k eq for gases

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K EQ FOR GASES Kp

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K eq for GASES. Kp. Equilibrium and Pressure. 2SO 2 (g) + O 2 (g) 2SO 3 (g) If reactants and products are gases, partial pressures of each species can be used to determine the equilibrium The total pressure = P SO 2 + P O 2 + P SO 3. Concentration of Gases. - PowerPoint PPT Presentation

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Page 1: K eq  for GASES

KEQ FOR GASES

Kp

Page 2: K eq  for GASES

Equilibrium and Pressure

2SO2(g) + O2(g) 2SO3(g)

If reactants and products are gases, partial pressures of each species can be used to determine the equilibrium

The total pressure = PSO2 + PO2+ PSO3

Page 3: K eq  for GASES

Concentration of Gases

For reactions that contain gases:PV = nRTP = (n/V)RTP = CRT “C” is concentration in mol/LC = P/RT

The concentration of a gas is directly proportional to the pressure exerted by the gas.

Page 4: K eq  for GASES

Problems Involving Pressure

2 NO2 (g) N2O4 (g)

KP = 7.63atm at 300K.

What are the equilibrium pressures of NO2 and N2O4 if a flask initially contains 1.000 atm of NO2?

What is the value of the Kc at 300K.

Page 5: K eq  for GASES

Relating Kp and Kc

Kp = Kc (RT)Δn

Kp = partial pressure constant using atm

Kc = molar concentration constant using M

R = ideal gas constant, 0.0821 L-atm/mol-KT = absolute temperature in KΔn= moles of product – moles of reactants

Page 6: K eq  for GASES

DIFFERENT TYPE OF K

Page 7: K eq  for GASES

Ksp - Solubility Product of a Salt

KCl (s) K+ (aq) + Cl- (aq)

CaCl2 (s) Ca+2 (aq) + 2 Cl- (aq)

Page 8: K eq  for GASES

Ksp values

Acid Formula Ka value

Potassium nitrate

KNO3 Very large

Calcium sulfate CaSO4 3.0 x 10 -5

Lead (II) chloride

PbCl2 1.7 x 10 -5

Lead (II) iodide PbI2 1.0 x 10 -8

Silver iodide AgI 1.0 x 10 -16

Silver sulfide Ag2S 1.0 x 10 -49

Why are some substances more soluble than others?

Is any substance truly insoluble?

Page 9: K eq  for GASES

Problem Involving Ksp

The Ksp of lead (II) chloride at 298K is 1.7 x 10-5.

Write the dissociation of lead (II) chloride in water.

Write the expression for the solubility product, Ksp, of lead (II) chloride.

Determine the concentration of the chloride ions and the lead ions of a saturated solution at 298K.

Page 10: K eq  for GASES

Kw - Autoionization of WaterKw is 1.00 x 10-14 at room temperature.

What would be the Keq expression?

How does the [H3O+] compare to the [OH-]?

What would be the [H3O+]? What would be the pH?

Like any other equilibrium constant, the value of Kw varies with temperature.

Page 11: K eq  for GASES

Ka – Acid Dissociation ConstantArrhenius acids dissociate to produce hydrogen ions in solution.

What is the difference between a strong and weak acid?

Page 12: K eq  for GASES

Ka values

Acid Formula Ka value

Perchloric acid HClO4 Very large

Hydrochloric acid

HCl Very large

Nitric acid HNO3 Very large

Sulfuric acid H2SO4 Very large

Phosphoric acid H3PO4 7.5 x 10 -3

Citric acid H3C6H5O7 7.1 x 10 -4

Hydrofluoric acid

HF 3.5 x 10 -4

Page 13: K eq  for GASES

Problem Involving Ka

Acetic acid, HC2H3O2, has a Ka of 1.7 x 10-5 at 298K. Write the expression for the acid dissociation of acetic

acid. Determine the concentration of the hydrogen ions and

the acetate ions at 298K of a 1.0 M acetic acid.

Page 14: K eq  for GASES

Kb – Base Dissociation ConstantArrhenius bases dissociate to produce hydroxide ions in solution.

Some bases dissociate completely; other dissociate only partially. This determines if a base is strong or weak.

Page 15: K eq  for GASES

Kb values

Based Formula Ka value

Potassium hydroxide

KOH Very large

Sodium hydroxide

NaOH Very large

Zinc hydroxide Zn(OH)2 9.6 x 10 -4

Methylamine CH3NH2 3.7 x 10 -4

Ammonia NH3 1.8 x 10 -5

Page 16: K eq  for GASES

Problem Involving Kb

1.0 moles of pure ammonia gas is dissolved to make a 1 Liter solution at 298K. Ammonia partially dissociates to produce the ions shown below. Ammonia, NH3, has a Kb of 1.8 x 10-5 at 298K.

Write the expression for the Kb of ammonia. What is the hydroxide concentration at equilibrium? What is the pH of a 1.0 M NH3(aq) solution?

Page 17: K eq  for GASES

DIFFERENT TYPE OF KMULTISTEP REACTIONS

Page 18: K eq  for GASES

Overall Keq

A + B C Keq = K1

C D + E Keq = K2

A + B D + E Keq = K1K2

Page 19: K eq  for GASES

Practice Problem

Mg(OH)2 (s) Mg+2(aq) + 2 OH- (aq) Ksp = 5.6 x 10-11

MgF2 (s) Mg+2(aq) + 2 F- (aq) Ksp = 5.8 x 10-10

NET REACTION: 2 F- (aq) + Mg(OH)2 (s) 2 OH- (aq) + MgF2 (s)

Manipulate the equations for the dissolving of Mg(OH)2 and MgF2 to obtain the net ionic reaction.

Calculate the Keq of the net ionic reaction.