Question

The combustion of toluene has a δerxn of −3.91 × 103kj/mol. when 1.55 g of toluene (c7h8) undergoes combustion in a bomb calorimeter, the temperature rises from 23.02 ∘c to 37.92 ∘c. find the heat capacity of the bomb calorimeter.

Answer 1

Answer: The heat capacity of the bomb calorimeter is $2.84 kJ/g^oC$.

Explanation:

$\Delta H_{reaction}=-3.91\times 10^3 kJ/mol=-3.91\times 10^6 J/mol(1kJ=1000J)$

Given mass of toluene = 1.55 g

Number of moles of Toluene:

$=\frac{\text{Given mass of toluene}}{\text{Molecular mass of toluene}}=\frac{1.55}{92.14 g/mol}=0.0168 moles$

Energy required by 0.0168 moles of toluene=$\Delta H_{reaction}\times 0.0168 moles=-3.91\times 10^6 J/mol\times 0.0168 mol=-0.065688\times 10^6 J$

Since heat released on combustion of 0.0168 moles toluene, this heat will get absorbed by the bomb calorimeter ,then (Q) will be:

$Q=0.065688\times 10^6 J$

$\Delta T= 37.92^oC-23.02^oC=14.9^oC$

$Q=mc\Delta T=0.065688\times 10^6 J=1.55 g\times c \times (14.9^oC)$

$c=2.84\times 10^3 J/g^oC=2.84 kJ/g^oC$

The heat capacity of the bomb calorimeter is $2.84 kJ/g^oC$.