The Foundations of Chemistry

EXAMPLE 17-11 Applying a Stress to a System at Equilibrium

Some hydrogen and iodine are mixed at 229°C in a 1.00-liter container. When equilibrium is

established, the following concentrations are present: [HI]0.490 M,[H 2 ]0.080 M,and

[I 2 ]0.060 M.If an additional 0.300 mol of HI is then added, what concentrations will be

present when the new equilibrium is established?

H 2 (g)I 2 (g)^34 2HI(g)

Plan

We use the initial equilibrium concentrations to calculate the value of Kc. Then we determine

the new concentrations after some HI has been added and calculate Q.The value of Qtells us

which reaction is favored. Then we can represent the new equilibrium concentrations. We

substitute these representations into the Kcexpression and solve for the new equilibrium

concentrations.

Solution

Calculate the value of Kcfrom the first set of equilibrium concentrations.

Kc 50

When we add 0.300 mol of HI to the 1.00-liter container, the [HI] instantaneously increases

by 0.300 M.

H 2 (g)  I 2 (g) 34 2HI(g)

equilibrium 0.080 M 0.060 M 0.490 M

mol/L added 0 M 0 M 0.300 M

new initial conc’n 0.080 M 0.060 M 0.790 M

Substitution of these new initialconcentrations into the reaction quotient gives

Q 130

Because Q Kc, the reaction proceeds to the left to establish a new equilibrium. The new equi-

librium concentrations can be determined as follows. Let xmol/L of H 2 formed; so

xmol/L of I 2 formed, and 2xmol/L of HI consumed.

H 2 (g)  I 2 (g) 34 2HI(g)

new initial conc’n 0.080 M 0.060 M 0.790 M

change due to rxn x M x M  2 x M

new equilibrium (0.080x) M (0.060x) M (0.790 2 x) M

Substitution of these values into Kcallows us to evaluate x.

Kc 50 

0.247.0x 50 x^2 0.6243.16x 4 x^2

46 x^2 10.2x0.38 0

Solution by the quadratic formula gives x0.032 and 0.25.

Clearly, x0.25 is the extraneous root, because xcannot be less than zero in this case.

This reaction does not consume a negative quantity of HI, because the reaction is proceeding

toward the left. Thus, x0.032 is the root with physical meaning, so the new equilibrium

concentrations are

0.6243.16x 4 x^2



0.00480.14xx^2

(0.790 2 x)^2



(0.080x)(0.060x)

(0.790)^2



(0.080)(0.060)

[HI]^2



[H 2 ][I 2 ]

(0.490)^2



(0.080)(0.060)

[HI]^2



[H 2 ][I 2 ]

Equal concentrations of H 2 and I 2
must be formed by the new progressof
the reaction.

To “consume a negative quantity of
HI” would be to form HI. The value
x0.25 would lead to [H 2 ]
(0.080x) M(0.0800.25) M
0.17 M.A negative concentration
is impossible, so x0.25 is the
extraneous root.

730 CHAPTER 17: Chemical Equilibrium

It is obvious that adding some HI
favors the reaction to the left. If more
than one substance is added to the
reaction mixture, it might not be
obvious which reaction will be favored.
Calculating Qalways lets us make the
decision.