Question #0799c

#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#.

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)#?