimeter, the temperature rises 1.643 ∘C . When a 0.275-g sample of caffeine, C8H10O2N4, is burned, the temperature rises 1.584 ∘C . Using the value 26.38 kJ/g for the heat of combustion of benzoic acid, calculate the heat of combustion per mole of caffeine at constant volume.

Answer 1

Answer : The heat of combustion per mole of caffeine at constant volume is 76197.18 kJ/mole

Explanation :

First we have to calculate the specific heat calorimeter.

Formula used :

$Q=m\times c\times \Delta T$

where,

Q = heat of combustion of benzoic acid = 26.38 kJ/g = 26380 J/g

m = mass of benzoic acid = 0.235 g

c = specific heat of calorimeter = ?

$\Delta T$ = change in temperature = $1.643^oC$

Now put all the given value in the above formula, we get:

$26380J/g=0.235g\times c\times 1.643^oC$

$c=68323.38J/^oC$

Thus, the specific heat of calorimeter is $68323.38J/^oC$

Now we have to calculate the heat of combustion of caffeine.

Formula used :

$Q=c\times \Delta T$

where,

Q = heat of combustion of caffeine = ?

c = specific heat of calorimeter = $68323.38J/^oC$

$\Delta T$ = change in temperature = $1.584^oC$

Now put all the given value in the above formula, we get:

$Q=68323.38J/^oC\times 1.584^oC$

$Q=108224.23J=108.2kJ$

Now we have to calculate the moles of caffeine.

$\text{Moles of caffeine}=\frac{\text{Mass of caffeine}}{\text{Molar mass of caffeine}}$

Mass of caffeine = 0.275 g

Molar mass of caffeine = 194.19 g/mole

$\text{Moles of caffeine}=\frac{0.275g}{194.19g/mole}=0.00142mol$

Now we have to calculate the heat of combustion per mole of caffeine at constant volume.

$\text{Heat of combustion per mole of caffeine}=\frac{108.2kJ}{0.00142mol}=76197.18kJ/mole$

Therefore, the heat of combustion per mole of caffeine at constant volume is 76197.18 kJ/mole