chemical equilibrium. up to this point we have mostly been considering reactions “to...
TRANSCRIPT
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CHEMICAL EQUILIBRIUM
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Up to this point we have mostly been considering reactions “to completion”, where all the reactants change into product.
However, most reactions are reversible = occurs in both the forward and the reverse directions.
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N2(g) + 3H2(g) 2NH3 (g) forward
N2(g) + 3H2(g) 2NH3 (g) reverse
N2(g) + 3H2(g) 2NH3 (g)
orN2(g) + 3H2(g) ↔ 2NH3 (g)
Combined in one equation using double arrows
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Chemical Equilibrium
A state in which the forward and reverse reactions balance each other and when the forward reaction proceeds at the same rate as the reverse reaction.
Concentrations of reactants and products are constant at equilibrium.
(Constant ≠ equal)
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Rate vs. Time
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Question
Can we change the equilibrium position thereby increasing the amount of products in a reaction?
Yes – by adding stress to a system in equilibrium.
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LE CHÂTELIER’S PRINCIPLE
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Le Châtelier’s Principle
If a stress is applied to a system at equilibrium, the system shifts in the direction that relieves the stress.
Stress is anything that upsets equilibrium – changes in concentration, pressure, or temperature.
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Le Châtelier’s Principle
If a stress is applied to a system at equilibrium, the system shifts in the direction that relieves the stress. Concentration Pressure TemperatureNote: Only temperature
affects K. The larger the value of K, the more product at equilibrium.
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Concentration
Measure of molarity (moles/L) If you ↑concentration of a
reactant, equilibrium will shift toward the products.
If you ↓concentration of a reactant, equilibrium will shift toward the reactants.
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CO(g) + 3H2(g) ↔ CH4(g) + H2O(g)
CO(g) + 3H2(g) ↔ CH4(g) + H2O(g)
CO(g) + 3H2(g) ↔ CH4(g) + H2O(g)
CO(g) + 3H2(g) ↔ CH4(g) + H2O(g)
CO(g)
H2O(g)
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Changes in Concentration
A + B <=> C + D
Increasing the concentration of “A” will shift the reaction to the right we need to get rid of excess “A”
Decreasing the concentration of “C” will shift the reaction to the right there is a deficit, so more “C” needs to be
made
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Temperature Increasing the temperature shifts
the reaction away from the side that contains the “heat”
Decreasing the temperature shifts the reaction toward the side that contains the “heat”
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heat
CO(g) + 3H2(g) ↔ CH4(g) + H2O(g) + heat
CO(g) + 3H2(g) ↔ CH4(g) + H2O(g) + heat
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Sample Reaction
heat + NH4Cl (s) <=> NH3 (g) + HCl (g)
In the above endothermic reaction, increasing the temperature will drive the reaction to the right (in other words, forward)
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Pressure
Ideal Gas Law: PV=nRT If ↑P then ↑n, which means
more number of atoms. If ↑P, then the equilibrium will
shift toward the side with fewer moles of gas.
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CO(g) + 3H2(g) ↔ CH4(g) + H2O(g)
4 moles of gas 2 moles of gas
Note: If moles of gaseous reactant = moles of gaseous product, then no shift in equilibrium will occur from a change in pressure
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Equilibrium constant
K is called the equilibrium constant. It is a ratio of the concentrations of products to the concentration of reactants.
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Equilibrium Constant Expression
aA + bB ↔ cC + dD
Keq= [C]c[D]d
[A]a[B]b
A & B = molar [ ] of reactantsC & D = molar [ ] of productsExponents a, b, c, and d = coefficients in the balanced equation.
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Equilibrium Constant
If Keq > 1: products are favored at equilibrium
If Keq < 1: reactants are favored at equilibrium
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Important!
Only substances that are gases and aqueous get factored into the equilibrium expression
Pure liquids and solids do not appear in the expression.
