Question

A 2.50 mol sample of benzene (C6H6, 78.11 g/mol) was burned in a bomb calorimeter with a heat capacity of 800 J/°C. The calorimeter contained 100g of water (4.18 J/g°C), and the temperature increased by 4°C. What is the molar heat of combustion of benzene in kJ/mol of benzene?

Answer 1

Answer : The molar heat of combustion of benzene is 1.95 kJ/mol

Explanation :

Heat released by the reaction = Heat absorbed by the calorimeter + Heat absorbed by the water

$q=[q_1+q_2]$

$q=[c_1\times \Delta T+m_2\times c_2\times \Delta T]$

where,

q = heat released by the reaction

$q_1$ = heat absorbed by the calorimeter

$q_2$ = heat absorbed by the water

$c_1$ = specific heat of calorimeter = $800J/^oC$

$c_2$ = specific heat of water = $4.18J/g^oC$

$m_2$ = mass of water = 100 g

$\Delta T$ = change in temperature = $4^oC$

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

$q=[(800J/^oC\times 4^oC)+(100g\times 4.18J/g^oC\times 4^oC)]$

$q=4872J$

Now we have to determine the molar heat of combustion of benzene in kJ/mol.

As, 2.50 mole of benzene produces heat on combustion = 4872 J

So, 1 mole of benzene produces heat on combustion = $\frac{4872}{2.50}=1948.8J/mol=1.95kJ/mol$

conversion used : (1 kJ = 1000 J)

Therefore, the molar heat of combustion of benzene is 1.95 kJ/mol