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

$q_{rxn}=-q_{cal}$

$q_{cal}=c_{cal}\times \Delta T$

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:

$q_{cal}=(97100J/^oC)\times (2.282^oC)$

$q_{cal}=221582.2J=221.6kJ$

As, $q_{rxn}=-q_{cal}$

So, $q_{rxn}=-221.6kJ$

Thus, the heat of the reaction is -221.6 kJ