Question

A chemical compound has a molecular weight of 89.05 g/mole. 1.400 grams of this compound underwent complete combustion under constant pressure conditions in a special calorimeter. This calorimeter had a heat capacity of 2980 J °C.1 (Note that the calorimeter was made of a metal shell, a water "substitute" - a special oil, and a thermocouple). The temperature went up by 11.95 degrees.
Required:
Calculate the molar heat of combustion of the compound.

Answer 1

Answer:

$\Delta _{comb}H=-2,265\frac{kJ}{mol}$

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:

$Q _{comb}=-C_{calorimeter}\Delta T_{calorimeter}$

In such a way, we can compute the total released heat due to the combustion considering the calorimeter specific heat and the temperature raise:

$Q _{comb}=-2980\frac{J}{\°C} *11.95\°C\\\\Q _{comb}=-35,611J$

Next, we compute the molar heat of combustion of the compound by dividing by the moles, considering 1.400 g were combusted:

$n=1.400g*\frac{1mol}{89.05g} =0.01572mol$

Thus, we obtain:

$\Delta _{comb}H=\frac{Q_{comb}}{n}=\frac{-35,611J}{0.01572mol} \\\\\Delta _{comb}H=-2,265,331\frac{J}{mol}*\frac{1kJ}{1000J} \\\\\Delta _{comb}H=-2,265\frac{kJ}{mol}$

Best regards!