175k views
0 votes
A 6.55 g sample of aniline (C_{6}H_{5}NH_{2}, molar mass = 93.13 g/mol) was combusted in a bomb calorimeter. If the temperature rose by 32.9°C, use the information below to determine the heatcapacity of the calorimeter.

1 Answer

3 votes

This is an incomplete question, here is a complete question.

A 6.55 g sample of aniline
(C_6H_5NH_2) molar mass = 93.13 g/mol) was combusted in a bomb calorimeter. If the temperature rose by 32.9°C, use the information below to determine the heat capacity of the calorimeter.


4C_6H_5NH_2(l)+35O_2(g)\rightarrow 24CO_2(g)+14H_2O(g)+4NO_2(g)

ΔH°rxn = -1.28 × 10⁴ kJ

Answer : The heat capacity of the calorimeter is,
6.84kJ/^oC

Explanation :

First we have to calculate the moles of aniline.


\text{Moles of aniline}=\frac{\text{Mass of aniline}}{\text{Molar mass of aniline}}


\text{Moles of aniline}=(6.55g)/(93.13g/mol)


\text{Moles of aniline}=0.0703mol

Now we have to calculate the heat releases.

As, 4 mole of aniline on combustion to releases heat =
1.28* 10^4kJ

So, 0.0703 mole of aniline on combustion to releases heat =
(0.0703)/(4)* 1.28* 10^4kJ=224.96kJ

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


\text{Heat capacity of the calorimeter}=\frac{\text{Heat releases}}{\text{Change in temperature}}


\text{Heat capacity of the calorimeter}=(224.96kJ)/(32.9^oC)


\text{Heat capacity of the calorimeter}=6.84kJ/^oC

Thus, the heat capacity of the calorimeter is,
6.84kJ/^oC

User Jmbucknall
by
4.3k points