Calculating Equilibrium Concentrations

Question

Consider the equilibrium system involving the decomposition of hydrogen iodide.

2HI(g) H2(g) + I2(g)	K =	[H2] [I2]	= 2.23×10-2 at 978 K
[HI]2

A flask originally contains 0.284 M hydrogen iodide. Calculate the equilibrium concentrations of the three gases.

[HI]	=	M
[H2]	=	M
[I2]	=	M

Answer 1

2HI(g) <===> H₂(g) + I₂(g)

K = [H₂][I₂] / [HI]² = 2.23x10^-2

Set up an ICE table:

2HI(g) <===> H₂(g) + I₂(g)

0.284.............0............0...........Initial

-2x................+x..........+x..........Change

0.284-2x........x.............x..........Equilibrium

Plug the equilibrium values into the K expression:

K = 2.2x20^-2 = [H₂][I₂] / [HI]² = (x)(x) / (0.284-x)²

2.2x10^-2 = x² / x² - 0.568x + 0.0807

x² = 2.2x10^-2x² - 0.0125x + 1.78x10^-3

0.978x² + 0.0125x - 1.78x10^-3 = 0

x = 0.0367 M

Equilibrium concentrations:

[HI] = 0.284 - 2x = 0.284 - 0.0734 = 0.211 M

[H₂] = x = 0.0367 M

[I₂] = x = 0.0367 M