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

Dec 21, 2015

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, $\text{BrCl}$.

As you know, the standard enthalpy change of formation for a compound, $\Delta {H}_{f}^{\circ}$, 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 $\text{1 atm}$.

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

$\frac{1}{2} {\text{Br"_text(2(g]) + 1/2"Cl"_text(2(g]) -> "BrCl}}_{\textrm{\left(g\right]}}$

Here $\Delta {H}_{\text{rxn}}^{\circ} = \Delta {H}_{f}^{\circ}$.

Now, the reaction given to you

${\text{Br"_text(2(g]) + "Cl"_text(2(g]) -> color(red)(2)"BrCl}}_{\textrm{\left(g\right]}}$

features the formation of $\textcolor{red}{2}$ moles of bromine monochloride. This means that the enthalpy change for this reaction will be twice the value of $\Delta {H}_{f}^{\circ}$, 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

$\Delta {H}_{f}^{\circ} = + \text{14.6 kJ/mol}$
$\Delta {H}_{\text{rxn"^@ = 2 xx "14.6 kJ/mol" = +"29.2 kJ/mol}}$