Question

The enthalpy of combustion of benzoic acid (C6H5COOH) is commonly used as the standard for calibrating constant-volume bomb calorimeters; its value has been accurately determined to be −3226.7 kJ/mol. When 2.8161 g of benzoic acid are burned in a calorimeter, the temperature rises from 21.84°C to 24.67°C. What is the heat capacity of the bomb? (Assume that the quantity of water surrounding the bomb is exactly 2250 g.)

Answer 1

Answer : The heat capacity of the bomb calorimeter
`15.633kJ/^oC`

Explanation :

First we have to calculate the heat released by the combustion.

`q=n* \Delta H`

where,

q = heat released by combustion = ?

n = moles of benzoic acid =
`\frac{\text{Mass of benzoic acid}}{\text{Molar mass of benzoic acid}}=(2.8161g)/(122.122g/mole)=0.02306mole`

`\Delta H` = enthalpy of combustion = 3226.7 kJ/mole

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

`q=(0.02306mole)* (3226.7kJ/mole)=74.407kJ`

Now we have to calculate the heat capacity of the bomb calorimeter.

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

`q=[q_1+q_2]`

`q=[c_1* \Delta T+m_2* c_2* \Delta T]`

where,

q = heat released by the reaction = 74.4077 kJ = 74407.7 J

`q_1` = heat absorbed by the calorimeter

`q_2` = heat absorbed by the water

`c_1` = specific heat of calorimeter = ?

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

`m_2` = mass of water = 2550 g

`\Delta T` = change in temperature =
`T_2-T_1=24.67-21.84=2.83^oC`

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

`74407J=[(c_1* 2.83^oC)+(2550g* 4.18J/g^oC* 2.83^oC)]`

`c_1=15633.226J/^oC=15.633kJ/^oC`

Therefore, the heat capacity of the bomb calorimeter
`15.633kJ/^oC`