Question

Using standard heats of formation, calculate the standard enthalpy change for the following reaction.
H2(g) + C2H4(g)--------C2H6(g)

Answer 1

Answer:

$\Delta _rH=-136.27kJ/mol$

Explanation:

Hello,

In this case, for the given reaction, the enthalpy of reaction is computed in terms of the enthalpies of formation as:

$\Delta _rH=\Delta _fH_{C_2H_6}-\Delta _fH_{C_2H_4}-\Delta _fH_{H_2}$

Whereas hydrogen, ethene and ethane enthalpies of formation are 0 kJ/mol, 52.47 kJ/mol and -83.8kJ/mol respectively. Therefore, we compute:

$\Delta _rH=-83.8kJ/mol-52.47kJ/mol-0kJ/mol\\\\\Delta _rH=-136.27kJ/mol$

Best regards.