Question #0799c

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Question #0799c

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Answer 1 · 2017-05-20T08:33:10

$P_{B r C l} = 0.222 bar$ $bb(P_(BrCl) = "0.222 bar")$

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

$\frac{P_{i}^{2}}{0.345} = 0.0172$ $P_i^2/(0.345) = 0.0172$

But if that were the case, $P_{i} = 0.173 bar$ $P_i = "0.173 bar"$ , and $\frac{{0.173}^{2}}{0.345} \neq 0.0172$ $0.173^2/0.345 ne 0.0172$ .

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

Furthermore, $Br$ $"Br"$ and $Cl$ $"Cl"$ exist as diatomic molecules in nature. So...

$2 BrCl (g) ⇌ {Br}_{2} (g) + {Cl}_{2} (g)$ $2"BrCl"(g) rightleftharpoons "Br"_2(g) + "Cl"_2(g)$

Therefore:

$K_{P} = \frac{P_{i}^{2}}{0.345 - 2 P_{i}} = 0.0172$ $K_P = P_i^2/(0.345 - 2P_i) = 0.0172$

where $P_{i}$ $P_i$ is the partial pressure of either the bromine or chlorine product, and $0.345 bar$ $"0.345 bar"$ is the NON-EQUILIBRIUM partial pressure of $BrCl$ $"BrCl"$ .

A nice trick @anor277 loves to use is a recursive small $x$ $x$ approximation. Since $\frac{K_{P}}{P_{B r C l}} << 1$ $K_P/P_(BrCl) "<<" 1$ , we have our initial guess as the usual small $x$ $x$ approximation:

$P_{i} \approx \sqrt{K_{P} \cdot 0.345 bar} \approx 0.0770 bar$ $P_i ~~ sqrt(K_Pcdot"0.345 bar") ~~ "0.0770 bar"$

As it is, the approximation fails badly, because

$\frac{P_{i}}{P_{B r C l}} \times 100 % = \frac{0.0770 bar}{0.345 bar} \times 100 % > 5 %$ $P_i/P_(BrCl) xx 100% = "0.0770 bar"/"0.345 bar" xx 100% > 5%$ ...

But, if we take the converged $P_{i}^{*}$ $P_i^"*"$ to be the equilibrium partial pressure of ${Br}_{2} (g)$ $"Br"_2(g)$ or ${Cl}_{2} (g)$ $"Cl"_2(g)$ and recursively use the obtained $P_{i}$ $P_i$ in

$P_{i}^{*} = \sqrt{K_{P} (0.345 bar - 2 P_{i})}$ $P_i^"*" = sqrt(K_P("0.345 bar" - 2P_i))$ ,

we get:

$P_{i}^{*} \to \to 0.0617 bar$ $P_i^"*" -> -> "0.0617 bar"$

And to check:

$0.0172 ? = \frac{{0.0617}^{2}}{0.345 - 2 \cdot 0.0617} = 0.0172$ $0.0172 stackrel(?" ")(=) (0.0617^2)/(0.345 - 2*0.0617) = 0.0172$

So, the equilibrium partial pressure of $BrCl (g)$ $"BrCl"(g)$ is

$P_{B r C l} = 0.345 bar - 2 \cdot 0.0617 bar = 0.222 bar$ $color(blue)(P_(BrCl)) = "0.345 bar" - 2*"0.0617 bar" = color(blue)("0.222 bar")$

I assume you can determine the equilibrium partial pressures of ${Br}_{2} (g)$ $"Br"_2(g)$ and ${Cl}_{2} (g)$ $"Cl"_2(g)$ ?

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Question #0799c

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