# Question cd6d2

Apr 15, 2016

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:

• $\Delta {H}_{\text{form" ("CH"_4) = -17.9 " kJ/mol}}$, from the reaction:
${\text{C" + 2 "H"_2 rightarrow "CH}}_{4}$

• $\Delta {H}_{\text{form" ("H"_2"O") = -241.82 " kJ/mol}}$, from the reaction:
$\text{H"_2 + 1/2 "O"_2 rightarrow "H"_2 "O}$

• $\Delta {H}_{\text{form" ("CO"_2) = -393.5 " kJ/mol}}$, from the reaction:
${\text{C" + "O"_2 rightarrow "CO}}_{2}$

• $\Delta {H}_{\text{form" ("O"_2) = 0 " kJ/mol}}$, like every element at its natural state.

So, now:

$\Delta {H}_{\text{reaction}} =$
= sum n cdot Delta H_"form" ("products") - sum m cdot Delta H_"form" ("reagents") =#
$= \left[\left(- 393.5\right) + 2 \cdot \left(- 241.82\right)\right] - \left[\left(- 17.9\right) + 2 \cdot 0\right] = - 859.24 \text{ kJ/mol}$

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.

Apr 24, 2016