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.

Explanation:

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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:

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

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:

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"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"

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