Wilkes University Chemical Equilibrium Questionnaire

9/1/2021Chapter 6: Chemical Equilibrium
Equilibrium Review
• Equilibrium Constant (K)
• Thermodynamics (K related to H, S, G)
• Le Châtelier’s Principle (K vs. Q)
More Equilibrium Fun
• Acids-Bases (Ka, Kb)
• Solubility Product (Ksp)
The Equilibrium Constant
Consider the generic equation:
Stoichiometric
coefficient
Chemical
Species
Equilibrium Constant =
[A] = concentration of A in standard state
a = stoichiometric coefficient of A
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The Equilibrium Constant
The Equilibrium Constant
• K > 1, [Products] > [Reactants] “favors products”
• K is dimension-less! (NO UNITS)
• K = 1, [Products] = [Reactants]
• All [A] must be expressed in standard state
If A is a …
• K < 1, [Products] < [Reactants] “favors reactants” 3 M (molarity, mol/L) Gas bar (pressure units of bar) Pure solid (or liquid) OMIT Solvent (ex. H2O) OMIT 4 Manipulating Equilibrium Constants If the equation is reversed: Manipulating Equilibrium Constants If the equation is multiplied by 2: What happens to K? What happens to K? 1 = Kc = Kc 2 INVERSE of original K 5 Express in K as … Solute original K is SQUARED 6 1 9/1/2021 Manipulating Equilibrium Constants If two equations are added together: N2(g) + O2(g) ⇌ 2 NO(g) K1 = Rxn 2 2 NO(g) + O2(g) ⇌ 2 NO2(g) K2 = Rxn 3 N2(g) + 2 O2(g) ⇌ 2 NO2(g) Rxn 1 + NO If the equation is reversed: INVERSE of original K N2 O2  NO2  2 NO O2  2 “Flip” the reaction, “Flip” the K If the equation is multiplied by a value: original K is raised to that power Multiply the reaction by “x,” “x” becomes the exponent NO × NO2  = NO2  N  2 O2  NO2 O2  N2 O2 2 2 K 3 = K1 × K 2 = Manipulating Equilibrium Constants 2 2 If equations are added together: MULTIPLY the equilibrium constants 2 If equations are added together, MULTIPLY the K’s 7 8 Equilibrium and Thermodynamics • Equilibrium controlled by thermodynamics of a reaction • Enthalpy and Entropy contribute independently to the degree the reaction is favored or disfavored Equilibrium and Thermodynamics Enthalpy (H) = heat absorbed or released ∆H = heat absorbed/released when reaction takes place under constant applied pressure • ∆H° = standard enthalpy change (standard state) • ∆H° < 1, exothermic (heat released) • ∆H° > 1, endothermic (heat absorbed)
exothermic (heat released)
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Equilibrium and Thermodynamics
Entropy (S) = measure of “disorder”
• greater disorder, greater entropy
• S gas > S liquid > S solid
• ∆S° = change in entropy (products – reactants) when
all species in standard state
▪ ∆S° < 1, products less disordered than reactants ▪ ∆S° > 1, products more disordered than reactants
Equilibrium and Thermodynamics
• With -∆H and +∆S, chemical reaction will be favored
• With +∆H and -∆S, chemical reaction will NOT be favored
• What happens when we have BOTH +∆H and +∆S,
or -∆H and -∆S?
The change in Gibb’s Free Energy (G)
determines whether H or S prevails!
• ∆G° < 0 (-), Reaction favored • ∆G° > 0 (+), Reaction NOT favored
dissolved ions more disordered than solid salt
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Equilibrium and Thermodynamics
So why do we care about G and thermodynamics so
much when it comes to chemical equilibrium?
Chemical Equilibrium constant (K) and
Gibb’s Free Energy (G) are related
R = gas constant = 8.314472 J/(K × mol)
T = temperature (Kelvin)
Chapter 6: Chemical Equilibrium
Equilibrium Review
• Equilibrium Constant (K)
• Thermodynamics (K related to H, S, G)
• Le Châtelier’s Principle (K vs. Q)
More Equilibrium Fun
• Solubility Product (Ksp)
• Acids-Bases (Ka, Kb)
The more negative (-) the ∆G°, the larger the K
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Equilibrium and Le Châtelier’s Principle
Equilibrium and Le Châtelier’s Principle
Let’s start with a familiar concept … a see saw!
Reactants
Products
R and P “balanced”
Equilibrium!
Chemistry works the same way!
• System moves to stability
• Goal: equilibrium!
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Equilibrium and Le Châtelier’s Principle
Reactants
Reactants
Products
R and P “unbalanced”
Equilibrium and Le Châtelier’s Principle
Le Châtelier’s principle – If a system at equilibrium is disturbed, the
direction the system proceeds back to equilibrium is so the
change is partially offset
What kind of disturbances?
