Question

A sealed container was filled with 0.3 mol H₂(g), 0.4 mol I₂(g) and 0.2 mol HI(g) at 800 K and total pressure 1.00 bar. Calculate the amounts of the components in the mixture at equilibrium given that K = 870 for the reaction, \[\ce{A2 (g) + B2 (g) <=> 2 AB (g)}\].

Answer 1

\[\ce{A2 (g) + B2 (g) <=> 2 AB (g)}\]

	A2	B2	AB
Initial no. of moles	1	1	-
No. of moles reached	x	x	-
no. of. moles at equilibrium	1 - x	1 - x	2x

Total no. of moles = 1 - x + 1 - x + 2x = 2

`"K"+"P" = (["P"_"AB"]^2)/(["P"_("A"_2)]["P"_("B"_2)])`

`= ((2x)/2 xx "P")^2/(((1 - "x")/2 xx "P")((1 - "x")/2 xx "P"))`

`= (4"x"^2)/(1 - x)^2`

Given that, K_p = 1;

`= (4"x"^2)/(1 - x)^2` = 1

⇒ 4x² = (1 – x)² = 1

⇒ 4x² = 1 + x² – 2x

3x² + 2x – 1 = 0

x = `(- 2 +- sqrt(4 - 4 xx 3 xx -1))/(2(3))`

`= (- 2 +- sqrt(4 + 12))/6`

`= (- 2 +- sqrt16)/6`

`= (- 2 + 4)/6; (- 2 - 4)/6`

`= 2/6  ;  (- 6)/6`

x = 0.33 – 1(not possible)

∴ [A₂]_eq = 1 – x = 1 – 0.33 = 0.67

[B₂]_eq = 1 – x = 1 – 0.33 = 0.67

[AB₂]_eq = 2x × 0.33 = 0.66