Question

The equilibrium constant, K, for the following reaction is 1.80×10^-2 at 698 K.

2HI(g) <----><H₂(g) + I₂(g)

An equilibrium mixture of the three gases in a 1.00 L flask at 698 K contains 0.307 M HI,   4.13×10^-2 M H₂ and 4.13×10^-2 M I₂. What will be the concentrations of the three gases once equilibrium has been reestablished, if 0.173 mol of HI(g) is added to the flask?

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

Answer 1

First, let us define the equilibrium constant for any species:

The equilibrium constant will relate product and reactants distribution. It is similar to a ratio

The equilibrium is given by

rReactants -> pProducts

Keq = [products]^p / [reactants]^r

For a specific case:

aA + bB = cC + dD

Keq = [C]^c * [D]^d / ([A]^a * [B]^b)

Keq = [H2][I2] / [HI]^2

in equilbirium

Keq = (4.13*10^-2)(4.13*10^-2)/(0.307^2)

Keq = 0.01809

if we add

0.173 mol ot V = 1L --> 0.173 M of HI

initially

[H2] = 0.0413

[I2] = 0.0413

[HI] = 0.173

in equilbirum

[H2] = 0.0413+x

[I2] =  0.0413+x

[HI] = 0.173 -2x

substiotuite in K

Keq = [H2][I2] / [HI]^2

0.01809= (0.0413+x)(0.0413+x) / (0.173 -2x)^2

sqrt(0.01809) =  (0.0413+x) / (0.173 -2x)

0.13449*(0.173 -2x) = 0.0413+x

0.13449*(0.173) -2*0.13449*x = 0.0413+x

0.13449*(0.173) - 0.0413= (1 + 2*0.13449)x

x = -0.018033 / ((1 + 2*0.13449) )

x = -0.01421

[H2] = 0.0413+-0.01421 = 0.02709 M

[I2] =  0.0413+-0.01421 = 0.02709M

[HI] = 0.173 -2*(0.01421) = 0.14458 M