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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

Respuesta :

Answer :

The specific heat of calorimeter is [tex]1.78kJ/^oC[/tex]

The energy of combustion per mole of vanillin is [tex]-4.18\times 10^3kJ/mol[/tex]

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

[tex]\Delta E_{rxn}=q_{rxn}=-q_{cal}[/tex]

[tex]q_{rxn}=q_{cal}=-c_{cal}\times \Delta T[/tex]

where,

[tex]q_{rxn}[/tex] = heat released by the reaction = -4.293 kJ

[tex]q_{cal}[/tex] = heat absorbed by the calorimeter

[tex]c_{cal}[/tex] = specific heat of calorimeter = ?

[tex]\Delta T[/tex] = change in temperature = [tex]2.41^oC[/tex]

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

[tex]-4.293kJ=-c_{cal}\times 2.41^oC[/tex]

[tex]c_{cal}=1.78kJ/^oC[/tex]

Thus, the specific heat of calorimeter is [tex]1.78kJ/^oC[/tex]

Part 2 :

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

[tex]q_{rxn}=q_{cal}=-c_{cal}\times \Delta T[/tex]

[tex]q_{cal}=c_{cal}\times \Delta T[/tex]

where,

[tex]q_{rxn}[/tex] = heat released by the reaction = ?

[tex]q_{cal}[/tex] = heat absorbed by the calorimeter

[tex]c_{cal}[/tex] = specific heat of calorimeter = [tex]1.78kJ/^oC[/tex]

[tex]\Delta T[/tex] = change in temperature = [tex]3.19^oC[/tex]

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

[tex]q_{cal}=c_{cal}\times \Delta T[/tex]

[tex]q_{cal}=1.78kJ/^oC\times 3.19^oC[/tex]

[tex]q_{cal}=5.68kJ[/tex]

[tex]\Delta E_{rxn}=q_{rxn}=-q_{cal}[/tex]

[tex]\Delta E_{rxn}=-5.68kJ[/tex]

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

[tex]\text{Moles of vanillin}=\frac{\text{Mass of vanillin}}{\text{Molar mass of vanillin}}[/tex]

Molar mass of vanillin = 152.15 g/mole

Mass of vanillin = 0.2070 g

[tex]\text{Moles of vanillin}=\frac{0.2070g}{152.15g/mole}=0.00136mole[/tex]

[tex]\Delta E_{rxn}=\frac{-5.68kJ}{0.00136mole}=-4.18\times 10^3kJ/mol[/tex]

Thus, the energy of combustion per mole of vanillin is [tex]-4.18\times 10^3kJ/mol[/tex]