Question

Use the following bond enthalpies to calculate the heat of reaction?

Answer 1

For

`2"CH"_3"OH"(g) + 3"O"_2(g) -> 2"CO"_2(g) + 4"H"_2"O"(g),`

we have to determine which bonds have been broken and made. It really helps to draw the Lewis structures. Good thing your reaction is balanced.

We assume that the stoichiometric coefficients have scaled the reaction to a per-mol basis. For instance, we assume that there are `"2 mol"`s of `"CH"_3"OH"`.

LEWIS STRUCTURE REPRESENTATIONS

Oxygen reacts in a combustion reaction as `"O"="O"`.

BOND-BREAKING/FORMING ACCOUNT

Reactants side

`2"CH"_3"OH"`:

• 6 `"mol"`s of `"C"-"H"` sigma bonds broken
• 2 `"mol"`s of `"C"-"O"` sigma bonds broken
• 2 `"mol"`s of `"O"-"H"` sigma bonds broken

`3"O"_2`:

• 3 `"mol"`s of `"O"="O"` double bonds broken

Products side

`2"CO"_2`:

• 4 `"mol"`s of `"C"="O"` [sigma+pi] bonds formed

`4"H"_2"O"`:

• 8 `"mol"`s of `"O"-"H"` sigma bonds formed

SIGN CONVENTIONS

• Bonds broken `->` `\mathbf(Delta"H"_"bond" > 0)` because potential energy is stored into the bond to break the bond.

• Bonds formed `->` `\mathbf(Delta"H"_"bond" < 0)` because potential energy is released from the bond upon forming the bond.

RESULTS

You weren't told what mass of methanol you started with, so I'm assuming you're supposed to calculate the ENTHALPY of reaction, NOT the heat flow of the reaction (these are NOT the same!).

Thus, the approximate enthalpy of reaction, having already accounted for the stoichiometries, is:

`color(blue)(Delta"H"_"rxn")`

`~~ sum Delta"H"_"broken" - sum Delta"H"_"formed"`

`= [6Delta"H"_("C"-"H") + 2Delta"H"_("C"-"O") + 2Delta"H"_("O"-"H") + 3Delta"H"_("O"-"O")] - [4Delta"H"_("C"="O") + 8Delta"H"_("O"-"H")]`

`= [6(413) + 2(358) + 2(463) + 3(495)] - [4(799) + 8(463)]` `"kJ/mol"`

`= color(blue)(-"1295 kJ/mol")`

which makes sense to be negative, because it's a combustion reaction.