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:
[tex]Q_{cyclop}=-Q_c[/tex]
Which can be also written as:
[tex]n_{cyclop}\Delta _cH_{cyclop}=-C_c(T_f-T_i)[/tex]
In such a way, we can compute the enthalpy of combustion of cyclopentane:
[tex]\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[/tex]
Next, since the combustion of cyclopentane is:
[tex]C_5H_1_0+\frac{15}{2} O_2\rightarrow 5CO_2+5H_2O[/tex]
And the enthalpy of combustion is computed thermochemically as:
[tex]\Delta _cH_{cyclop}=5\Delta _fH_{CO_2}+5\Delta _fH_{H_2O}-\Delta _fH_{C_5H_1_0}[/tex]
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:
[tex]\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[/tex]
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.
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