forward and reverse reactions taking place at equal rates it is a dynamic state - reactions are...
TRANSCRIPT
Forward and reverse reactions taking place at equal rates
It is a dynamic state - reactions are constantly occurring
Chemical Equilibrium
(a) Start: 10 goldfish in the left tank and 10 guppies in the right.(b) Equilibrium state. with 5 of each kind of fish in each tank. The equilibrium is dynamic;
an averaged state and not a static condition . The fish do not stop swimming when they have become evenly mixed.
(c) If we were to observe one single fish (here a guppy among goldfish). we would find that it spends half its time in each tank .
Brightstorm videosChemical Equilibrium Definition 5:01http://www.youtube.com/watch?v=FYc_SoW2M40&list=PL06C3C4E3F84C6A24&index=42
Crash course chemistryhttp://www.youtube.com/watch?v=g5wNg_dKsYY 10:56 Equilibriumhttp://www.youtube.com/watch?v=DP-vWN1yXrY 9:28 Equilibrium equationsYou don’t need to know how to do the RICE table starting at 4:40
Isaacs Teach http://www.youtube.com/watch?v=g4TKRInLdPA 10:09 EquilibriumGood basic explanation!
http://www.youtube.com/watch?v=4z4_rc6nsKU 12:46 What is the equilibrium constant, Keq? Also very good explanation
Equilibrium constant expressions
aA + bB cC + dDKeq = [C]c[D]d
[A]a[B]b
General information about the Keq expression
• Equilibrium [ ] of products are placed in the numerator.
• Equilibrium [ ] of reactants are placed in the denominator.
• Each [ ] term is raised to an exponent equal to its coefficient in the balanced equation.
• If there is more than 1 product or reactant, the terms are multiplied.
• Solids and liquids (pure substances) are not included in the Keq expression. This is because their [ ] are their densities. The density of a substance does not change with changing temperatures.
• Keq is constant for a given reaction at a given temperature. There are no units associated with the value of Keq.
• The value of Keq is independent of the: – individual [ ] of reactants and products– original [ ] of reactants and products– volume of the container.
• The value of Keq is dependent on temperature.
• What does the value of Keq tell you about a reaction?
Keq >1: more products than reactants at equilibrium
Keq < 1: more reactants than products at equilibrium
Using equilibrium constantsCalculating equilibrium concentrations:
Example: At 1405 K, hydrogen sulfide, also called rotten egg gas (because of its bad odor), decomposes to form hydrogen and a diatomic sulfur molecule,S2. Keq = 2.27 x 10-3.
(a) Write the balanced equation for the reaction described above. Write out the Keq
expression.
(b) Calculate the concentration of hydrogen gas if [S2] = 0.0540 M and [H2S] = 0.184 M.
Solving the problem – part (a)
2H2S (g) 2H2 (g) + S2 (g)
Keq = [H2]2[S2]
[H2S]2
Solution – part (b)
[H2]2= Keq [H2S]2 =
[S2]
(2.27 x 10-3)(0.184 M)2 =
0.0540 M
[H2]2= 1.42 x 10-3 M
[H2] = 3.77 x 10-2 M
Le Châtelier’s principle:1884 - Henri Le Châtelier
When a stress is applied to a system at
equilibrium, the system shifts in the direction
that relieves the stress.
Δ in concentration• Adding more of a reactant or product: the
reaction will shift in the direction to consume a portion of what was added. – more reactant shifts right – more product shifts left
• Removing some of a reactant or product: the reaction will shift in a direction to restore part of what was removed. – reactants removed reaction shifts left (i.e. the
reverse reaction)– products removed reaction shifts right (i.e. the
forward reaction).
Δ in volumeRelevant when discussing gaseous equilibria and when
the number of moles of gaseous reactants differ from the
number of moles of gaseous products. The change in
volume is a result of a change in pressure of the gaseous
system.• When P↓, the reaction will shift in a direction to↑ number of
moles of gas.
PCl5 (g) PCl3 (g) + Cl2 (g) 1 mol 2 mol
2NH3(g) N2(g) + 3H2(g) 2 mol 4 mol
• When P↑, the reaction will shift in a direction to ↓ number of moles gas.
PCl5 (g) PCl3 (g) + Cl2 (g) 1 mol 2 mol
2NH3(g) N2(g) + 3H2(g) 2 mol 4 mol
Δ in temperature View changes in temperature as reactants or products.
When the temperature of an equilibrium system is ↑ the reaction that is endothermic (ΔH>0) will take place.* forward rxn is endothermic more product (shifts to the right). * reverse rxn is endothermic less product (shifts to the left)
When the temperature of an equilibrium system is ↓, the rxn which is exothermic (ΔH<0) will take place.* forward rxn is exothermic – more product (shifts to the right). * reverse rxn is exothermic – less product (shifts to the left)
General rule: if the forward rxn is endothermic,↑K.If the forward rxn is exothermic ↓K.
Animation demonstration http://www.learnerstv.com/animation/animation.php?ani=120&cat=chemistry
http://www.youtube.com/watch?v=PciV_Wuh9V8 7:00 Le Chatelier’s Principle; good explanations with visuals and excellent discussion on how to increase yield of a reaction
http://www.youtube.com/watch?v=dd5p0VZ-MZg 5:36 Equilibrium disturbancesThis one will help you with the lab we’re doing. He also discusses the effect of disturbances (changes) in an equilibrium system and how they affect the value of K (the equilibrium constant)
Reactions that go to completion
Formation of a gas
H2CO3 (aq) H2O (l) + CO2 (g)
Formation of precipitate (Double displacement reactions)
Formation of a slightly ionized product; often times H2O (i.e. in a neutralization reaction)
H3O+ + OH- 2H2O (l)
Solubility equilibria
http://www.youtube.com/watch?v=YJ-dyEtB66A&feature=topics
The Solubility Product Constant, Ksp
• Many important ionic compounds are only slightly soluble in water and equations are written to represent the equilibrium between the compound and the ions present in a saturated aqueous solution.
