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

Answer:

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:

$Q_{cyclop}=-Q_c$

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:

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

Next, since the combustion of cyclopentane is:

$C_5H_1_0+\frac{15}{2} O_2\rightarrow 5CO_2+5H_2O$

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.