Cyclopentene is a cyclic hydrocarbon like the ones used in the experiment. In another bomb calorimetry experiment 0.8278 g of cyclopentene is burned and the temperature of the calorimeter increased from 19.341C to 22.955C. The heat capacity of the calorimeter is 10.56 kJ C−1. Calculate the enthalpy of formation of cyclopentene in kJ/mol cyclopentene. Compare this to the accepted value of +32.6 kJ mol−1.

Computed value is 45.3 kJ/mol whereas the accepted one is 32.6 kJ/mol which means there is a significant difference between them.

Explanation:

Hello.

In this case, for this problem we write the following equation, representing that the heat released due to the combustion of cyclopentane equals the heat gained by the calorimeter:

Which can be also written as:

In such a way, we can compute the enthalpy of combustion of cyclopentane:

Next, since the combustion of cyclopentane is:

And the enthalpy of combustion is computed thermochemically as:

Since the enthalpy of formation of CO2 and H2O are -395.5 and -241.8 kJ/mol respectively, we can compute the enthalpy of formation of cyclopentane based on the previously computed enthalpy of combustion on the calorimeter part:

In such a way, we see a significant difference between the computed value 45.3 kJ/mol and the accepted value 32.6 kJ/mol.

Best regards.

thats how its supposed to be