Question

The combustion of 0.1625 g benzoic acid increases the temperature of a bomb calorimeter by 2.41°C. Calculate the heat capacity of this calorimeter. (The energy released by combustion of benzoic acid is 26.42 kJ/g.)

kJ/°C
A 0.2070 g sample of vanillin (C8H8O3) is then burned in the same calorimeter, and the temperature increases by 3.19°C. What is the energy of combustion per gram of vanillin?
kJ/g
What is the energy of combustion per mole of vanillin?
kJ/mol

Answer 1

The specific heat of calorimeter is
`1.78kJ/^oC`

The energy of combustion per mole of vanillin is
`-4.18* 10^3kJ/mol`

Explanation :

Part 1 :

First we have to calculate the energy released for 0.1625 g of benzoic acid.

Energy released = Energy released × Mass of benzoic acid

Energy released = (26.42 kJ/g) × (0.1625g)

Energy released = -4.293 kJ

Now we have to calculate the specific heat of calorimeter.

Heat released by the reaction = Heat absorbed by the calorimeter

`\Delta E_(rxn)=q_(rxn)=-q_(cal)`

`q_(rxn)=q_(cal)=-c_(cal)* \Delta T`

where,

`q_(rxn)` = heat released by the reaction = -4.293 kJ

`q_(cal)` = heat absorbed by the calorimeter

`c_(cal)` = specific heat of calorimeter = ?

`\Delta T` = change in temperature =
`2.41^oC`

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

`-4.293kJ=-c_(cal)* 2.41^oC`

`c_(cal)=1.78kJ/^oC`

Thus, the specific heat of calorimeter is
`1.78kJ/^oC`

Part 2 :

First we have to calculate the energy released by the reaction.

`q_(rxn)=q_(cal)=-c_(cal)* \Delta T`

`q_(cal)=c_(cal)* \Delta T`

where,

`q_(rxn)` = heat released by the reaction = ?

`q_(cal)` = heat absorbed by the calorimeter

`c_(cal)` = specific heat of calorimeter =
`1.78kJ/^oC`

`\Delta T` = change in temperature =
`3.19^oC`

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

`q_(cal)=c_(cal)* \Delta T`

`q_(cal)=1.78kJ/^oC* 3.19^oC`

`q_(cal)=5.68kJ`

`\Delta E_(rxn)=q_(rxn)=-q_(cal)`

`\Delta E_(rxn)=-5.68kJ`

Now we have to calculate the energy of combustion per mole of vanillin.

`\text{Moles of vanillin}=\frac{\text{Mass of vanillin}}{\text{Molar mass of vanillin}}`

Molar mass of vanillin = 152.15 g/mole

Mass of vanillin = 0.2070 g

`\text{Moles of vanillin}=(0.2070g)/(152.15g/mole)=0.00136mole`

`\Delta E_(rxn)=(-5.68kJ)/(0.00136mole)=-4.18* 10^3kJ/mol`

Thus, the energy of combustion per mole of vanillin is
`-4.18* 10^3kJ/mol`