Question

Question #cd6d2

Answer 1

To calculate `Delta H` for a reaction, you must substract enthalpy of formation of products to enthalpy of formation of reagents.

Explanation:

Let us see it with an example:

`"CH"_4 ("g") + 2 "O"_2 ("g") rightarrow "CO"_2 ("g") + 2 "H"_2 "O" ("g")`

We must know enthalpy of formation of all compounds and elements:

• `Delta H_"form" ("CH"_4) = -17.9 " kJ/mol"`, from the reaction:
  `"C" + 2 "H"_2 rightarrow "CH"_4`

• `Delta H_"form" ("H"_2"O") = -241.82 " kJ/mol"`, from the reaction:
  `"H"_2 + 1/2 "O"_2 rightarrow "H"_2 "O"`

• `Delta H_"form" ("CO"_2) = -393.5 " kJ/mol"`, from the reaction:
  `"C" + "O"_2 rightarrow "CO"_2`

• `Delta H_"form" ("O"_2) = 0 " kJ/mol"`, like every element at its natural state.

So, now:

`Delta H_"reaction" =`
`= sum n cdot Delta H_"form" ("products") - sum m cdot Delta H_"form" ("reagents") =`
`= [(-393.5) + 2 cdot (-241.82)] - [(-17.9) + 2 cdot 0] = -859.24 " kJ/mol"`

where `m, n` represent the coefficients in the reaction.

Note: albeit there are other methods to obtain enthalpy (such as Hess's law, for example), I think this is the one which better adjusts to what you are looking for.

Answer 2

Comment - the link wouldn't show up in the Comments section.

Explanation:

Handy tables of compounds here:
http://bilbo.chm.uri.edu/CHM112/tables/thermtable.htm