Question

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.

Answer 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.

Step-by-step 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:

`Q_(cyclop)=-Q_c`

Which can be also written as:

`n_(cyclop)\Delta _cH_(cyclop)=-C_c(T_f-T_i)`

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

`\Delta _cH_(cyclop)=(-C_c(T_f-T_i))/(n_(cyclop)) =(-10.56(kJ)/(\°C) (22.955-19.341)\°C)/(0.8278g*(1mol)/(70.1g) )=-3231.8kJ/mol`

Next, since the combustion of cyclopentane is:

`C_5H_1_0+(15)/(2) O_2\rightarrow 5CO_2+5H_2O`

And the enthalpy of combustion is computed thermochemically as:

`\Delta _cH_(cyclop)=5\Delta _fH_(CO_2)+5\Delta _fH_(H_2O)-\Delta _fH_(C_5H_1_0)`

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:

`\Delta _fH_(C_5H_1_0)=5\Delta _fH_(CO_2)+5\Delta _fH_(H_2O)-\Delta _cH_(cyclop)\\\\\Delta _fH_(C_5H_1_0)=5*-395.5+5*-241.8-(-3231.8)\\\\\Delta _fH_(C_5H_1_0)=45.3kJ/mol`

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.