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Transcript of Reactions are reversibleReactions are reversible A + B C + D ( forward) C + D A + B (reverse) ...
![Page 1: Reactions are reversibleReactions are reversible A + B C + D ( forward) C + D A + B (reverse) Initially there is only A and B so only the forward.](https://reader036.fdocuments.net/reader036/viewer/2022062423/5697bfc71a28abf838ca7fb4/html5/thumbnails/1.jpg)
Chemical Equilibrium
Reactant and Product Concentrations at Equilibrium
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Reactions are reversible
A + B C + D ( forward)
C + D A + B (reverse)
Initially there is only A and B so only the forward reaction is possible
As C and D build up, the reverse reaction speeds up while the forward reaction slows down.
Eventually the rates are equal
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React
ion
Rate
Time
Forward Reaction
Reverse reaction
Equilibrium
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What is equal at Equilibrium?
Rates are equal
Concentrations are not.
Rates are determined by concentrations and activation energy.
The concentrations do not change at equilibrium.
or if the reaction is verrrry slooooow.
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The units for K
Are determined by the various powers and units of concentrations.
They depend on the reaction.
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K is CONSTANT
At any temperature.
Temperature affects rate.
The equilibrium concentrations don’t have to be the same only K.
Equilibrium position is a set of concentrations at equilibrium.
There are an unlimited number.
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Calculate K
N2 + 3H2 3NH3
Initial At Equilibrium
[N2]0 =1.000 M [N2] = 0.921M
[H2]0 =1.000 M [H2] = 0.763M
[NH3]0 =0 M [NH3] = 0.157M
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Equilibrium and Pressure
Some reactions are gaseous
PV = nRT
P = (n/V)RT
P = CRT
C is a concentration in moles/Liter
C = P/RT
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Equilibrium and Pressure
2SO2(g) + O2(g) 2SO3(g)
Kp = (PSO3)2
(PSO2)2 (PO2)
K = [SO3]2
[SO2]2 [O2]
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The Reaction Quotient
Tells you the directing the reaction will go to reach equilibrium
Calculated the same as the equilibrium constant, but for a system not at equilibrium
Q = [Products]coefficient
[Reactants] coefficient
Compare value to equilibrium constant
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What Q tells us
If Q<K Not enough products Shift to right
If Q>K Too many products Shift to left
If Q=K system is at equilibrium
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Checking the assumption
The rule of thumb is that if the value of X is less than 5% of all the other concentrations, our assumption was valid.
If not we would have had to use the quadratic equation
Our assumption was valid.
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Using KC to determine [products] and [reactants] at equilibrium
Only known KC and initial concentrations of reactants
Changes in concentrations (ΔC) of chemical compounds related to STOICHIOMETRIC rations in chemical equation.
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Steps for Complex Problems
1) Make an ICE Chart
2) Set up the equilibrium expression
3) Solve for ΔC
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Ice Chart
“ICE” Reactants Products
Initial Concentrations (I):
Change in Concentration (ΔC or x):
Equilibrium Concentrations (E):
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Example 3:
H2O(g) is present in a rigid container at 25°C with an initial partial pressure of 0.784 atm. What are the partial pressures of H2(g) and O2(g) at equilibrium? (KP = 2.0 x 10 -42)
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Example 3: continued
If Keq < 1x10-4, remove the “x or ΔC” value in denominator. ΔC is very small compared to initial
concentration so subtraction would not be a huge difference.
Only works when adding or subtracting ΔC If concentrations or partial pressures are
very small where their magnitude is approximately equal to Keq, CANNOT discount ΔC value.
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Example 4:
Sulfur trioxide decomposes to form sulfur dioxide and oxygen at 300°C°. Calculate the concentrations of all chemical compounds at equilibrium with an initial SO3 concentration of 0.100M and KC = 1.6 x 10-10.
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Example 5:
0.194 mol of COCl2 comes to equilibrium in a 5.8L container at 25°C (KC = 7.27 x 10-38). Find the equilibrium concentrations of all chemical compounds in the following equation. (Hint: first find the initial [COCl2] )
COCl2 (g) CO(g) + Cl2 (g)
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% Ionization
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Le’Chatlier’s Principle
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Le Chatelier’s Principle
If a stress is applied to a system at equilibrium, the position of the equilibrium will shift to reduce the stress.
3 Types of stress
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Change amounts of reactants and/or products
Adding product makes Q>K
Removing reactant makes Q>K
Adding reactant makes Q<K
Removing product makes Q<K
Determine the effect on Q, will tell you the direction of shift
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Variables disrupting equilibrium
1) Addition or removal of chemical compounds from the reaction.
2) Pressure
3) Temperature/Heat
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Basic Concept
Main concept in chemical equilibrium
Change in a variable that alters the equilibrium of a system (chemical reaction) products a shift in the OPPOSITE direction Reaction shifts to counteract the variable’s influence
Reaction tries to get back to equilibrium state-----SO reaction shifts
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Basic Concept (cont.)
What happens to a chemical reaction when equilibrium shifts
When one side of a chemical reaction is stressed, the reaction shifts to the side of LEAST stress !
