k eq for gases
DESCRIPTION
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 PresentationTRANSCRIPT
KEQ FOR GASES
Kp
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
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.
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.
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
DIFFERENT TYPE OF K
Ksp - Solubility Product of a Salt
KCl (s) K+ (aq) + Cl- (aq)
CaCl2 (s) Ca+2 (aq) + 2 Cl- (aq)
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?
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.
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.
Ka – Acid Dissociation ConstantArrhenius acids dissociate to produce hydrogen ions in solution.
What is the difference between a strong and weak acid?
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
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.
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.
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
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?
DIFFERENT TYPE OF KMULTISTEP REACTIONS
Overall Keq
A + B C Keq = K1
C D + E Keq = K2
A + B D + E Keq = K1K2
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.