# Consider the following reaction: "P"_(4(s)) + 10"Cl"_(2(g)) -> 4"PCl"_(5(g)), DeltaH =-"1776 kJ", what is the standard enthalpy of formation for "PCl"_5?

## Give the standard enthalpy of formation for PCl5(g).

Apr 29, 2016

$\Delta {H}_{f}^{\circ} = - {\text{444 kJ mol}}^{- 1}$

#### Explanation:

The most important thing to remember about standard enthalpies of formation is that they represent the enthalpy change of reaction when one mole of a compound is formed from its constituent elements in their standard state.

This means that in order to expresses the standard enthalpy of formation, $\Delta {H}_{f}^{\circ}$, of a given compound, a thermochemical equation must always have one mole of that compound on the products' side.

In your case, the thermochemical equation that describes the synthesis of phosphorus pentachloride, ${\text{PCl}}_{5}$, from white phosphorus, ${\text{P}}_{4}$ and chlorine gas, ${\text{Cl}}_{2}$, looks like this

$\text{P"_ (4(s)) + 10"Cl"_ (2(g)) -> color(red)(4)"PCl"_ (5(g))" " DeltaH_"rxn"^@ = -"1776 kJ}$

This thermochemical equation tells you that when $1$ mole of white phosphorus reacts with $10$ moles of chlorine gas, $\textcolor{red}{4}$ moles of phosphorus pentachloride are formed and $\text{1776 kJ}$ of heat are being released.

In order to have a thermochemical equation that expresses the enthalpy change that occurs when $1$ mole of ${\text{PCl}}_{5}$ is formed, i.e. $\Delta {H}_{f}^{\circ}$, divide all the stoichiometric coefficients by $\textcolor{red}{4}$

1/color(red)(4)"P"_ (4(s)) + 5/2"Cl"_ (2(g)) -> "PCl"_ (5(g))" " DeltaH_f^@ = ?

So, if you give off $\text{1776 kJ}$ of heat when $\textcolor{red}{4}$ moles of phosphorus pentachloride are formed, it follows that you will give off

1 color(red)(cancel(color(black)("mole PCl"_5))) * "1776 kJ"/(color(red)(4)color(red)(cancel(color(black)("moles PCl"_5)))) = "444 kJ"

when one mole of phosphorus pentachloride is formed. This means that the standard enthalpy change of formation for this compound will be

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\Delta {H}_{f}^{\circ} = - {\text{444 kJ mol}}^{- 1}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Remember, the minus sign is used to symbolize heat lost.