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Using KC to determine [products] and [reactants] at equilibrium
Only known KC and initial concentrations of reactants
Changes in concentrations (ΔC) of chemical compounds related to STOICHIOMETRIC rations in chemical equation.
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Change Pressure
By changing volume
System will move in the direction that has the least moles of gas.
Because partial pressures (and concentrations) change a new equilibrium must be reached.
System tries to minimize the moles of gas.
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Change in Pressure
By adding an inert gas
Partial pressures of reactants and product are not changed
No effect on equilibrium position
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Change in Temperature
Affects the rates of both the forward and reverse reactions.
Doesn’t just change the equilibrium position, changes the equilibrium constant.
The direction of the shift depends on whether it is exo- or endothermic
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Example 5:
Predict the influence of certain changes to the following chemical reaction. 2SO3(g) 2SO2(g) + O2(g) ΔH°= 197.84 kJ/mol
A) increasing reaction’s temperature
B) increasing reaction’s pressure
C) adding more O2 at equilibrium
D) Removing O2 at equilibrium
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Free Energy and Equilibrium
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ΔG° ΔG
Concentration and partial pressures can change for reactants and products.
ΔG = ΔG° + RTlnQ
What happens when we leave the standard state conditons?
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At equilibrium, ΔG = 0, so reaction quotient (Q) = equilibrium constant (K)
At equilibrium ΔG° = - RTlnK
Enables the reaction’s equilibrium constant (K) to be calculated from the change in free energy (ΔG°)
ΔG = ΔG° + RTlnQ
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The magnitude of ΔG° indicates how far the chemical reaction in its standard state is from equilibrium. ΔG° = 0 , equilibrium ΔG°= large value, far from equilibrium ΔG° = small value, close to equilibrium
The sign (+, - ) indicates which direction the reaction needs to shift to achieve equilibrium Positive (+) -------- shift to left, no reaction Negative (-) -------- shift to right, reaction goes to completion
What is the relationship between free energy(ΔG) and K?
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Calculate the equilibrium constant for the following reaction given ΔG° = -33 kJ/mol.
N2 + 3H2 2NH3
Example 1
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Weak Acid/Base Calculations
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Reaction between H2O and Acid
HA + H2O H3O+ + A-
Write the equilibrium expression
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Acid Dissociation Constant (Ka)
Quantifies how much an ACID dissociates in water
Describes the acid strength
Ka = [H3O+] [A-]
[HA]
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Acid Dissociation Constant
(Ka) Strong Acids > 1
Weak Acids <<<< 1
Examples: HCl Ka = 1x106
CH3CO2H Ka = 1.8 x 10-5
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Base Ionization Constant (Kb)
Quantifies how much a BASE dissociates in water
Describes base strength
B + H2O BH+ + OH-
Kb = [BH+] [OH-]
[B]
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Other helpful equations
pKa = -logKa
pKb = -logKb
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Weak Acid pH Calculations
Weak acids only partially dissociate
Calculating: Use ICE method Compare reaction quotient (Qa) to equilibrium
constant (Ka)----- % Ionization When can we ignore change in initial acid
concentration (ΔC) ?
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When can we ignore change in initial acid concentration (ΔC)?
If ΔC < 5% of initial acid concentration
Ka < 1x10-4
THEN, ΔC can be disregarded for equilibrium acid concentration
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Example 1:
Determine the H3O+ ion concentration and pH of a 1.0M acetic acid solution with a Ka = 1.8 x 10-5
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Example 2:
Calculate the pH of a 0.10M NH3 solution with a
pKb of 4.74
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Ka and Kb
MUST determine Kb from Ka value
Use relationship between an acid and its conjugate base
KaKb = Kw
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Example 5:
Find the pH of a 3.5M NH3 solution with a Kb value of 1.8 x 10-5
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Acid/Base Mixtures
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NaOAc
HOAc
Now we would have 2 sources for OAc - ion 1) Acetic Acid Dissociation
2) Salt Hydrolysis/Dissociation
**Acetate ion is COMMON to both reactions**
What would happen to the pH if we added….
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Common ion Ion present in both acid/base dissociation and
salt ionization
Common ion effect Seen with weak acids and bases Hinders the ionization of a weak acid or base due
to the presence of a common ion between the acid/base and a salt
Common Ion Effect
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HF + H2O(l) H3O+ + F-
Addition of KF
KF K+ + F-
Result is less [H3O+] so pH ----more BASIC
Example 1:
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NH3 + H2O(l) OH- + NH4+
Addition of NH4Cl
NH4Cl NH4+ + Cl-
Result is less [OH-] so pH ----more ACIDIC
Example 2:
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450ml of 1.0M HCO2H is mixed with 250ml of 0.40M NaHCO2 at 25°C. Find the pH of the solution (Ka = 1.8 x 10-4).
