Question #a82de

1 Answer
Stefan V.
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(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")#

Related questions
See all questions in Enthalpy
Impact of this question
1866 views around the world
You can reuse this answer
Creative Commons License