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Chap 14 Equilibrium
Calendar 2013
M 4/8 FilmB-1 4/9-10 14.1 Equil 14.2 k expressionB-2 4/11-12 14.3 LeChatM 4/15 KspB-1 4/16-17 Lab kspB-2 4/18-19 ReviewM 4/22 Test
Equilibrium Calendar 2014-15
M 3/30 Rev test/14.1 Equilib Syst 3/31-4/12 break M 4/13 14.1 Rev/Eq constant Keq B-1 4/14-5 14.2 Using Eq Const B-2 4/16-7 Quiz Keq 14.2 Ksp M B-1 4/20-1 Lab Ksp W = a 4/22 Quiz /rev Lab/14.3 LeChat B-2 4/23-4 Review LeChat/ M 4/27 Rev Chapter B-1 4/28-9 Test
Chemical Equilibrium
A + B C + D Complete reactions Go to completion in one direction
All reactants are converted to products Reaction proceeds in one direction only Many involve to formation of gas or ppt
Many rx don’t go to completion Called reversible
Chemical Equilibrium
A + B C + D Reversible reactions
As C and D form and increase in conc, they collide and reforming A and B
Opposite reactions occur at the same rate Forward reaction and reverse reaction
Amounts of reactants and products are constant Can be more reactant or product
Equilibrium Conc of R and P
N2(g) + 3H2(g) 2NH3(g)
Examples
Solute and solvent in a saturated solution
2 phases at phase change temp Ice and water Liquid and vapor
Acid base indicators
Equilibrium Constant At equilib the concentration [M] of reactants and
products are constant (NOT equal) The ratio of conc of products to reactants, each
raised to its coefficient as an exponent is a constant
aA bB Keq = [B]b
[A]a
Called equilibrium constant expression Solids and liquids do not show up in the expression
Equilibrium Constant
At equilib the concentration [M] of reactants and products are constant (NOT equal)
The ratio of conc of products to reactants, each raised to its coefficient as an exponent is a constant
aA + bB cC + dD [ ] means Molarity
Keq =[C]c[D]d
[A]a[B]b
Called equilibrium constant expressionSolids and liquids do not show up in the expression
Write the Keq Expression
Equilibrium Constant Exp Write the the equilibrium constant expression for: 2NO2 (aq) N2 (aq) + 2O2 (aq)
ZnO (aq) + CO (aq) Zn (s) + CO2 (aq)
Ni(s) + 4CO (aq) Ni(CO)4 (aq)
4H+(aq) + 2Cl-(aq) + MnO2 (aq) Mn2+
(aq) + 2H2O (l) + Cl2 (aq)
Size of Keq =
At equilibrium: aA + bB cC + dD If k is about 1 there are about equal amounts of
reactants and products If k is greater 1 there are more products than
reactants at equilib The higher value K = more products less reactants
If k is less than 1 there are more reactants than products at equilib
The lower the value K = more reactants less products
[C]c[D]d
[A]a[B]b
Value of k
Values of k can only be determined by experiment – run the reaction in the lab
2 types of math problems involve k 1 - given [eq] of all r and p calc k 2 - given k and [eq] of all r and p but
one, solve for its [eq]
H2CO3(aq) + H2O(l) <--> H30+(aq) + HC03
-
(aq)
Determine the value of K for the reaction above if H2CO3(aq) = [3.3 x 10-2] H30+ = [1.2 x 10-4] HC03
- = [1.2 x 10-4]
4.36 x 10-7
Determination of K
Determination of K
Calc k if an equilib mixture has [1.2 x 10-3] HCl [3.8 x 10-4] O2
[5.8x 10-2] H2O [5.8 x 10-2] Cl2 4HCl (aq) + O2 (aq) <--> 2H2O (aq) + 2Cl2
(aq)
1.4 x 1010
H2 and I2 gas react to form HI. The equation for this reaction is
H2(g) + I2(g) ↔ 2HI(g) If the equilibrium concentrations of gases
are [H2] = 0.81 M; [I2] = 0.44 M; [HI] = 0.58 M, calc the value of k
0.94 M
Write k expression andCalc k for the eq system:
2 NO(g) + 2 H2(g) N2(g) + 2
H2O(g) [NO] = 0.100 M [H2] = 0.100 M [N2] = 0.0500 M [H2O] = 0.100 M
Using K
For the equilibrium:
2IBr (aq) I2 (aq) + Br2 (aq)
K is 4.13 x 10-2
The equilibrium conc of is IBr [0.0124]. Calc the conc of I2 and Br2 at equilibrium?
