Question

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

Answer 1

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"`