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 0.322 M HI, 4.32×10^-2 M H₂ and 4.32×10^-2 M I₂. What will be the concentrations of the three gases once equilibrium has been reestablished, if 3.12×10^-2 mol of I₂(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

Keq = 0.018

if we add I2 initially, then, the revers reaciton occurs

H2 + I2 = 2HI

Krev = (1/Ke) = 1/(0.018) = 55.556

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

initially

[HI] = 0.322

[I2] = 0.0432 +0.0312 = 0.0744

[H2] = 0.0432

in equilibirum

[HI] = 0.322 + 2x

[I2] =0.0744 - x

[H2] = 0.0432 - x

substitute

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

55.556 = ( 0.322 + 2x)^2 / (0.0744 - x)(0.0432 - x)

55.556 *(0.0744*0.0432 ) - 55.556*(0.0744 +0.0432)x + x^2 = 0.322 ^2 + 2*0.322 x + 4x^2

0.17856 - 6.5333856x + x^2 = 0.103684 + 0.644x + 4x^2

3x2 + (0.644+6.5333856)x + (0.103684 -0.17856 ) = 0

3x^2 + 7.1773x -0.074876 = 0

x = 0.010377

[HI] = 0.322 + 2*0.010377 = 0.342754 M

[I2] =0.0744 - 0.010377 = 0.064023

[H2] = 0.0432 - 0.010377= 0.032823 M