• Addition of a reactant or product
• Removal of a reactant or product
• Change in volume (and pressure!) of system
• Change in temperature
BUT – reactants can
form products …
“Shift” reaction to products
Reactants
Reactants
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Products
R and P “balanced”
Equilibrium!
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Chemical Equilibrium: Reactants/Products
At equilibrium:
Chemical Equilibrium: Reactants/Products
At equilibrium:
What if we remove some NH3?
What happens when we increase [N2] to 3.51 M?
Reaction proceeds towards products, NH3
(i.e. “use up N2”, Shift reaction to the right)
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Reaction proceeds towards products, NH3
(i.e. “use up N2”, Shift reaction to the right)
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Chemical Equilibrium: Reactants/Products
At equilibrium:
Chemical Equilibrium: Reactants/Products
At equilibrium:
What happens when we increase [NH3]?
What happens when we remove N2 from original mix?
Reaction proceeds towards reactants, N3 and H2
(Shift reaction to the left)
Reaction proceeds towards reactants, N3 and H2
(Shift reaction to the left)
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Chemical Equilibrium: Volume (and Pressure!)
When volume DECREASES,
equilibrium shifts in the
direction to MINIMIZE total
number of moles of gas.
Chemical Equilibrium: Temperature
For exothermic reactions, consider heat as a “product”
CoCl4–2 + 6H2O → Co(H2O)6+2 + 4Cl– + heat
Volume ↓, moles gas ↓
When volume INCREASES,
equilibrium shifts in the
direction to MAXIMIZE total
number of moles of gas.
Volume ↑, moles gas ↑
Add heat
CoCl4–2
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Remove heat
Mixture
Co(H2O)6+2
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Chemical Equilibrium: Temperature
Le Châtelier’s Principle Examples
HA
For endothermic reactions, consider heat as a “reactant”
Colorless
H+ + A-
What happens if we make
the solution more acidic?
Brown
Mixture
NH4+
NH3(aq) + H+
Add heat
What happens if we add
(NH4)2SO4 and dissolve it
in this solution?
Ok … how do we prove it???
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K vs. Q: Predicting the Direction of a Reaction
Equilibrium constant (K) :
Reaction Quotient (Q):
K vs. Q: Predicting the Direction of a Reaction
Calculated using
equilibrium
concentrations
Q vs. K determines direction of reaction
Q = K : system is at equilibrium
Q < K : Reaction will proceed FORWAD (left to right) Why? Numerator (product) too small, OR denominator (reactants) too big Q > K : Reaction will proceed REVERSE (right to left)
Why? Numerator (product) too big, OR
denominator (reactants) too small
Calculated
using initial
concentrations
Only when the reaction is AT EQUILIBRIUM, K = Q
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Chapter 6: Chemical Equilibrium
Equilibrium Review
• Equilibrium Constant (K)
• Thermodynamics (K related to H, S, G)
• Le Châtelier’s Principle (K vs. Q)
Protic Acids and Bases
Bronsted Lowry Acids and Bases
Acids: Proton (H+) Donors
Bases: Proton (H+) Acceptors
More Equilibrium Fun
• Acids-Bases (Ka, Kb)
• Solubility Product (Ksp)
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Autoprotolysis
Autoprotolysis of water (Kw)
Autoprotolysis:
When a substance ionizes itself, and acts as BOTH an acid and a base
The equilibrium constant for autoprotolysis of H2O: Kw.
H2O
Protic solvents have a reactive H+, and all undergo autoprotolysis
H+ + OH−
KW = [H+][OH−]
• Kw varies with temperature (Table 6-1)
• At 25.0°C
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KW = 1.01 × 10−14
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pH
pH
The equilibrium constant for autoprotolysis of H2O: Kw.
H+ + OH−
H2O
KW = [H+][OH−]
• Kw varies with temperature (Table 6-1)
• At 25.0°C
KW = 1.01 × 10−14
• pH = −log[H+]
• pH + pOH = −log(KW) = 14.00 at 25°C
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Acidic solutions
• [H+] > [OH−]
• pH < 7 34 Strong Acids and Bases Strong acid or base is COMPLETELY dissociated in aqueous solution HCl (aq) → H+ (aq) + Cl- (aq) Formula Name Hydrochloric acid HBr Hydrobromic acid HI Hydroiodic acid a Sulfuric acid H2 SO4 HNO3 Nitric acid HClO4 Perchloric acid a. For H2 SO4, only the first proton ionization is complete. Dissociation of the second proton has an equilibrium constant of 1.0 × 10−2 . 35 Weak Acids and Bases Weak Acids (HA) react with water, donate a proton to H2O KOH (aq) → K+ (aq) + OH- (aq) Formula More commonly: Name Bases Acids HCl Basic solutions • [H+] < [OH−] • pH > 7
LiOH
NaOH
KOH
RbOH
CsOH
Lithium hydroxide
Sodium hydroxide
Potassium hydroxide
Rubidium hydroxide
Cesium hydroxide
Quaternary ammonium
R 4 NOHb
hydroxide
b. This is a general formula for any
hydroxide salt of an ammonium cation
containing four organic groups.
