Equilibrium. For equilibrium to occur:. System must be closed. Temperature must be constant. Reactions must be reversible (do not go to completion). H 2 (g) + Cl 2 (g) 2HCl(g) + energy No visible change…… A dynamic equilibrium exists.
Rate f = Rate r
N2O4(g) + q 2 NO2 (g)
What happened here?
Le Chatelier’s Principle: when a stress is applied to a system at equilibrium…. The reaction will “shift” in a direction to minimize the stress.
2H2(g) + O2(g) 2H2O(g) + energy
“shift” = rxn speeds up in a particular direction.
E.g. shifts right…. Forward rxn speeds up. More products “appear”. We say the reaction is “favored” to the right or the “position of the equilibrium” shifts right.
Use it up!
Make more pressure
What would a catalyst do to the equilibrium position?
Nothing! It would simply be reached sooner.
What would happen if we added He gas ?
Nothing! It does not affect the partial press of other gases.
Describe what happens when your instructor removes the tube from the freezer, containing the system described above. Explain your observation(s) using LeChatelier’s Principle, and all of the appropriate terminology.
If we divide Equilibrium product values by reactant, each factor raised to the correct power (explained in a moment)…. What do you notice????
N2O4(g) + q 2 NO2(g)
Given the general equation:
aA(aq)* + bB(aq)cC(aq) + dD(aq)
*also if g = gas
equilibrium constant expression
@ constant Temp. ( Keq can also be Kp , Kc , Ksp , Ka , Kb , etc.)
K eq =
At Equilibrium… there is more product than reactant!
At Equilibrium… there
is more reactant than product!
HC2H3O2 (+ H2O) H+ (aq) + C2H3O2-(aq)
= 1.8 x 10-5
What is there more of……reactant or product?
HC2H3O2 only 3% ionizes. There is almost no product present.
This is why it is a weak acid. Strong acids 100% ionize….Ka is LARGE.
Insoluble salts: PbI2 (s) (+ H2O) Pb+2 (aq) + 2 I-(aq) Ksp = [Pb+2][I-]2 = 8.4 x 10-9 What does this value tell us?
This solid is very insoluble; mostly solid PbI2 present.
Keqis used to calculate concentrations of species at equilibrium.
Given: N2(g) + O2(g) = 2NO(g). At 25°C the
Kc = 1.0 x 10-30. [N2] =0.040 & [O2] = 0.010.
What is the concentration of NO?
M.A.E. : Kc =
= 1.0 x 10-30
Small K value…….very little product!
[NO]2 = [N2] [O2] • 1.0 x 10-30
= (0.040) (0.010) (1.0 x 10-30) = 4.0 x 10-34
[NO] = 2.0 x 10-17mol/L
For the system: CO2 + H2 = CO + H2O Kc = 0.64 @ 900°C.
The initial concentrations of reactants are both 0.100mol/L. When the system reaches equilibrium what are the concentrations of reactants and products?
Orig. Conc (mol/L) Change in Conc. Equilib. Conc.
Kc = 0.64 =
x = 0.044; [CO] = [H2O] = 0.044mol/L
[CO2] = [H2] = 0.100 - 0.044 = 0.056 mol/L
CO2(g) + H2(g) CO(g) + H2O (l)
Thus, Keq =
Solubility & K: (read pgs.597-612)
PbSO4(s) = Pb+2(aq) + SO4-2 (aq)
Given, PbSO4 , Ksp= 1.6 x 10-8. Calculate its solubility, [Pb+2] & [SO4-2].
Ksp = [Pb+2][SO4-2] = 1.6 x 10-8
x2 = 1.6 x 10-8 (since x = [Pb+2] & [SO4-2])
x = 1.3 x 10-4. [Pb+2] & [SO4-2] = 1.3 x 10-4M
Mg(OH)2(s) = Mg+2(aq) + 2OH -1(aq)
Ksp = 1.5 x 10 -11 . Calculate the [Mg+2], [OH-1], and the [Mg(OH)2] at equilibrium.
If X = [Mg+2]
So… how do we set up the mass action expression?
(x) (2x)2 =
1.5 x 10-11
x = 1.6 x10 -4 = [Mg+2]; [OH-1] = 3.2 x10 -4
[Mg(OH)2] = 1.6 x10 -4
1. Bromine(I) chloride gas is formed in an endothermic reaction. At 400°C, after the reaction reaches equilibrium, the mixture contained 0.82M BrCl, 0.20M Br2(g)and 0.48M Cl2(g).
a. write the equation for this reaction:
1 Cl2 (g) + 1Br2(g) + q = 2 BrCl (g)
b. Write the equilibrium Expression, the MAE
c. Calculate the value for Keq:
d. What direction is favored?
2. Will a precipitate form when the following two solutions are combined? (assume volumes are additive.)
0.0025 M Pb(NO3)2 + 0.000036 M NaI
The Ksp for PbI2 = 8.4 x 10-9
[Pb+2] = 2.5 x 10-3 , [I-] = 3.6 x 10-5
Reaction Quotient: (2.5 x 10-3) (3.6 x 10-5)2 = 3.24 x 10-12
Since the rx. Quotient < the Ksp , No ppt. Forms!