Answer:
[tex]\Delta _rH=-136.27kJ/mol[/tex]
Explanation:
Hello,
In this case, for the given reaction, the enthalpy of reaction is computed in terms of the enthalpies of formation as:
[tex]\Delta _rH=\Delta _fH_{C_2H_6}-\Delta _fH_{C_2H_4}-\Delta _fH_{H_2}[/tex]
Whereas hydrogen, ethene and ethane enthalpies of formation are 0 kJ/mol, 52.47 kJ/mol and -83.8kJ/mol respectively. Therefore, we compute:
[tex]\Delta _rH=-83.8kJ/mol-52.47kJ/mol-0kJ/mol\\\\\Delta _rH=-136.27kJ/mol[/tex]
Best regards.
