Question

The heat capacity of a bomb calorimeter was determined by burning 6.91 g of methane (energy of combustion = −803 kJ/mol CH4) in the bomb. The temperature changed by 11.1°C. (a) What is the heat capacity of the bomb? kJ/°C (b) A 14.0-g sample of acetaldehyde (CH3CHO) produced a temperature increase of 11.3°C in the same calorimeter. What is the energy of combustion of acetaldehyde (in kJ/mol)?

Answer 1

(a) The heat capacity of calorimeter
`31.25kJ/^oC`

(b) The energy of combustion of acetaldehyde is, 1109.8 kJ/mol

Explanation :

First we have to calculate the heat produced.

`\Delta H=(q)/(n)`

where,

`\Delta H` = enthalpy change = -803 kJ/mol

q = heat released = ?

m = mass of
`CH_4` = 6.91 g

Molar mass of
`CH_4` = 16 g/mol

`\text{Moles of }CH_4=\frac{\text{Mass of }CH_4}{\text{Molar mass of }CH_4}=(6.91g)/(16g/mole)=0.432mole`

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

`-803kJ/mol=(q)/(0.432mole)`

`q=-346.896kJ`

(a) Now we have to calculate the heat capacity of calorimeter.

`q=c* (\Delta T)`

where,

q = heat produced = 346.896 kJ = 346896 J

c = heat capacity of calorimeter = ?

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

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

`346896J=c* (11.1^oC)`

`c=31251.8J^oC=31.25kJ/^oC`

(b) Now we have to calculate the moles of acetaldehyde.

Mass of
`CH_3CHO` = 14.0 g

Molar mass of
`CH_3CHO` = 44 g/mol

`\text{Moles of }CH_3CHO=\frac{\text{Mass of }CH_3CHO}{\text{Molar mass of }CH_3CHO}=(14.0g)/(44g/mole)=0.3182mole`

Now we have to calculate the heat produced in combustion.

`q=c* (\Delta T)`

where,

q = heat produced = ?

c = heat capacity of calorimeter =
`31.25kJ/^oC`

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

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

`q=31.25kJ/^oC* (11.3^oC)`

`q=353.125kJ`

Now we have to calculate the energy of combustion of acetaldehyde.

`\Delta H=(q)/(n)`

where,

`\Delta H` = enthalpy change = ?

q = heat released = 353.125 kJ

n = moles of
`CH_3CHO` = 0.3182mole

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

`\Delta H=(353.125 kJ)/(0.3182mole)`

`\Delta H=1109.8kJ/mol`