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Example #1
Write the equilibrium expression for the following reaction:
2 CO (g) + O2 (g) ↔ 2 CO2 (g)
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Answer
Keq = [CO2]2 / ([CO]2[O2])
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Example #2
Write the equilibrium expression for the following reaction:
CO (g) + 3 H2 (g) ↔ CH4 (g) + H2O (g)
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Answer
Keq = [CH4][H2O] / ([CO][H2]3)
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Drill #14 May 19 & 20, 2014 What are the 3 types of stress that can
affect the equilibrium of a system? Which states of matter get factored
into an equilibrium expression? If Keq > 1: are favored at
equilibrium If Keq < 1: are favored at
equilibrium
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Drill #16 May 27 & 28, 2014 An equilibrium mixture of N2, O2, and NO
gases at 1500 K is determined to consits of 6.4 x 10-3 mole/L of N2, 1.7 x 10-3 mol/L of O2 and 1.1 x 10-5 mo/L of NO. What is the equilibrium constant for the system at this temperature?
Given: the molarity of the three gases Unknown: K Write the balanced equation Substitute the given values for the [ ]s
into the equilibrium expression.
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Answer to Drill
N2 + O2 2NO
K = [NO]2
[N2][O2]
K = (1.1 x 10-5 mo/L)2 (6.4 x 10-3 mole/L )(1.7 x 10-3
mol/L )
K = 1.1 x 10-5 The value of K is < 1 therefore reactants are favored at equilibrium.
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Equilibrium Constant
If Keq > 1: products are favored at equilibrium
If Keq < 1: reactants are favored at equilibrium
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Temperature• Think of heat as a reactant or a product.
CO(g) + 3H2(g) ↔ CH4(g) + H2O(g) + heat
Is this Exothermic or Endothermic?
Reversible reactions are exothermic (energy as heat shown on product side) in one direction and endothermic (energy as heat shown on reactant side) in the other.
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Drill #17 May 28 & 29, 2014
1. Concentration Changed 2A (g) + 3B (g) 2C (g) + D (g)Action Effect?
Increase [A] Shift to the ? Increase [C] Shift to the ? Decrease [B] Shift to the ? Decrease [C] Shift to the ?
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Drill #17 May 28 & 29, 2014
1. Concentration Changed 2A (g) + 3B (g) 2C (g) + D (g)Action Effect?
Increase [A] Shift to the
right Increase [C] Shift to the
left Decrease [B] Shift to the
left Decrease [C] Shift to the
right
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Drill #17 May 28 & 29, 2014
2. Volume Changed 2A (g) + 3B (g) 2C (g) + D (g)Action Effect?
Volume decreased Shifts to the ? Volume increased Shifts to the ?
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Drill #17 May 28 & 29, 2014
2. Volume Changed 2A (g) + 3B (g) 2C (g) + D (g)Action Effect?
Volume decreased Shifts to right (side with
fewest moles) Volume increased Shifts to left (side with
most moles)
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Drill #17 May 28 & 29, 2014
3. Temp. Changed for Endothermic Rxn.
A (g) + B (g) C (g) + D (g)Action Effect?
Increase temp Shift to
the ? Decrease temp Shift to
the ?
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Drill #17 May 28 & 29, 2014
3. Temp. Changed for Endothermic Rxn.
A (g) + B (g) C (g) + D (g)Action Effect?
Increase temp Shifts to right
(endothermic rxn.) Decrease temp Shifts to left
(exothermic rxn.)
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Agenda
Quiz Finish Notes on Ice Table Finish Worksheet and new worksheet on
Ice Table Chemical Energy notes
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INTRO TO ICE BOXES
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Equilibrium Problems
The equilibrium constant can also be used to calculate equilibrium concentrations from initial concentrations
Use an ICE box! To start with, we will use a generic
equation A(g) + B(g) 2 C(g)
[A] = .100 M and [B] = .100 M
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A(g) + B(g) 2 C(g)
A B 2 C
Initial .100 M .100 M 0
Change -x -x +2x
Equilibrium
.100 – x .100 – x 2x
A and B are assigned –x values because they are being reacted, while C gets a +x because products are being
formed. The 2x comes from the coefficient in front of the C
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ICE BOX PRACTICE PROBLEMSWe will complete in class examples and then you will try on your own!
