# Question 0799c

May 20, 2017

$\boldsymbol{{P}_{B r C l} = \text{0.222 bar}}$

As a note, I assume by "final" you mean after the vessel pumping was stopped. If the equilibrium partial pressure of $\text{BrCl}$ is $\text{0.345 bar}$, then $2 {P}_{i} = \text{ 0.345 bar}$, which doesn't make sense, because:

${P}_{i}^{2} / \left(0.345\right) = 0.0172$

But if that were the case, ${P}_{i} = \text{0.173 bar}$, and ${0.173}^{2} / 0.345 \ne 0.0172$.

We don't know the reaction, but it doesn't matter. We know that the number of bromine and chlorine atoms are $1 : 1$ on the left side of the reaction, so they must be $1 : 1$ on the right side of the reaction.

Furthermore, $\text{Br}$ and $\text{Cl}$ exist as diatomic molecules in nature. So...

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

Therefore:

${K}_{P} = {P}_{i}^{2} / \left(0.345 - 2 {P}_{i}\right) = 0.0172$

where ${P}_{i}$ is the partial pressure of either the bromine or chlorine product, and $\text{0.345 bar}$ is the NON-EQUILIBRIUM partial pressure of $\text{BrCl}$.

A nice trick @anor277 loves to use is a recursive small $x$ approximation. Since ${K}_{P} / {P}_{B r C l} \text{<<} 1$, we have our initial guess as the usual small $x$ approximation:

P_i ~~ sqrt(K_Pcdot"0.345 bar") ~~ "0.0770 bar"

As it is, the approximation fails badly, because

P_i/P_(BrCl) xx 100% = "0.0770 bar"/"0.345 bar" xx 100% > 5%...

But, if we take the converged ${P}_{i}^{\text{*}}$ to be the equilibrium partial pressure of ${\text{Br}}_{2} \left(g\right)$ or ${\text{Cl}}_{2} \left(g\right)$ and recursively use the obtained ${P}_{i}$ in

P_i^"*" = sqrt(K_P("0.345 bar" - 2P_i)),

we get:

${P}_{i}^{\text{*" -> -> "0.0617 bar}}$

And to check:

0.0172 stackrel(?" ")(=) (0.0617^2)/(0.345 - 2*0.0617) = 0.0172

So, the equilibrium partial pressure of $\text{BrCl} \left(g\right)$ is

color(blue)(P_(BrCl)) = "0.345 bar" - 2*"0.0617 bar" = color(blue)("0.222 bar")#

I assume you can determine the equilibrium partial pressures of ${\text{Br}}_{2} \left(g\right)$ and ${\text{Cl}}_{2} \left(g\right)$?