Question

The equilibrium constant for the following reaction is 1.6 ×10⁵ at 1024K

\[\ce{H2(g) + Br2(g) ⇌ 2HBr(g)}\]

Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a sealed container at 1024K.

Answer 1

Given,

`"K"_"P"` for the reaction i.e.,

\[\ce{H_{2(g)} + Br_{2(g)} ↔ 2HBr_{(g)}}\] is `1.6 xx 10^5`

Therefore, for the reaction

\[\ce{2HBr_{(g)} ↔ H_{2(g)} + Br_{2(g)}}\]  the equilibrium constant will be,

`"K"_"P"^' = 1/"K"_"P"`

`= 1/(1.6 xx 10^5)`

`= 6.25 xx 10^(-6)`

Now, let p be the pressure of both H2 and Br2 at equilibrium.

	2HBr(g)	↔	H2(g)	+	Br2(g)
Initial conc.	10		0		0
At equilibrium	10 - 2p		p		p

Now, we can write,

`("p"_("H"_2"Br"_2) xx "p")/"p"_("HBr")^2 = "K"_"P"^'`

`("p"xx"p")/(10 - 2"p")^2 = 6.25 xx 10^(-6)`

`"p"/(10-2p) = 2.5 xx 10^(-3)`

`"p" = 2.5 xx 10^(-2) - (5.0 xx 10^(-3))"p"`

`(1005 xx 10^(-3))"p" = 2.5 xx 10^(-2)`

`"p" = 2.49 xx 10^(-2) bar = 2.5 xx 10^(-2)` bar (approximately)

Therefore, at equilibrium,

`["H"_2] = ["Br"_2] = 2.49 xx 10^(-2)` bar

[HBr] = `10 - 2 xx (2.49 xx 10^(-2))` bar

= 9.95 bar

= 10 bar (approximately)