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Drill #18 June 2 & 3, 2014 What is energy? What are the two different types of
energy?
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What is Energy?The ability to do work or produce heat.
Potential Energy – stored energy. Kinetic Energy – energy of motion.
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Agenda
Equilibrium Constant for Acids and Bases Worksheets Chemical Energy Diagrams Final Exam Review Project
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Strong Acids/Bases vs. Weak Acids/Bases
STRONG Ionize almost
100% in water (react to completion)
WEAK Do not ionize
completely, achieve an equilibrium
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HOW CAN I TELL THE DIFFERENCE BETWEEN STRONG AND WEAK ACIDS?
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Two different ways:
Study the list of strong acids and bases Look at the arrow in an acid/base
balanced equation If arrow is one direction it is STRONG If arrow is two directions it is WEAK
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Introduction
Strong Acid Example HNO3(aq) + H2O(l) H3O+
(aq) + NO3-(aq)
Weak Acid Example CH3COOH(aq) + H2O(l) H3O+
(aq) + CH3COO-(aq)
Strong Base Example: NaOH(aq) + H2O(l) Na+
(aq) + OH-(aq)
Weak Base Example: NH3(aq) + H2O(l) NH4
+(aq) + OH-
(aq)
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INSTEAD OF USING KEQ WE USE A DIFFERENT CONSTANT FOR WEAK ACIDS AND BASES
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Ionization Constant of a Weak Acid Ka – acid ionization constant Ka varies at different temperatures Generic Example:
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Weak Acid Example
CH3COOH(aq) + H2O(l) H3O+(aq) + CH3COO-
(aq)
Weak acids have a Ka < 1 Leads to small [H3O+] pH of weak acids is calculated by solving
the expression for [H3O+] at equilibrium and then taking the –log[H3O+]
–3 3
a3
[H O ][CH COO ]K
[CH COOH]
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Ionization Constant of a Weak Base Kb – base ionization constant Weak bases = Kb < 1 Leads to a small [OH-] To find pOH take -log[OH-]
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Weak Acid Example
HA(aq) + H2O(l) H3O+(aq) + A-
(aq)
Initially, you have 1.00 M HA. Calculate the equilibrium concentrations of HA, H3O+, A-, and calculate the pH. Ka = 1.80 x 10-5
Use an ICE table!
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Weak Acid Example
HA H3O+ A-
I 1.00 0 0
C -x +x +x
E 1.00 – x +x +x
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Weak Acid Example
Write Ka expression:
Assume x is small because Ka is very small, therefore x can be ignored
Ka = 1.80 x 10-5 = [H3O+][A-] = x2
[HA] 1.00-x
Ka = 1.80 x 10-5 = x2
1.00
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Weak Acid Example
x = [H3O+] = [A-] = 4.20 x 10-3 M pH = - log [H3O+] = 2.37
Ka = 1.80 x 10-5 = x2
1.00
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Weak Base Example
NH3(aq) + H2O(l) NH4+
(aq) + OH-(aq)
Initially, you have a concentration of 0.010 M NH3. Find the concentrations at equilibrium. Calculate the pH. Kb = 1.80 x 10-5
Use an ICE table!