Ka = acid dissociation constant
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Weak Acids and Bases
Weak Acids and Bases
Weak Bases (B) react with water, accept a proton from H2O
Common Weak Acids
Common Weak Bases
Kb = base hydrolysis constant
(Amines)
(Ammonium Ions)
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Polyprotic Acids and Bases
Polyprotic Acids and Bases
Polyprotoic acids/bases: compounds that can donate or
accept more than one proton
Polyprotoic acids/bases: compounds that can donate/accept
more than one proton
Kb1 = 2.3 × 10−2
Ka1 = 5.62 × 10−2
Kb2 = 1.60 × 10−7
Ka2 = 5.42 × 10−5
Kb3 = 1.42 × 10−12
Different K for 1st and 2nd H+ lost
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Different K for all H+ accepted
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Polyprotic Acids and Bases
Conjugate acid-base pair Ka and Kb are RELATED!
Di-protic system
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Tri-protic system
Chapter 6: Chemical Equilibrium
Equilibrium Review
• Equilibrium Constant (K)
• Thermodynamics (K related to H, S, G)
• Le Châtelier’s Principle (K vs. Q)
More Equilibrium Fun
• Acids-Bases (Ka, Kb)
• Solubility Product (Ksp)
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Recall from General Chemistry
Recall from General Chemistry
Soluble: ionic compounds that dissolve a LOT in H2O (≥ 0.01M )
Precipitate:
an insoluble
solid product
that separates
from a solution
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Recall: Solubility Guides for Ionic Compounds
Slightly Soluble = ionic compounds that do NOT dissolve much in H2O
→ most remains as a solid
→ often form precipitates
How can we predict which ionic compounds will be soluble?
There are RULES you need to MEMORIZE!
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Solubility Equilibria: The REAL Solubility Rules
Slightly Soluble = ionic compounds that do NOT dissolve much in H2O
→ most remains as a solid
→ often form precipitates
•
Any ionic compound will dissolve to some extent
•
a TINY bit of compound dissociates into ions
• Equilibrium is established between the
dissociated ions and the ionic compound
AgCl is not very soluble, but what does
dissolve dissociates completely
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Solubility Equilibria: The REAL Solubility Rules
If it’s in equilibrium …
we can write a K expression
Solubility product constant (Ksp) – equilibrium constant
that indicates to what extent a slightly soluble
compound dissociates in water
• Large Ksp = more soluble
• Small Ksp = less soluble
Solubility Example
Calculate the molar solubility of copper(II) hydroxide in pure water.
(Ksp = 2.2 x 10–20)
1. Balanced equation
Cu(OH)2(s)
2. K expression
Ksp = [Cu2+] [OH-]2
3. ICE table, solve for “x”
Cu(OH)2(s)
I
C
E
→ Solid is ALWAYS a reactant (not included in K expression)
→ Ions ALWAYS products
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Cu2+(aq) + 2 OH-(aq)
Cu2+(aq) + 2 OH-(aq)
0
0
x
x
2x
2x
Ksp = [Cu2+] [OH-]2
2.2×10–20 = (x)(2x)2
2.2×10–20 = 4×3
1.8×10-7= x
1.8×10-7= [Cu2+]
1.8×10-7 mol Cu2+ × 1 mol Cu(OH)2 = 1.8×10-7 M Cu(OH)
2
1L
1 mol mol Cu2+
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Predicting Precipitation: Q vs Ksp
Compare Q with Ksp to determine if a precipitate will form!
If Q < Ksp Too much reactant More solid will dissolve NO precipitate Common Ion Effect Consider the following: Why does KCl precipitate? If Q = Ksp Equilibrium No additional accumulation of solid or ions NO precipitate If Q > Ksp Too much product
Ions will combine to form solid
YES, precipitate forms!
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Common Ion Effect
Common Ion Effect: A salt will be less soluble if one of its
constituent ions is already present in the solution
It’s really just an application of Le Chatelier’s principle:
Common Ion Example
What is the concentration of Hg22+ in equilibrium with 0.10 M
solution of KCl containing excess, undissolved Hg2Cl2(s)?
For Hg2Cl2(s), Ksp=1.2×10-18
KCl (s) ↔ K+ (aq) + Cl- (aq)
Push Equil. to LEFT, and
precipitate reactant
Increase either product
Another way to think about it ….
