[H 2 ] (0.080x) M (0.0800.032) M 0.112 M[I 2 ] (0.060x) M (0.0600.032) M 0.092 M[HI] (0.790 2 x) M(0.7900.064) M 0.726 MIn summary,
Original Equilibrium Stress Applied New Equilibrium
[H 2 ]0.080 M [H 2 ]0.112 M
[I 2 ]0.060 M Add 0.300 MHI [I 2 ]0.092 M
[HI]0.490 M
[HI]0.726 MWe see that some of the additional HI is consumed, but not all of it. More HI remains after
the new equilibrium is established than was present before the stress was imposed. The new
equilibrium [H 2 ] and [I 2 ] are substantially greater than the original equilibrium concentrations,
however.
You should now work Exercise 64.
We can also use the equilibrium constant to calculate new equilibrium concentrations
that result from decreasing the volume (increasing the pressure) of a gaseous system that
was initially at equilibrium.
EXAMPLE 17-12 Applying a Stress to a System at Equilibrium
At 22°C the equilibrium constant, Kc, for the following reaction is 4.66 10 ^3. (a) If 0.800 mol
of N 2 O 4 were injected into a closed 1.00-liter container at 22°C, how many moles of each gas
would be present at equilibrium? (b) If the volume were halved (to 0.500 L) at constant temper-
ature, how many moles of each gas would be present after the new equilibrium has been
established?
N 2 O 4 (g)^34 2NO 2 (g) Kc
[
[N
NO2 O2 ]
42]4.66 10 ^3
Plan
(a) We are given the value of Kcand the initial concentration of N 2 O 4. We write the reaction
summary, which gives the representation of the equilibrium concentrations. Then we substi-
tute these into the Kcexpression and solve for the new equilibrium concentrations.
(b) We obtain the new initialconcentrations by adjusting the equilibrium concentrations from
part (a) for the volume change. Then we solve for the newequilibrium concentrations as we
did in part (a).
Solution
(a) Let xmol/L of N 2 O 4 consumed and 2xmol/L of NO 2 formed.
N 2 O 4 (g) 34 2NO 2 (g)
initial 0.800 M 0 M
change due to rxn x M 2 x M
equilibrium (0.800x) M 2 x M17-8 Application of Stress to a System at Equilibrium 731