Question

Question #0799c

Answer 1

`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"` is `"0.345 bar"`, then `2P_i = " 0.345 bar"`, which doesn't make sense, because:

`P_i^2/(0.345) = 0.0172`

But if that were the case, `P_i = "0.173 bar"`, and `0.173^2/0.345 ne 0.0172`.

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

`2"BrCl"(g) rightleftharpoons "Br"_2(g) + "Cl"_2(g)`

Therefore:

`K_P = P_i^2/(0.345 - 2P_i) = 0.0172`

where `P_i` is the partial pressure of either the bromine or chlorine product, and `"0.345 bar"` is the NON-EQUILIBRIUM partial pressure of `"BrCl"`.

A nice trick @anor277 loves to use is a recursive small `x` approximation. Since `K_P/P_(BrCl) "<<" 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^"*"` to be the equilibrium partial pressure of `"Br"_2(g)` or `"Cl"_2(g)` and recursively use the obtained `P_i` in

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

we get:

`P_i^"*" -> -> "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 `"BrCl"(g)` 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 `"Br"_2(g)` and `"Cl"_2(g)`?