Question

When 1.04g of cyclopropane was burnt in excess oxygen in a bomb calorimeter, the temperature rose by 3.69K. The total heat capacity of the calorimeter and it's contents was 14.01kJ/K. Determine the enthalpy of combustion of cyclopropane.

Answer 1

Answer:

$\Delta _{comb}H=-2,093\frac{kJ}{mol}$

Explanation:

Hello!

In this case, since these calorimetry problems are characterized by the fact that the calorimeter absorbs the heat released by the combustion of the substance, we can write:

$Q_{rxn}+Q_{cal}=0$

Thus, given the temperature change and the total heat capacity, we obtain the following total heat of reaction:

$Q_{rxn}=-14.01kJ/K*3.69K\\\\Q_{rxn}=-51.70kJ$

Now, by dividing by the moles in 1.04 g of cyclopropane (42.09 g/mol) we obtain the enthalpy of combustion of this fuel:

$n=\frac{1.04g}{42.09g/mol}=0.0247mol\\\\\Delta _{comb}H=\frac{Q_{rxn}}{n}\\\\ \Delta _{comb}H=-2,093\frac{kJ}{mol}$

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