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Weak Base Example
NH3 NH4+ OH-
I 0.010 0 0
C -x +x +x
E 0.010 – x +x +x
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Weak Base Example
Write Kb expression:
Assume x is small because Kb is very small, therefore x can be ignored
Kb = 1.80 x 10-5 = [NH4][OH-] = x2
[NH3] 0.010-x
Kb = 1.80 x 10-5 = x2
0.010
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Weak Base Example
x = [NH4+] = [OH-] = 4.20 x 10-4 M
pOH = - log [OH-] = 3.37 pH + pOH = 14 Therefore, 14 - 3.37 = 10.6 = pH
Kb = 1.80 x 10-5 = x2
0.010
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DIFFERENT TYPES OF EQUILIBRIUM
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Concentration Equilibrium Kc (or Keq)
nA + mB ↔ xC + yD
Kc = [C]x[D]y
[A]n[B]m
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Remember…equilibrium is where the rates of forward and reverse reactions are the same. It means that the concentrations do not change, NOT that they are identical.
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Because equilibrium expressions have to do with concentration (in molarity) we do not include items that are not in solution so NO LIQUID or SOLID states! They are in excess so can be ignored.
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Acid Equilibrium
Acid + H2O ↔ H3O+ + Acid Ion
or HA + H2O ↔ H3O+ + A-
Ka = [H3O+][A-] Ka = [H3O+][A-]
[HA][H2O] [HA]
Because water is a solvent (liquid) and its conc. greatly exceeds the acid, we can assume that the conc. of water does not change.
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Base Equilibrium
Base + H2O ↔ OH- + Base Ion
or B + H2O ↔ OH- + HB+
Kb = [OH-][HB+] Kb = [OH-][HB+]
[B][H2O] [B]
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Acid Strength
A stronger acid will react completely to form ions (strong electrolytes) and hydronium ions (H3O+) in water (dissociation)
Hydrochloric acid (all HCL molecules are ionized into hydronium and chloride ions)
HCl + H2O H3O+ + Cl-
Acetic Acid (not all A.A. molecules are ionized into hydronium and ions)
CH3COOH + H2O H3O+ + CH3COO-
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Examples of Strong & Weak Acids Strong Weak
HCl CH3COOH
HBr H2CO3
H2SO4 HClO
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Base Strength
A strong base has the strongest affinity for H+ ions and dissociates entirely into metal ions and OH-.
For ex. Calcium oxide (CaO); oxygen strongly attracts H+ ions.
Strong bases: CaO, NaOH, KOH Weak base: ammonia (NH3)
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Sample Problem
Write an equation to show the dissociation of hydrochloric acid in water.
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Answer
HCl + H2O → H3O+ + Cl−
HCl is a STRONG ACID
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Property Strong Acid Weak AcidKa value Ka is large Ka is small
Position of equilibrium
Far to the right (a lot of dissociation)
Far to the left (little dissociation)
[H+] compared to original [HA]
[H+]≈[HA]o [H+]<<[HA]o
Strength of conjugate base
A- is much weaker
A- is much stronger
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Graphic Representation of the Behavior of Acids of
Different Strengths in Aqueous Solution
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Solubility Equilibrium
Salt (s) ↔ Cation (aq) + Anion (aq)
Solids are not included in equilibrium equations! So…
Ksp = [Cation][Anion]
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Solubility Equilibrium Example CaF2 (s) ↔ Ca+2 (aq) + 2F- (aq)
Solids are not included in equilibrium equations! So…
Ksp = [Ca+2][F-]2
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Drill #15 May22 & 23, 2014
1. Write the equilibrium expression for the reaction:
H2 (g) + I2 (s) 2HI (g)
2. How would the following shift the equilibrium in the equation (forward, reverse or no change):H+ (aq) + Cl- (aq) ↔ HCl (aq) + 10.3 kJ
a) Increasing temperatureb) Increasing pressurec) Adding Cl-
d) Removing HCl
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Answers: May22 & 23, 2014
1. Write the equilibrium expression for the reaction:
H2 (g) + I2 (s) 2HI (g)
Keq = [HI]2
[H2]
2. How would the following shift the equilibrium in the equation (forward, reverse or no change):
H+ (aq) + Cl- (aq) ↔ HCl (aq) + 10.3 kJa) Increasing temperature (reverse)b) Increasing pressure (forward)c) Adding Cl- (forward)d) Remove HCl (forward)