Question

Question #a82de

Answer 1

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(s)) + 10Cl_(2(g)) -> color(red)(4)PCl_(5(g))`

Notice that your reaction produces `color(red)(4)` moles of phosphorus pentachloride. In order for the change in enthalpy, `DeltaH`, to match the standard enthalpy of formation, `DeltaH_f^0`, you need your reaction to produce 1 mole of phosphorus pentachloride.

Divide all the stoichiometric coefficients by 4 to get

`1/4P_(4(s)) + 10/4Cl_(2(g)) -> color(red)(4)/4PCl_(5(g))`

This is equivalent to

`1/4P_(4(s)) + 5/2Cl_(2(g)) -> PCl_(5(g))`

The change in enthalpy for this reaction will be

`DeltaH_"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")`