Question

The equilibrium constant for the following reaction is 2.84 x 10-2 at 430 oC.
2 HI(g)H2(g) + I2(g)

A lab technician places 0.200 mol of hydrogen iodide in a 1.00 L reaction vessel and allows to reach equilibrium, use an I.C.E. table to determine the equilibrium concentrations of all species.

Answer 1

Conditions of question:

reaction 2 HI(g) = H2(g) + I2(g)

equilibrium constant Kc=2.84*10^(-2) at 430^oC

n(HI(g))=0.200 mol, V=1 l.

To find: the equilibrium concentrations of all species.

Solution: $Kc=([H2]*[I2])/[HI]^2$

Since the volume of the reaction vessel is 1 liter, the concentrations will be equal to the number of moles of substances, because $C=n/V$ . Now we compile an ICE table consisting of the initial concentration, change in concentration and equilibrium concentration of each participant in the reaction. In this table, X is equal to the amount of reacted HI. The ICE table is presented in the attached screenshot.

Now we substitute this data into the expression for the equilibrium constant and solve the quadratic equation.

$Kc=(0.5*X*0.5*X)/(0.2 - X)^2$

Kc=2.84*10^(-2)

2.84*10^(-2)=$(X^2*0.25)/0.04+X^2-0.4*X$

$0.25*X^2= 0.001136+0.0284*X^2 - 0.01136*X$

$0.2216*X^2 +0.01136*X-0.001136 = 0$

D=$0.01136^2 - 4*0.2216*0.001136=0.001136$

x₁= (-0.01136+0.001136^2)/2*0.2216=0.0504

x2= (-0.01136-0.001136^2)/2*0.2216=-0.1017

x2 does not fit the meaning of the task, then X = 0.0504.

The equilibrium concentrations of all species: [HI] = 0.200 - 0.0504 = 0.1496 M

[H2] = [I2] = 0.5*0.0504 = 0.0252 M.

Answer: [HI] = 0.1496 M, [H2] = [I2] = 0.0252 M.