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

Explanation:

Hello,

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

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

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the number of valence electrons for each molecule or ion is shown beneath the structure. remember hydrogen will not have more than two electrons. this hydrogen is part of a covalent bond (sharing two electrons). a double bond here would cause hydrogen to share four electrons with phosphorus.

it forms an anion that has a larger radius. anions have a negative charge and are larger than the original atom.