0.0025
Using k H2 and I2 gas react together to form HI gas. The equation for this reaction is H2(g) + I2(g) ↔ 2HI(g)
The equilib constant for this reaction is 7.1 x 102 at 25 °C. If the final concentrations of gases are [H2] = 0.42 M; [I2] = 0.20 M; calculate the [HI] at equilibrium.
7.72 M
Sample Prob Carbonic acid is a weak acid found in carbonated beverages. Its
value for keq at 25°C is 4.3 x 10-7. Carbonic acid is added to water and creates an eq system:
H2CO3 (aq) H+ (aq) + HCO3- (aq)
Calculate the concentrations of [H+] and [HCO3-], if the equilibrium
concentration of [H2CO3] = 0.027 M.
1.07 x 10-4 ??
Determine k
Sample problems 14a, and 14b pg 526
Do practice problems 1, 2, and 3 on pg 528
Solubility product constant
Ksp
Learn how to write the solubility constant expression Ksp
Learn how to determine solubility from Ksp
Learn how to determine Ksp from solubility
Three parts to this section:
The Solubility Product Constant, Ksp
Many ionic compounds are only slightly soluble in water. equations are written to represent the
equilibrium between the solid compound and the ions present in a saturated aqueous solution.
The solubility product constant, Ksp, is the product of the conc of the ions produced, each raised to their coefficient as an exponent
Solubility Product ConstantJust like any other k value = [P]/[R]
Written for the equilibrium of the solution of slightly soluble salts AaBb(s) aA+
(aq) + bB-(aq)
ksp = [A+]a [B-]b
CaS(s) Ca2+(aq ) + S2-
(aq)
ksp = [Ca2+] [S2-] Ag2CO3(s) 2Ag1+
(aq) + CO32-
(aq)
ksp = [Ag+1]2 [CO32-]
Solid and liquids do not show up in the ksp expression-like any other equilibrium equation
Write ksp expression for each of the following:
AgOH AgOH(s) Ag1+ (aq) + OH1-
(aq)
Cu(OH)2 Cu(OH)2(s) Cu2+ (aq) + 2OH1-
(aq)
PbI2 PbI2(s) Pb2+ (aq) + 2I1-
(aq)
MgCO3 MgCO3(s) Mg2+(aq) + CO3
2-(aq)
Ca3(PO4)2 Ca3(PO4)2(s) 3Ca2+(aq) + 2PO4
3-
(aq)
Solubility Product Constant = ksp
Ksp shows the solubility of the salt High ksp = more soluble Low ksp = less soluble
Types of problems: Calc ksp given molar concentrations of all
salts What is the conc of one ion if know ksp and
the conc of the other ion? Given ksp what is the solubility (mol/L) of a
slightly soluble salt?
Table 3 pg 508
Values For Ksp At 25 oCLarge Ksp = more solubleSmall Ksp = less soluble
Relationship Between Ksp and Solubility
Based on number of moles of cations and anions that dissolved (molar solubility)
s = M the molarity in mol/L 1:1 Cation : anion ratio
(NaCl, KF, CaSO4, NH4Cl) Ksp = s2 (s = solubility which is Molarity!)
2:1 or 1:2 ratio Cation : anion ratio (CaF2, KS2, Ba(NO3)2, (NH4)2CO3
Ksp = 4s3 (s = solubility which is Molarity!)
The Solubility Equilibrium Equation and Ksp
For every mol NaF that breaks apart 1 Na+ and 1 F- are formed NaF(s) Na+
(aq) + F-(aq)
s sKsp = [Na+][F-] or Ksp = [s][s] or Ksp = s2
For every mol Ag2CO3 that breaks apart 2 mol Ag+ and 1 mol CO3
2- are formed
Ag2CO3(s) 2Ag+(aq) + CO3
2-(aq)
2s s
Ksp = [Ag+]2[CO32-] or Ksp = [2s]2[s] or Ksp = 4s3
Calculate Molar Solubility From Ksp
Calculate the solubility of AgI in water if Ksp = 1.8 x 10-10
AgI Ag+ + I-
s sKsp = [Ag+][I-] = s2 1.8 x 10-10 = s2
s = √1.8 x 10-10 = 1.3 x 10-5
Ksp of MgCO3 is 6.8 x 10-6. Calculate the solubility of
MgCO3.
