Answer:
[tex]\Delta _{comb}H=-2,265\frac{kJ}{mol}[/tex]
Explanation:
Hello!
In this case, for such calorimetry problem, we can notice that the combustion of the compound releases the heat which causes the increase of the temperature by 11.95 °C, it means that we can write:
[tex]Q _{comb}=-C_{calorimeter}\Delta T_{calorimeter}[/tex]
In such a way, we can compute the total released heat due to the combustion considering the calorimeter specific heat and the temperature raise:
[tex]Q _{comb}=-2980\frac{J}{\°C} *11.95\°C\\\\Q _{comb}=-35,611J[/tex]
Next, we compute the molar heat of combustion of the compound by dividing by the moles, considering 1.400 g were combusted:
[tex]n=1.400g*\frac{1mol}{89.05g} =0.01572mol[/tex]
Thus, we obtain:
[tex]\Delta _{comb}H=\frac{Q_{comb}}{n}=\frac{-35,611J}{0.01572mol} \\\\\Delta _{comb}H=-2,265,331\frac{J}{mol}*\frac{1kJ}{1000J} \\\\\Delta _{comb}H=-2,265\frac{kJ}{mol}[/tex]
Best regards!
