Question

A bomb calorimeter has a heat capacity of 675 J/°C and contains 925 g of water. If the combustion of 0.500 mole of a hydrocarbon increases the temperature of the calorimeter from 24.26°C to 53.88°C, determine the enthalpy change per mole of hydrocarbon.

Answer 1

Answer : The enthalpy change per mole of hydrocarbon is, 269 kJ/mole

Explanation :

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

`q_1` = heat absorbed by the calorimeter

`q_2` = heat absorbed by the water

`c_1` = specific heat of calorimeter =
`675J/^oC`

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

`m_2` = mass of water = 925 g

`\Delta T` = change in temperature =
`T_2-T_1=(53.88-24.26)=29.62^oC`

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

`q=[(675J/^oC* 29.62^oC)+(925g* 4.18J/g^oC* 29.62^oC)]`

`q=134519.23J=134.5kJ`

Now we have to calculate the enthalpy change per mole of hydrocarbon.

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

where,

`\Delta H` = enthalpy change = ?

q = heat released = 134.5 kJ

n = moles of hydrocarbon = 0.500 mol

`\Delta H=(134.5kJ)/(0.500mole)=269kJ/mole`

Therefore, the enthalpy change per mole of hydrocarbon is, 269 kJ/mole

Answer 2

Answer: The enthalpy of the reaction is 269.4 kJ/mol

Step-by-step explanation:

To calculate the heat absorbed by the calorimeter, we use the equation:

`q_1=c\Delta T`

where,

q = heat absorbed

c = heat capacity of calorimeter = 675 J/°C

`\Delta T` = change in temperature =
`T_2-T_1=(53.88-24.26)^oC=29.62^oC`

Putting values in above equation, we get:

`q_1=675J/^oC* 29.62^oC=19993.5J`

To calculate the heat absorbed by water, we use the equation:

`q_2=mc\Delta T`

where,

q = heat absorbed

m = mass of water = 925 g

c = heat capacity of water = 4.186 J/g°C

`\Delta T` = change in temperature =
`T_2-T_1=(53.88-24.26)^oC=29.62^oC`

Putting values in above equation, we get:

`q_2=925g* 4.186J/g^oC* 29.62^oC=114690.12J`

Total heat absorbed =
`q_1+q_2`

Total heat absorbed =
`[19993.5+114690.12]J=134683.62J=134.7kJ`

To calculate the enthalpy change of the reaction, we use the equation:

`\Delta H_(rxn)=(q)/(n)`

where,

q = amount of heat absorbed = 134.7 kJ

n = number of moles of hydrocarbon = 0.500 moles

`\Delta H_(rxn)` = enthalpy change of the reaction

Putting values in above equation, we get:

`\Delta H_(rxn)=(134.7kJ)/(0.500mol)=269.4kJ/mol`

Hence, the enthalpy of the reaction is 269.4 kJ/mol