Question

4.41 g of propane gas (C3H8) is injected into a bomb calorimeter and ignited with excess oxygen, according to the reaction below. The calorimeter (including the water) has a heat capacity of 97.1 kJ/°C. C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O() (a) If the temperature rose from 25.000°C to 27.282°C, what is the heat of the reaction, qrxn?

Answer 1

Answer : The heat of the reaction is -221.6 kJ

Explanation :

Heat released by the reaction = Heat absorbed by the calorimeter

`q_(rxn)=-q_(cal)`

`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 =
`97.1kJ/^oC=97100J/^oC`

`\Delta T` = change in temperature =
`(T_(final)-T_(initial))=(27.282-25.000)=2.282^oC`

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

`q_(cal)=(97100J/^oC)* (2.282^oC)`

`q_(cal)=221582.2J=221.6kJ`

As,
`q_(rxn)=-q_(cal)`

So,
`q_(rxn)=-221.6kJ`

Thus, the heat of the reaction is -221.6 kJ