# Question a82de

May 21, 2015

The standard enthalpy of formation for any substance is the change in enthalpy that accompanies the formation of 1 mole of that substance from its contituent elements in their most stable form.

Take a look at the balanced chemical equation for your reaction

${P}_{4 \left(s\right)} + 10 C {l}_{2 \left(g\right)} \to \textcolor{red}{4} P C {l}_{5 \left(g\right)}$

Notice that your reaction produces $\textcolor{red}{4}$ moles of phosphorus pentachloride. In order for the change in enthalpy, $\Delta H$, to match the standard enthalpy of formation, $\Delta {H}_{f}^{0}$, you need your reaction to produce 1 mole of phosphorus pentachloride.

Divide all the stoichiometric coefficients by 4 to get

$\frac{1}{4} {P}_{4 \left(s\right)} + \frac{10}{4} C {l}_{2 \left(g\right)} \to \frac{\textcolor{red}{4}}{4} P C {l}_{5 \left(g\right)}$

This is equivalent to

$\frac{1}{4} {P}_{4 \left(s\right)} + \frac{5}{2} C {l}_{2 \left(g\right)} \to P C {l}_{5 \left(g\right)}$

The change in enthalpy for this reaction will be

$\Delta {H}_{\text{new" = (DeltaH)/4 = "-1776 kJ"/4 = "-444 kJ}}$

Since this represents the enthalpy change when 1 mole of phosphorus pentachloride is formed, you can write

DeltaH_"new" = DeltaH_f^0 = color(green)("-444 kJ/mol")#