Ksp = [K+][Cl-]
If one concentration increased, other must
DECREASE so that multiplication still = Ksp
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Common Ion Example
Separation by Precipitation
Hg22+ in
What is the concentration of
equilibrium with 0.10 M
solution of KCl containing excess, undissolved Hg2Cl2(s)?
For Hg2Cl2(s), Ksp=1.2×10-18
Precipitation can be used to separate ions
Consider a mixed solution of lead and mercury
ions, both at 0.010 M concentrations:
PbI2 (s)
Pb2+ + 2IKsp = 7.9 × 10-9
Hg2I2 (s)
Hg22+ + 2I-
Ksp = 4.6 × 10-29
Smaller Ksp = LESS soluble
Is it possible to lower the concentration of Hg 22+ by 99.990%
by selective precipitation with I-, without precipitating Pb2+?
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Separation by Precipitation
Mixed solution of lead and mercury ions, both 0.010 M
Ksp = 7.9 × 10-9
PbI2 (s)
Pb2+ + 2IHg2I2 (s)
Hg22+ + 2I-
Separation by Precipitation
Mixed solution of lead and mercury ions, both 0.010 M
Ksp = 7.9 × 10-9
PbI2 (s)
Pb2+ + 2I-
Ksp = 4.6 × 10-29
Is it possible to lower the concentration of Hg22+ by 99.990%
by selective precipitation with I-, without precipitating Pb2+?
Step 1: Figure out [Hg2+] if reduced by 99.990%
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Hg2I2 (s)
Hg22+ + 2I-
Ksp = 4.6 × 10-29
Is it possible to lower the concentration of Hg 22+ by 99.990%
by selective precipitation with I-, without precipitating Pb2+?
Step 2: Determine the [I-] at this reduced [Hg2+]
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Separation by Precipitation
Mixed solution of lead and mercury ions, both 0.010 M
Ksp = 7.9 × 10-9
PbI2 (s)
Pb2+ + 2I-
Hg2I2 (s)
Hg22+ + 2I-
Separation by Precipitation
Mixed solution of lead and mercury ions, both 0.010 M
Ksp = 7.9 × 10-9
PbI2 (s)
Pb2+ + 2I-
Ksp = 4.6 × 10-29
Hg22+ by
Hg2I2 (s)
Hg22+ + 2I-
Ksp = 4.6 × 10-29
Is it possible to lower the concentration of
99.990%
by selective precipitation with I-, without precipitating Pb2+?
Is it possible to lower the concentration of Hg22+ by 99.990%
by selective precipitation with I-, without precipitating Pb2+?
Step 3: Using [I-] at reduced [Hg2+], solve for the Q for PbI2
Step 4: Compare the Q to Ksp
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Separation by Precipitation
Mixed solution of lead and mercury ions, both 0.010 M
Ksp = 7.9 × 10-9
PbI2 (s)
Pb2+ + 2IHg2I2 (s)
Hg22+ + 2I-
Ksp = 4.6 × 10-29
Is it possible to lower the concentration of Hg22+ by 99.990% by
selective precipitation with I-, without precipitating Pb2+?
Step 1: Figure out what the [Hg2+] would be if reduced by 99.990%
Step 2: Determine the [I-] at this reduced [Hg2+]
Step 3: Using the [I-] at reduced [Hg2+], solve for the Q for lead
Step 4: Compare the Q for lead to the Ksp for lead
We wish life were so easy to ensure separation.
BUT, sometimes we get co-precipitation
ONLY an experiment can tell if the separation actually worked!
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Homework Chapter 6
1. Find [Cu2+] in a solution saturated with Cu4(OH)6(SO4) if [OH-] is fixed at 1.0 x 10-6 M.
Cu4(OH)6(SO4) (s) ↔ 4Cu2+ (aq) + 6OH- (aq) + SO42- (aq)
Ksp = 2.3 x 10-69
2. A solution contains 0.0500 M Ca2+ and 0.0300 M Ag+.
a. Can 99% of Ca2+ be precipitated by sulfate without precipitating Ag+?
(Hint: Ksp values can be located in Appendix F of your textbook)
b. What will be the concentration of Ca2+ when Ag2SO4 begins to precipitate?
3. Locate Appendix G in your textbook – here you will find the acid dissociation constants
(Ka) for many common weak acids. When multiple values are listed, the first (top) value
is the Ka1, under it is Ka2, then Ka3, etc.
a. Write the Ka3 chemical reaction of phosphoric acid. What is the Ka3 value for
this reaction?
b. Write the Kb2 chemical reaction of sodium carbonate. What is the Kb2 value
for this reaction? (Hint: Appendix G only lists acids, so you will need to think
about the conjugate acid of carbonate!)