Question

What is the heat of hydrogenation in a hydrogenation reaction?

Answer 1

Hydrogenation reactions consist of the addition of (guess what?) hydrogen to a molecule. For example...

`"Ethene + "H_2 "" stackrel("Pd/C")(->) "Ethane"`

The heat of whatever event at constant pressure, `q_p`, is simply the enthalpy of such an event, `DeltaH`. For a hydrogenation reaction, the enthalpy of hydrogenation is simply the enthalpy of reaction, or `DeltaH_"rxn"`.

This enthalpy could be broken down into which bonds were broken or made. One might call those `DeltaH_"broken"` and `DeltaH_"made"`.

Whatever the case, the heat of hydrogenation is fundamentally based on which bonds were broken, which were made, and the overall differences in them throughout a hydrogenation reaction typically on a per-`"mol"` basis, commonly of alkenes, sometimes of alkynes.

In the example I listed above, you break:

• `1` `C=C` bond

and make:

• `1` `C-C` bond

since you had a double bond and then you just have a single bond. Then you break:

• `1` `H-H` bond

before you allow `H_2` to make:

• `2` `C-H` bonds

These enthalpies are:

`C=C` bond: `"~602 kJ/mol"`
`C-C` bond: `"~346 kJ/mol"`
`H-H` bond: `"~436 kJ/mol"`
`C-H` bond: `"~413 kJ/mol"`

Breaking a bond takes outside energy and puts it into the bond, and is thus positive. Making a bond releases energy into the atmosphere and is thus reported as negative. Overall, you get about:

`DeltaH_"rxn" = sum_i DeltaH_"broken,i" - sum_i DeltaH_"made,i"`

`= overbrace([(underbrace(602)_(broken) - underbrace(346)_(made)) + underbrace(436)_(broken)])^("Step 1") - overbrace([underbrace(413 + 413)_(made)])^("Step 2")`

`= overbrace((602+436))^(broken) - overbrace((346 + 413 + 413))^(made)`

`~~ color(blue)("-134 kJ/mol")`

The enthalpy or heat of hydrogenation of ethene into ethane is exothermic.