MgCO3(s) Mg2+(aq) + CO3
2-
(aq)
Calculate Ksp From Molar Solubility
It is found that 1.2 x 10-3 mol of lead (II) iodide, PbI2, dissolves in 1.0 L of aqueous solution at 25oC. What is the Ksp at this temperature?
PbI2(s) Pb2+(aq) + 2I1-(aq)s 2s
Ksp = [Pb2+][I-]2
= 4s3
Ksp = 4(1.2 x 10-3)3
Ksp = 6.9 x 10-9
The solubility of silver bromide (Ag2S) in water is 0.0072 M. Calculate it’s Ksp.
Lab
Write equation for dissolution of Sr(OH)2
Write ksp expression
If 49.0 ml of saturated Sr(OH)2 leaves 0.77 g of solid upon drying, calc ksp
Ksp and Solubility – Equation Form
s = solubility of the salt
Common Ion Effect
Methanol can be synthesized by the reaction of CO and hydrogen gas according to the reaction:
CO (g) + 2H2 (g) <--> CH3OH (g) Keq = 290
Calculate the concentration of hydrogen gas when [CO] = 0.0098 M, [CH3OH] = 0.0098 M.
LeChatelier’s Principle
When an equilibrium system’s conditions are changed, the equilibrium system shifts to the right or left to relieve the stress One reaction occurs more than the other
Equil shifts toward (produces more) reactant or product
3 types of changes affect equilibrium [concentration]
[ ] means molarity (mol/L) Increase or decrease
temperature pressure – for gas systems only
Changes in k
Changes in conc DO NOT change k adding or removing reactant or product [R] and [P] change but k remains the same
Changes in the pressure of the system DO NOT change k Placing in a bigger or smaller container
Changes in temp DO cause k to change [R] or [P] gets bigger or smaller
Depends on the way the equil shifts
Increase in Concentration A + B C + D
Increase in the amt of reactants in number of collision between reactants Forward rx occurs more than the reverse
rx Conc of all r and p change K remains the same
An increase in conc of a subst pushes the eq away from the side of the increase
Same thought process for an increase in the conc of the products
Decrease in Concentration A + B C + D
Decrease in the amt of reactants in number of collision between reactants Reverse rx occurs more than the forward rx
Conc of all r and p change K remains the same
Decreases in conc of a substance pulls the equilib towards the side of the decrease
Same thought process for a decrease in the conc of the products
Examples
2NO2(g) N2O4(g)
What happens to k if the conc of NO2
is increased? What happens to conc of N2O4 if the
conc of NO2 is decreased? What happens to k if the conc of N2O4
is increased? What happens to conc of NO2 if the
conc of N2O4 is increased?
LeChatelier’s Principle: Changing the pressure on the container.
Increase in the pressure. Rx will shift toward the side with the fewer moles
of gas. Decreasing the pressure.
Rx will shift toward the side with more moles of gas.
Changing the pressure will make no difference if there are = number of moles of gas on each side
k does NOT change
LeChat Pressure
N2(g) + 3H2(g) <---> 2NH3(g)
Which way will the eq shift if the pressure is increased?
Which direction will the eq shift if the container size is increased?
Change in Temperature
Do you know the forms for endo and exo reactions??? A + B C + D + heat heat + A + B C + D
Heat, energy, or a number of KJ on reactant side = endo
Heat, energy, or a number of KJ on product side = exo
Changes in Temperature
When temp changes, write “heat” into the eq as a reactant or product (if not given) Which side is heat written?
Based on whether rx is endo or exo Treat “heat” as a reactant or product
in heat push rx away from the side “heat” is located in temp pull rx towards the side “heat” is located
K will get bigger or smaller depending on which direction the equil rx shifts
k get bigger k gets smaller
Change in Temperature
The rx 2NO2(g) N2O4(g)
gives off 57.2 KJ of energy. What happens to the conc of N2O4 if
the equilibrium system is heated? What happens to the value of K?
Rx is exo therefore 2NO2(g) N2O4(g) + heat (57.2 KJ)
increase in heat will push rx away from the side with heat
Change in Temperature
Co(H2O)62+
(aq) + 4Cl-(aq) <--> CoCl62-(aq) +
6H2O(aq)
Pink Blue
Based on the demo, is the forward reaction endothermic or exothermic?
5. Methanol can be synthesized by the reaction of CO and hydrogen gas according to the reaction:
CO (g) + 2H2 (g) CH3OH (g) At 700 K the value of Keq = 290. Calculate the concentration of hydrogen gas when [CO] = 0.0098 M, [CH3OH] = 0.0098 M. (0.094 M)
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