• The solubility product constant, Ksp, is the product of the concentrations of the ions involved in a solubility equilibrium, each raised to a power equal to the stoichiometric coefficient of that ion in the chemical equation for the equilibrium.
The Solubility Equilibrium Equation And Ksp
CaF2 (s) Ca2+ (aq) + 2F- (aq)
Ksp = [Ca2+][F-]2 Ksp = 5.3x10-9
As2S3 (s) 2As3+ (aq) + 3S2- (aq)
Ksp = [As3+]2[S2-]3 Ksp = 6 x 10-51
Ksp And Molar Solubility
• The solubility product constant is related to the solubility of an ionic solute, but Ksp and molar solubility - the molarity of a solute in a saturated aqueous solution - are not the same thing.
• Calculating solubility equilibria fall into two categories: – determining a value of Ksp from experimental data
– calculating equilibrium concentrations when Ksp is known.
Calculating Ksp From Molar Solubility
It is found that 1.2x10-3 mol of lead (II) iodide, PbI2, dissolves in 1.0 L of aqueous solution at 25 oC. What is the Ksp at this temperature?
Solution:
PbI2 (s) Pb2+ (aq) + 2I- (aq)
Ksp = [Pb2+] [I-]2
Ksp = (1.2 x 10-3 M) (2 x 1.2 x 10-3 M)2
Ksp = 6.9 x10-9
Calculating Molar Solubility From Ksp
Calculate the molar solubility of silver chromate,
Ag2CrO4, in water from Ksp = 1.1x10-12 for Ag2CrO4.
Solution:
Ag2CrO4 (s) 2Ag+ (aq) + CrO4 2- (aq)
Ksp = [Ag+]2 [CrO4 2-]
Ksp = (2x)2(x) = 1.1 x 10-12
4x3 = 1.1 x 10-12
X = 6.5 x 10-5 M
The Common Ion Effect In Solubility Equilibria
• The common ion effect also affects solubility equilibria.
• Le Châtelier’s principle is followed for the shift in concentration of products and reactants upon addition of either more products or more reactants to a solution.
The solubility of a slightly soluble ionic compound is lowered when a second solute that furnishes a common ion is added to the solution.
Solubility Equilibrium Calculation-The Common Ion Effect
What is the solubility of Ag2CrO4 in 0.10 M K2CrO4? Ksp = 1.1x10-12 for Ag2CrO4.
Ag2CrO4 (s) 2Ag+ (aq) + CrO4 2- (aq)
Ksp = [Ag+]2 [CrO4 2-]
Ksp = (2x)2(0.10) = 1.1 x 10-12
x = 1.65 x 10-6 M
Comparison of solubility of Ag2CrO4
In pure water: 6.5 x 10-5 M (prior slide)In 0.10 M K2CrO4: 1.7 x 10-6 MThe common ion effect!!
Determining Whether Precipitation Occurs
• Q is the ion product reaction quotient and is based on initial conditions of the reaction.
• Q can then be compared to Ksp.
• To predict if a precipitation occurs:
- Precipitation should occur if Q > Ksp.
- Precipitation cannot occur if Q < Ksp.
- A solution is just saturated if Q = Ksp.
DR lab: unexpected PPT according to solubility rules!
Ca(OH)2 (s) Ca2+ (aq) + 2OH- (aq)
Ksp = [Ca2+][OH-]2 Ksp = 6.5 x 10-6
Determining Whether Precipitation Occurs – An Example
The concentration of calcium ion in blood plasma is
0.0025 M. If the concentration of oxalate ion is
1.0x10-7 M, do you expect calcium oxalate to
precipitate? Ksp = 2.3x10-9.
Three steps:
(1) Determine the initial concentrations of ions.
(2) Evaluate the reaction quotient Q.
(3) Compare Q with Ksp.
SolutionCaC2O4 (s) Ca2+ (aq) + C2O4
2- (aq)
Ksp = [Ca2+] [C2O42-] = 2.3x10-9
Qsp = (2.5 x 10-3 M) (1.0x10-7 M) = 2.5 x 10-10
2.5 x 10-10 < 2.3x10-9
Q < Ksp therefore no ppt will be formed
Summary
• The solubility product constant, Ksp, represents equilibrium between a slightly soluble ionic compound and its ions in a saturated aqueous solution.
• The common ion effect is responsible for the reduction in solubility of a slightly soluble ionic compound.
• The solubilities of some slightly soluble compounds depends strongly on pH.
Equilibrium lab
Fe(OH)3 (s) Fe3+ (aq) + 3OH- (aq)
Ksp = [Fe3+][OH-]3 = 4 x 10-38
Q vs. Ksp
Q = [Fe3+][OH-]3 = (0.2M)(6.0M)3 = 43.2
Q >Ksp so a PPT forms to take the Fe3+ out of solution
Qualitative Inorganic Analysis
• Acid-base chemistry, precipitation reactions, oxidation-reduction, and complex-ion formation all come into sharp focus in an area of analytical chemistry called classical qualitative inorganic analysis.
• “Qualitative” signifies that the interest is in determining what is present, not how much is present.
• Although classical qualitative analysis is not as widely used today as instrumental methods, it is still a good vehicle for applying all the basic concepts of equilibria in aqueous solutions.