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)