Example 3:
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Salt Hydrolysis
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Salts
Ionic compound made up of CATION and ANION
Has acidic and basic properties Based on ions produced when salts dissociate No acid/base properties—group I/II cations
(ex. Na+, Li+, K+, Ca+2) No basic properties—conjugate bases from monoprotic
acids (ex. Cl-, Br-, NO3-)
Ex. NaCl, CaBr2
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1. Salt Formation from Strong Base and Weak
Acid Salt forms a BASIC solution.
Conjugate base ion reacts with water to give hydroxide (OH-) ions.
Ex. Potassium fluoride (KF) KF K+ + F-
F- + H2O HF + OH-
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2. Salt Formation from a Strong Acid and Weak
Base Salt forms an ACIDIC solution
Conjugate acid reactions with water to give hydronium ion (H3O+)
Ex. Ammonium nitrate (NH4NO3)
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3. Salt Formation from Strong Acid and Strong
Base Salt forms a NEUTRAL solution
Conjugate base resulting from salt dissociation is weak
Ex. Sodium chloride (NaCl)
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Example 1:
Calculate the concentration of HOAc, OAc- and OH- at equilibrium in a 0.10M NaOAc solution (Ka for HOAc = 1.8 x 10-5).
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Acid-Base Titrations
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Titration
Lab technique commonly utilized to determine an UNKNOWN concentration of a chemical compound with a KNOWN concentration of another chemical compound.
Chemical compounds combine with exact stoichiometric proportions
Analyte— Chemical compound with unknown concentration
Titrant— Chemical compound with known concentration Measured with volume and concentration Added to chemical compound with unknown concentration
in titration
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2 Types of Acid-Base Titrations
1) Strong Acid/Strong Base Titrations
2) Weak Acid/Strong Base Titrations
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1. Strong Acid/Strong Base Titrations
Low initial pH value
Sharp increase in pH before equivalence point
Equivalence point is pH = 7
Rapid pH increase after equivalence point
**Indicators with pH range 4-10 helpful for these titrations
**Neutralization reactions
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Strong Acid with Strong Base Titrant
Strong Base with Strong Acid Titrant
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2. Weak Acid/Strong Base Titrations
High initial pH value
pH = pKa at half-neutralization [weak acid] = [conjugate base]
Ka = [H3O+] [A -] SO Ka = [A-]/[HA] is 1:1
[HA]
Ka = [H3O+], SO pH = pKa
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2. Weak Acid/Strong Base Titrations
Equivalence point > 7 on pH scale
**Indicators with pH range > 7 helpful as pH equivalence point is basic
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Example 1: Weak Acid/Strong Base Titration Calculations
30 ml of 0.5M HC2H3O2 titrated with 0.50M NaOH. Ka for acetic acid is 1.8x10 -5.
a) Find the initial pH of 0.5M acetic acid.
HC2H3O2 + OH- C2H3O2- + H2O (titration
view)
HC2H3O2 H+ + C2H3O2- (in detail)
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Example 1: Weak Acid/Strong Base Titration Calculations
b) Find the pH after 15ml of NaOH were added
pH = pKa + log [C2H3O2-] / [HC2H3O2]
-used when dealing with buffer solutions
-buffer—mixture of weak acid/conjugate base
-more on this concept later
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Example 1: Weak Acid/Strong Base Titration Calculations
c) Find the pH at the equivalence point.
C2H3O2- + H2O HC2H3O2 + OH-
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Buffers
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Weak acid/conjugate base mixtures OR weak base/conjugate acid mixtures
“buffers” or reduces the affect of a change in the pH of a solution Absorbs slight changes in pH resulting from the
addition of small acid/base amounts to water.
Buffers
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HOAc + H2O(l) H3O+ + OAc –
What happens if an acid is added??? Reacts with OAc ion [HOAc] increases slight, [OAc] decreases
slightly, ratio mostly the same No pH change
Example 1: (cont.)
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HOAc + H2O(l) H3O+ + OAc –
What happens if a base is added??? Reacts with HOAc ion More OAc ion formed, removes excess
OH- from solution No pH change
Example 1: (cont.)
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1) Acidic Buffers Formed from mixing a weak acid and its conjugate
base pH < 7 Ex. HOAc and OAc –
2) Basic Buffers Formed from mixing a weak base and its conjugate
acid pH > 7 Ex. NH3 and NH4
+
Types of buffers
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Compare Ka and Kb from acid-conjugate base pair
Ka > Kb, (generally > 1x10-7) -------- acidic buffer
Ka < Kb, (generally > 1x10-7) -------- basic buffer
How do we tell acidic vs. basic buffers?
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1) Method applying “Common Ion Effect”
2) Henderson-Hasselbalch equation
Calculating the pH of buffers
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Easier method
pH = pKa + log[A-]/[HA]
pOH = pKb + log[HB+]/[B] [B] = molarity of weak base [HB+] = molarity of conjugate acid
Assumption: weak acids and conjugate bases do NOT change concentration with equilibrium.
Henderson-Hasselbalch Equation
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Find the pH of a buffered solution created by mixing 0.15mol NH4NO3 with 0.65L of a 0.25M NH3 solution. Assume that the volume change is negligible. (Kb = 1.8x10-5)
Example 1: