PMU Week 3 Chemical Equilibrium Reaction Lab Report this is week 3 knodlge check for chemistry i will be adding the quiestion in word doc and will be addin

PMU Week 3 Chemical Equilibrium Reaction Lab Report this is week 3 knodlge check for chemistry i will be adding the quiestion in word doc and will be adding the powerpoint slides just like last time. Preview – Look through the PP slides BEFORE the class (spend roughly 15-25 minutes to have a general sense of what
concepts I will be covering)
Attend Class – OR make sure you watch the video that I post.
Review After Class – This will help you connect the dots, reinforce the concepts you learned or teach you the concepts you
DID not learn.
Study
i) Form study groups (if possible in the current situation)
ii) Question “Am I understanding the material?” “What am I not understanding”
iii) Have I memorized what I am supposed to?
iv) Solve problems without looking at an example.
v) DO NOT just highlight or re-read the material.
vi) Explain your material to your friend or someone else.
Assess
i) Can you solve random ALEKS’ questions?
ii) Can you solve Exam Practice problems?
iii) Can you rework the problems I worked out in class without looking at the answers?
Chapter 13: Chemical Equilibrium
Learning Objectives
By the end of this section, you will be able to:
▪ Describe the nature of equilibrium systems
▪ Explain the dynamic nature of a chemical equilibrium
▪ Derive reaction quotients from chemical equations representing
homogeneous and heterogeneous reactions
▪ Calculate values of reaction quotients and equilibrium constants,
using concentrations and pressures
▪ Describe the ways in which an equilibrium system can be stressed
▪ Predict the response of a stressed equilibrium using Le Châtelier’s
principle
▪ Calculate equilibrium concentrations or pressures and equilibrium
constants, using various algebraic approaches
15-2
Chemical Equilibrium
• The point at which the forward and reverse reaction rates become the same so that
the net composition of the system no longer changes with time
2NO2(g) ⇌
Brown
N2O4(g)
Colorless
• Chemical Equilibrium is the phenomenon observed in reversible reactions.
15-3
Concept of Equilibrium
A two-person juggling act illustrates the dynamic aspect of chemical
equilibria. Each person is throwing and catching clubs at the same rate,
and each holds a (approximately) constant number of clubs.
15-4
The Concept of Chemical Equilibrium
15-5
Chemical Equilibrium in Biological Systems
Transport of carbon dioxide in the body involves several reversible
chemical reactions, including hydrolysis and acid ionization (among
others).
15-6
The Reaction Quotient Qc
What is Q?
Q measures the relative amounts of products and reactants present during a reaction at a particular point in time.
Qc: The ratio of the product concentrations (multiplied together) to the reactant
concentrations (also multiplied together), with each concentrations raised to the power
equal to the coefficients in a balanced chemical equation.
For the reaction, 2 + 3 2 ⇌ 2 3 ( ) =
[ ]
[ ][ ]
15-7
The Reaction Quotient Qp
Qp: The ratio of the partial pressure of products (multiplied together)
to the partial pressure of reactants (also multiplied together), with each
partial pressures raised to the power equal to the coefficients in a
balanced chemical equation.
• For the reaction, 2 + 3 2 ⇌ 2 3 ( )
=
( )
( ) ( )
Knowledge Check 1
• Suppose you run this reaction in a chemistry lab:
CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (g)
Write down the reaction quotient (Qp) for this reaction.
The Reaction Quotients Qc and Qp
The Equilibrium Constant K
15-11
ALEKS (Calculating an equilibrium constant from an
equilibrium composition)
Equilibrium Constant Practice (Q relation to K)
Consider the equation: 2NO2(g) ⇌ N2O4(g)
When 0.10 mol NO2 is added to a 1.0-L flask at
25oC, the concentration changes so that at
equilibrium, [NO2] = 0.016 M and [N2O4] =
0.042 M.
(a) What is the value of the reaction quotient
before any reaction occurs?
(b) What is the value of the equilibrium
constant for the reaction?
Think about these concepts
• What does a large and a small K mean in terms of amount of
reactants and products at equilibrium?
Knowledge Check 2
• For the question in the earlier slide, let us assume the concentration
changes so that at equilibrium, [NO2] = 0.042 M and [N2O4] = 0.016 M
(temperature and the volume of the flask are constant).
(a) What is the value of the reaction quotient before any reaction
occurs?
(b) What is the value of the equilibrium constant for the reaction?
Predicting the direction of reaction
Given below are the starting concentrations of reactants and products
for three experiments involving this reaction:
CO(g) + H2O(g) ⇌ CO2(g) + H2(g)
Kc = 0.64
Determine in which direction the reaction proceeds as it goes to
equilibrium in each of the three experiments shown.
Reactants/Products
Experiment 1
Experiment 2
Experiment 3
[CO]i
0.0203 M
0.011 M
0.0094 M
[H2O]i
0.0203 M
0.0011 M
0.0025 M
[CO2]i
0.0040 M
0.037 M
0.0015 M
[H2]i
0.0040 M
0.046 M
0.0076 M
15-16
Knowledge Check 3
• I worked through Experiment 1 in the earlier slide. Calculate the value
of the reaction quotient and then determine in which direction the
reaction proceeds as it goes to equilibrium in Experiment 3.
Homogeneous and Heterogeneous Equilibria
• Homogeneous equilibrium: An equilibrium in which the reactants and
products of an equilibrium reaction form a single phase, whether gas
or liquid
• C2H2(aq) + 2Br2(aq) ⇌ C2H2Br4(aq) =
• NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH−(aq)
[ 2 2 4 ]
[ 2 2 ][ 2 ]2
=
[ 4+ ][ − ]
[ 3 ]
• Heterogeneous equilibrium: An equilibrium in which the reactants of
an equilibrium reaction, the products, or both are in more than one
phase
• PbCl2(s) ⇌ Pb2+(aq) + 2Cl−(aq)
Kc = [Pb2+][Cl−]2
• TAKE HOME: Pure liquids and pure solids
are not included in equilibrium expressions
15-18
Relation between Kc and Kp
From ideal gas law, =
=

=
For the reaction, aA + bB → cC + dD
=
( ) ( )
( ) ( )
=
( × ) ( × )
( × ) ( × )
= ( )
+ −( + )
= ( )∆
=

×
+
+
= ( )∆
15-19
ALEKS (Interconverting Kp and Kc)
Knowledge Check 4
The two common chlorides of Phosphorus, PCl3 and PCl5, both important in the production of other phosphorous
compounds, coexist in equilibrium through:
PCl3 (g) + Cl2 (g) ⇌ PCl5 (g)
At 250 ∘C, an equilibrium mixture in a 2.50 L flask contains 0.105 g PCl5, 0.220 g PCl3, and 2.12 g Cl2. What are the
values of (a) Kc and (b) Kp for this reaction?
Learning Objectives
• 13.3 Shifting Equilibria: Le Châtelier’s Principle
• Describe the ways in which an equilibrium system can be stressed
• Predict the response of a stressed equilibrium using Le Chatelier’s principle
Shifting Equilibria: Le Châtelier’s Principle
Le Châtelier’s Principle: When a chemical system at equilibrium is
disturbed, it returns to equilibrium BY counteracting the disturbance.
• At equilibrium Q = K.
• The disturbance causes a change in Q.
• The reaction will shift to re-establish Q = K.
Conditions that affect the Equilibrium (Le Châtelier’s Principle)
1)
2)
3)
4)
Adding or Removing Reactant or product
Temperature
Catalysts
Pressure
1st factor: Adding or Removing a Reactant or Product
• If a chemical equilibrium is disturbed by adding a reactant or product,
the system will proceed in the direction that consumes part of the
added species.
• If a chemical equilibrium is disturbed by removing a reactant or
product, the system will proceed in the direction that restores part of
the removed species.
• The system responds in the way that restores equilibrium and
therefore allows Q = K again.
Adding or Removing a Pure Liquid or Solid
• Adding or removing a pure liquid or solid has no effect on the system
unless all of the liquid or solid is removed.
• This is because pure liquids and solids do not appear in the equilibrium
expression.
ALEKS (Le Chatelier’s Principle)
2nd factor: Changes in Temperature
• If you increase the temperature, you increase the heat. The equilibrium
responds in a fashion that consumes the added heat.
Temperature and the Equilibrium Constant
• Changing the temperature of a system results in a change in the value of
the equilibrium constant.
• If the forward reaction is exothermic, then K decreases as T increases.
• If the forward reaction is endothermic, then K increases as T increases.
3rd factor: Catalysts
• A catalyst speeds up the rate of a reaction.
• For reversible reactions, catalysts increase the rates of the forward and
reverse reactions.
• Result: A catalyst causes the system to reach equilibrium more quickly.
• But a catalyst does not affect the equilibrium concentrations or value of
the equilibrium constant.
Figure 13.8
Reaction diagrams for an elementary process in the absence (red) and presence (blue) of a catalyst.
The presence of catalyst lowers the activation energies of both the forward and reverse reactions but
does not affect the value of the equilibrium constant.
Figure 13.9
The figure shows a typical industrial setup for the commercial production of ammonia by the
Haber-Bosch process. The process operates under conditions that stress the chemical
equilibrium to favor product formation.
4th factor: Effect of Change in Pressure on Equilibrium
2NO (g) + O2 (g)
2NO2 (g)
If you increase the pressure (lower the volume),
the position of equilibrium will move in such a way
as to decrease the pressure again – if that is
possible.
Increasing pressure & Decreasing
pressure
It can do this by favoring the reaction which
produces the fewer molecules.
If there are the same number of molecules on
each side of the equation, then a change of
pressure makes no difference to the position of
equilibrium.
Using Le Châtelier’s Principle
• Write the expression for the equilibrium constant (Kc) for the reversible reaction:
2H2(g)+CO(g)⇌CH3OH(g)
ΔH=−90.2kJ
• What will happen to the concentrations of H2, CO, and CH3OH at equilibrium if more
H2 is added?
• What will happen to the concentrations of H2, CO, and CH3OH at equilibrium if the
temperature of the system is increased?
• What will happen to the concentrations of H2, CO, and CH3OH at equilibrium if more
catalyst is added?
15-34
Knowledge Check 5
For the reaction in the previous slide,
Condition 1: What will happen to the concentrations of H2, CO, and CH3OH at
equilibrium if CO is removed?
Condition 2: What will happen to the concentrations of H2, CO, and CH3OH at
equilibrium if the temperature of the system is decreased?
Solving Equilibrium Problems
• Two fundamental kinds of equilibrium problems:
Type A problem: The concentrations of the reactants and the products at
equilibrium are given
• Calculate the equilibrium constant for the reaction
Type B problem: The equilibrium constant and the initial concentrations of
reactants are given
• Calculate the concentration of one or more substances at equilibrium
15-36
Some practice problem for you with answer key
Type A: What is the value of the equilibrium constant at 500°C for the formation of NH3 according to the
following equation?
N2(g)+3H2(g)⇌2NH3(g)
An equilibrium mixture of NH3(g), H2(g), and N2(g) at 500 °C was found to contain 1.35 M H2, 1.15 M N2, and
4.12 × 10−1 M NH3.
(Ans: Kc = 6.00 × 10−2)
Type A: A 0.72-mol sample of PCl5 is put into a 1.00-L vessel and heated. At equilibrium, the vessel contains
0.40 mol of PCl3(g) and 0.40 mol of Cl2(g). Calculate the value of the equilibrium constant for the
decomposition of PCl5 to PCl3 and Cl2 at this temperature.
(Ans: 0.50)
Type B: A 1.00 mol sample of NOCl was placed in a 2.00 L reactor and heated to 227°C until the system
reached equilibrium. The contents of the reactor were then analyzed and found to contain 0.056 mol of Cl2.
Calculate Kc at this temperature.
2NOCl(g) ⇌ 2NO(g) + Cl2(g)
(Ans: 4.5 × 10-4)
15-37
Knowledge Check 6
• Type B problem:
A closed system initially containing 1.000 X 10-3 M H2 and 2.000 X10-3 M I2 at 100 oC
is allowed to reach equilibrium, and at equilibrium the HI concentration is 1.87 X
10-3 M. Calculate Kc at 100 0C for the reaction taking place.
H2 (g) + I2 (g)
2 HI (g)
ALEKS (Type B problem with a little twist)
Chapter 14: Acid –Base Equilibria
Learning Objectives
By the end of this section, you will be able to:
▪ Explain the theories used to define acids and bases
▪ Identify conjugate acid-base pairs from a given reaction
▪ Calculate acid ionization constants and pH of solutions
▪ Establish relationship between Ka, pKa and pH of solutions
▪ Compare the acidity based on bonding and structure
▪ Perform titration calculations using initial data
Definitions of Acids and Bases
Sinkholes are the result of reactions between acidic
groundwaters and basic rock formations, like limestone.
Corrosion of statues that are
carved from marble or limestone
Conjugate Acid–Base Pair
+][F−]
[H
O
HF(aq) + H2O(l) ⇌ H3O+(aq) + F-(aq) = 3
[HF]
−][NH +]
[OH
4
NH3(aq) + H2O(l) ⇌ OH-(aq) + NH4+(aq) =
[NH3]
What does the value of Ka
and Kb tell you about acid
and base strength?
Knowledge Check 7
Here are some Bronsted-Lowry acids. What are the conjugate bases of each acids?
H2SO4, NH4+ (for this look up the earlier slide), HCl
Knowledge Check 8
What are the conjugate acids of these bases?
NH3, OH-, HSO4-
Strength of acid and bases
pKa= -log Ka
= −
pKb= -log Kb
−
=
Autoionization of Water
Amphiprotic: Substances that can behave as either an acid or a base in a
chemical reaction
H2O(l) + H2O(l) ⇌ H3O+(aq) + OH-(aq)
= H3O+ OH−
Ionic Product of water (25°C): Kw = [H3O+][OH− ]= 1.0 × 10 − 14
Ka × Kb = Kw
Kb for NO2− is 2.17 × 10−11. Calculate the Ka for its conjugate acid.
Measuring the pH
• pH meters
• Indicators:
pH papers
Indicator solutions
Phenolphthalein
The Relationship among pH, pOH, and pKw
• The pH scale is a concise way of describing the H3O+ concentration and hence
the acidity or basicity of a solution
• pH = -log [H3O]+
pOH = -log[OH]• [H3O]+ = ?
[OH]- = ?
Kw = H3O+ OH− = 1.0 × 10 − 14
pKw = pH + pOH = 14
For Pure water at 25°C, [H3O]+ = [OH]- = 1.0 ×10-7M
Question 1: The Kw for water at 100°C is 4.99 × 10−13. Calculate pKw for water at this
temperature
Question 2: Calculate the pH and pOH of 0.15M HBr solution.
Knowledge Check 9
An aqueous solution at 25°C has a pOH of 6.67. Calculate
the pH. Round your answer to 2 decimal places.
ALEKS (Calculating the pH of a strong basic
solution)
A chemist dissolves 185. mg of pure NaOH in enough water to make up
140 mL of solution. Calculate the pH of the solution at 25oC.
Knowledge Check 10
In the ALEKS problem I solved in the earlier slide – What is the
concentration of the hydronium ion in the solution?
Hint: pH = -log (H30+)
ALEKS (Ionization of weak acids and weak bases)
Unlike strong acids and bases, weak acids and bases do not ionize completely in
aqueous solutions.
E.g. Acetic acid (CH3CO2H) ionizes to a small extent in water according to the equation:
CH3CO2H(aq)+H2O(l)⇌H3O+(aq)+CH3CO2−(aq)
[H3 O+ ][CH3 CO2 ]−
=
[CH3CO2H]
[H3 O+ ]
% =
[CH3CO2H]
Calculate the percent ionization of a 0.125-M solution of nitrous acid (a weak acid),
with a pH of 2.09.
(Ans: 6.5%)
ALEKS (Determination of Ka or Kb from the pH of a Weak Acid)
The pH of a 0.0516-M solution of nitrous acid, HNO2, is 2.34. What is its Ka?
HNO2(aq) + H2O(l) ⇌ H3O+(aq) + NO2−(aq)
Ans: 4.50 × 10-4
ALEKS (Determination of pH from Ka of Weak Acid)
Formic acid, HCO2H, is the irritant that
causes the body’s reaction to ant stings.
What is the concentration of hydronium ion and the pH in a 0.534-M solution of
formic acid? (opposite of what we did in the earlier slide)
HCO2H(aq) + H2O(l) ⇌ H3O+(aq) + HCO2−(aq)
Ka = 1.8×10−4
Ans: [H3O+] = 9.8 × 10−3M, pH = 2.01
Polyprotic Acids and Bases
• Polyprotic acids such as H2SO4, H3PO4, and H2CO3 contain more than one
ionizable proton
• The protons are lost in a stepwise manner
• The fully protonated species is always the strongest acid
• It is easier to remove a proton from a neutral molecule than from a negatively charged ion
• H2SO4 is more acidic than HSO4-
16-55
Molecular Structure and Acid–Base Strength
• Bond Strengths: The stronger the A–H or B–H+ bond
• The less likely the bond is to break to form H+ ions and thus the less acidic the substance
• The trend in bond energies is due to a steady decrease in overlap between the
orbital of hydrogen and the valence orbital of the halogen atom as the size of
the halogen increases
16-56
Acid Base Titrations
An acid–base titration is a method of quantitative analysis for determining the
concentration of an acid or base by exactly neutralizing it with a standard solution
of base or acid having known concentration.
End Point
• The point at which the
indicator changes its color
during titration
Equivalence Point
• The point in a titration where
a stoichiometric amount of
the titrant has been added
Properties of a good indicator:
• The color change must be
easily detected
• The color change must be
rapid
• The indicator molecule
must not react with the
substance being titrated
• The indicator must change
color within a specific
narrow pH range.
Figure: The Titration of (a) a Strong Acid with a Strong
Base and (b) a Strong Base with a Strong Acid
For the titration of a monoprotic
strong acid (HCl) with a monobasic
strong base (NaOH),
How much 0.20 M NaOH is required to completely neutralize 50.0 mL of a 0.10 M
solution of HCl?
Ans: 25 mL
16-58
Figure: The Titration of (a) a Weak Acid with a Strong Base and (b) a
Weak Base with a Strong Acid
ALEKS (Calculating the pH of a weak acid titrated with a strong base)
In a titration experiment, 5.00 mL of 0.200 M NaOH was added to 50.00 mL of 0.100 M CH3CO2H,
calculate the pH of the resulting solution. Ka for acetic acid is 1.74 × 10−5.
Step 1: Write the equation CH3CO2H(aq) + OH−(aq) ⇌ CH3CO2−(aq) + H2O(l)
Step 2: Using ICE-table, calculate the moles and finally the concentration of reactants and
products in the mixture after the addition is complete.
initial
change
final
[CH3CO2H]
5.00 mmol
−1.00 mmol
4.00 mmol
[OH−]
[CH3CO2−]
1.00 mmol 0 mmol
−1.00 mmol +1.00 mmol
0 mmol
1.00 mmol
Step 3: Use Ka for acetic acid to determine [H]+ ion concentration, and then pH.
Finally, pH = −log[H+]
= −log(6.95 × 10−5)
= 4.158
Buffers
• Solutions that maintain a relatively constant pH when an acid or a base is added.
• Buffers contain either a weak acid (HA) and its conjugate base (A−) OR a weak base (B)
and its conjugate acid (BH+).
• Buffer Capacity: The amount of strong acid or strong base that a buffer solution can
absorb before the pH changes
Example of buffers
• Example: Acetate buffer (a solution of 1.0 mole of acetic acid and 1.0 mole of
sodium acetate in water)
CH3COOH + H2O ⇌ CH3COO– + H3O+
Case 1: If we add a strong acid, such as HCl, the added H3O+ ions react with
acetate ions:
CH3COOH + H2O ← CH3COO– + H3O+
Case 2: If we add a strong base, such as NaOH, the added OH– ions react with
acetic acid:
CH3COOH + OH- → CH3COO– + H2O
Henderson-Hasselbalch equation
Used to estimate the pH of the buffer solution.
H = +
[ ]
log
[ ]
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