What is the enthalpy change for the reaction #Br_2 + Cl_2 -> 2BrCl#?

1 Answer
Dec 21, 2015

Answer:

Here's what I got.

Explanation:

From a conceptual point of view, the enthalpy change for this reaction will be equal to twice the standard enthalpy change of formation for bromine monochloride, #"BrCl"#.

As you know, the standard enthalpy change of formation for a compound, #DeltaH_f^@#, is the change in enthalpy when one mole of that compound is formed from its constituent elements in their standard state at a pressure of #"1 atm"#.

This means that the standard enthalpy change of formation will correspond to the change in enthalpy associated with this reaction

#1/2"Br"_text(2(g]) + 1/2"Cl"_text(2(g]) -> "BrCl"_text((g])#

Here #DeltaH_"rxn"^@ = DeltaH_f^@#.

Now, the reaction given to you

#"Br"_text(2(g]) + "Cl"_text(2(g]) -> color(red)(2)"BrCl"_text((g])#

features the formation of #color(red)(2)# moles of bromine monochloride. This means that the enthalpy change for this reaction will be twice the value of #DeltaH_f^@#, since

#2 color(red)(cancel(color(black)("moles BrCl"))) * (DeltaH_f^@)/(1color(red)(cancel(color(black)("mole BrCl")))) = color(green)(2 xx DeltaH_f^@)#

SIDE NOTE I was able to find a reference to the standard enthalpychange of formation for bromine monochloride here - page #615#, Gibbs energy, enthalpy and entropy

https://books.google.ro/books?id=jSq0m9GUm4EC

The value listed here is

#DeltaH_f^@ = +"14.6 kJ/mol"#

This means that the enthalpy change for the reaction given to you will be

#DeltaH_"rxn"^@ = 2 xx "14.6 kJ/mol" = +"29.2 kJ/mol"#