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#?

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, #DeltaH_f^@#, 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, #"PCl"_5#, from white phosphorus, #"P"_4# and chlorine gas, #"Cl"_2#, looks like this

#"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, #color(red)(4)# moles of phosphorus pentachloride are formed and #"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 #"PCl"_5# is formed, i.e. #DeltaH_f^@#, divide all the stoichiometric coefficients by #color(red)(4)#

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

So, if you give off #"1776 kJ"# of heat when #color(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

#color(green)(|bar(ul(color(white)(a/a)color(black)(DeltaH_f^@ = -"444 kJ mol"^(-1))color(white)(a/a)|)))#

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