# Given the following data, how do I find the mols at equilibrium for this reaction?

## ${\text{BrCl"(g) rightleftharpoons 1/2"Br"_2(g) + 1/2"Cl}}_{2} \left(g\right)$ DeltaG_f^@ ("Br"_2(g)) = "3.11 kJ/mol" DeltaG_f^@("BrCl"(g)) = -"0.98 kJ/mol" A) Find the mols of $\text{BrCl} \left(g\right)$ at equilibrium if the volume of the container is fixed at $\text{1.0 L}$. B) Find the mols of ${\text{Br}}_{2} \left(g\right)$ at equilibrium. C) Find the mols of ${\text{Cl}}_{2} \left(g\right)$ at equilibrium.

Aug 6, 2017

${n}_{B r C l \left(g\right) , e q} = \text{0.81 mols}$
${n}_{B {r}_{2} \left(g\right) , e q} = \text{0.29 mols}$
${n}_{C {l}_{2} \left(g\right) , e q} = \text{0.29 mols}$

Well, if we assume a rigid container, we can ignore the volume of the vessel and just use $\text{mols}$. The idea here is then:

1. Find $\Delta {G}_{r x n}^{\circ}$, i.e. at ${25}^{\circ} \text{C}$ and $\text{1 atm}$ using standard Gibbs' free energies of formation.
2. Use the equilibrium condition to find $K$.
3. Use $K$ to find ${n}_{B {r}_{2} \left(g\right) , e q}$ and ${n}_{C {l}_{2} \left(g\right) , e q}$.

A)

$\Delta {G}_{r x n}^{\circ} = {\sum}_{P} {\nu}_{P} \Delta {G}_{f , P}^{\circ} - {\sum}_{R} {\nu}_{R} \Delta {G}_{f , R}^{\circ}$,

where:

• $P$ and $R$ indicate products and reactants, respectively.
• $\nu$ is the stoichiometric coefficient.
• $\Delta {G}_{f}^{\circ}$ is the standard change in Gibbs' free energy (of formation), due to forming each substance from its elements in their elemental state (that is, at ${25}^{\circ} \text{C}$ and $\text{1 atm}$).
• $\Delta {G}_{f}^{\circ}$ for elements in their elemental state is thus $0$.

=> ul(DeltaG_(rxn)^@) = overbrace((1/2 cdot "3.11 kJ/mol" + 1/2 cdot "0 kJ/mol"))^"Products" - overbrace((1 cdot -"0.98 kJ/mol"))^"Reactants"

$=$ $\underline{\text{2.535 kJ/mol}}$

Normally, we could calculate $\Delta {G}_{r x n}$ at nonequilibrium conditions with:

$\Delta {G}_{r x n} = \Delta {G}_{r x n}^{\circ} + R T \ln Q$,

where $R T \ln Q$ accounts for the shift away from standard conditions and $Q$ would be the reaction quotient.

At chemical equilibrium, the reaction has no tendency to move either direction, so $\Delta G = 0$ and $Q \equiv K$. Thus,

$\Delta {G}_{r x n}^{\circ} = - R T \ln K$,

and the equilibrium constant (by dividing by $- R T$ and exponentiating both sides) is:

$\underline{K} = \text{exp} \left(- \Delta {G}_{r x n}^{\circ} / R T\right)$

= e^(-"2.535 kJ/mol"//("0.008314472 kJ/mol"cdot"K" cdot "298.15 K"))

$= \underline{0.3597}$

At this point, we can now determine the mols present of $\text{BrCl} \left(g\right)$ at equilibrium. Construct an ICE table using $\text{mols}$:

${\text{BrCl"(g) rightleftharpoons 1/2"Br"_2(g) + 1/2"Cl}}_{2} \left(g\right)$

$\text{I"" "1.40" "" "" "0" "" "" "" } 0$
$\text{C"" "-x" "" "+x//2" "" } + x / 2$
$\text{E"" "1.40 - x" "x//2" "" "" } x / 2$

$K = {\left({\left(x / 2\right)}^{\cancel{1 / 2}}\right)}^{\cancel{2}} / \left(1.40 - x\right) = \frac{x / 2}{1.40 - x} = 0.3597$

This $K$ is not small, but solving this is not that bad. Eventually we obtain the physical answer as:

$x = | \Delta {n}_{B r C l \left(g\right)} | = \text{0.5858 mols}$

$= 2 {n}_{B {r}_{2} \left(g\right) , e q} = 2 {n}_{C {l}_{2} \left(g\right) , e q}$

And that means...

$\textcolor{b l u e}{{n}_{B r C l \left(g\right) , e q}} = 1.40 - 0.5858 = \underline{\textcolor{b l u e}{\text{0.81 mols}}}$

Everything follows from here. Now the rest is easy.

B)

Refer to the ICE table above to realize that:

$\textcolor{b l u e}{{n}_{B {r}_{2} \left(g\right) , e q}} = \frac{x}{2} \approx \underline{\textcolor{b l u e}{\text{0.29 mols}}}$

C)

Refer to the ICE table above to realize that:

$\textcolor{b l u e}{{n}_{C {l}_{2} \left(g\right) , e q}} = \frac{x}{2} \approx \underline{\textcolor{b l u e}{\text{0.29 mols}}}$

And as a check, is $K$ still correct?

$K = {\left({\left(0.5858 / 2\right)}^{1 / 2}\right)}^{2} / \left(1.40 - 0.5858\right) \approx 0.3597$ color(blue)(sqrt"")