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Equilibrium. Equilibrium. The state where the concentrations of all reactants and products remain constant with time. Reactions are reversible. A + B C + D ( forward) C + D A + B (reverse) Forward and Reverse Rxns can be shown by double arrow. A + B C + D.

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equilibrium1
Equilibrium
  • The state where the concentrations of all reactants and products remain constant with time.
reactions are reversible
Reactions are reversible
  • A + B C + D ( forward)
  • C + D A + B (reverse)
  • Forward and Reverse Rxns can be shown by double arrow

A + B C + D

a b c d
A + B C + D
  • 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
    • So concentrations of the reactants and products no longer change with time
slide5

Forward Reaction

Reaction Rate

Equilibrium

Reverse reaction

Time

static or dynamic
Static or Dynamic?
  • At equilibrium, forward and reverse reaction rates are equal
  • May seem like no changes are occurring but there are changes
  • No NET changes
  • On the molecular level, there is frantic activity. Equilibrium is not static, but is a highly dynamic situation.
  • chemical reactions take place ,but concentrations of reactants and products remain unchanged
analogies and metaphors to think about
Analogies and Metaphors to think about
  • In a football game, the number of players on the field is constant although exchange of players (substitution) changes actual persons.
  • Connected fish bowl analogy . Two fish tanks are connected by a tube large enough to allow passage of fish. A number of fish are placed in one of the tanks. At equilibrium, the number of fish in each tank will eventually become unchanged.
  • Two jugglers analogy.
  • Drinking fountain line:
    • Ten students waiting in line to get a drink of water on a hot day. As each gets a drink, the same student reenters the line (equilibrium in a closed system).
    • (b) Same situation as "a," except as each student gets a drink and leaves, a new student enters the line (steady-state in an open system).
  • Picture a number of horses and wranglers in a corral. As each wrangler mounts a horse, the wrangler is bucked off. The equilibrium is:

Horse + Wrangler Mounted wrangler

molecular simulation
Molecular Simulation
  • In this simulation two gaseous reactants collide to produce a more dense solid.

A + B C

gaseous R dense P

  • http://www.absorblearning.com/media/attachment.action?quick=w8&att=2310
homo vs hetero
Homo vs Hetero
  • Homogeneous Equilibria all reacting species are in the same phase
    • gas phase
      • equilibrium constant can be expressed in terms of pressure or concentration, Kp or Kc
    • Solution (aqueous) phase
      • concentration term for the pure liquid does not appear in the expression for the equilibrium constant but aqueous substance concentrations do appear
  • Heterogeneous Equilibria  all reacting species are not in the same phase
      • concentration term for solid or liquid does not appear in the expression for the equilibrium constant
equilibrium summarized
Equilibrium Summarized
  • Forward and Reverse rates are equal
  • Concentrations are not.
  • Rates are determined by concentrations and activation energy.
  • Molecular Motion is frantic and constantly changing
  • Macroscopically no net change is occurring (We can’t observe any changes)
distinguishing between physical and chemical equilibrium
Distinguishing between Physical and Chemical Equilibrium
  • As with physical and chemical changes, physical and chemical equilibrium follow the same rules:
    • Physical  no changes to the chemical properties of the substances involved
      • Ex. equilibrium of water vapor with liquid water in a partly filled sealed bottle
    • Chemical  involve changes in the chemical composition of substances. Bond breaking and bond formation is involved.
      • Ex. dissociation of acetic acid water into acetate and hydronium ion
activity
Activity
  • Model Dynamic Equilibrium with Coins
  • Now lets plot the data using excel
law of mass action

c

c

Law of Mass Action
  • For a reaction: aA + bB ⇄ cC + dD
    • equilibrium constant: K
    • Pure liquids and pure solids have concentrations of 1.
playing with k

c

c

Playing with K
  • If we write the reaction in reverse.

cC + dD ⇄ aA + bB

  • Then the new equilibrium constant is
they are simply the inverse of one another
They are simply the inverse of one another.
  • Forward Reaction

aA + bB ⇄ cC + dD

So we call this K1

And K1= 1 = K2-1

K2

  • Reverse Reaction

cC + dD ⇄ aA + bB

So we call this K2

And K2= 1 = K1-1

K1

the units for k
The units for K
  • Are determined by the various powers and units of concentrations.
  • They depend on the reaction.
  • will always have the same value at a certain
  • temperature (Why will the T effect it?)
  • no matter what amounts are initially added
  • ratio at equilibrium will always be same
slide19
K
  • Has no units
  • Is constant at any given temperature.
  • Is affected by temperature.
  • Equilibrium constants are reaction, phase, temperature and pressure dependent
  • There is a K for each temperature.
  • Equilibrium constant values are thus established for a specific reaction in a specific system and will be unchanging (constant) in that system, providing the temperatures does not change.
what does the size of my k mean
What does the size of my K mean?
  • Large K > 1
    • products are "favored“
    • Ex. 1 x 1034
  • K = 1
    • neither reactants nor products are favored
  • Small K < 1
    • reactants are "favored“
    • Ex. 4 x 10-41
now let s calculate your k
Now Let’s Calculate Your K
  • Using your data from the simulation calculate K
different equilibrium constants all of these are known as k eq
Different Equilibrium Constants (All of these are known as Keq)
  • Kcis the most used general form with molar concentrations.
  • Kpcan be used with partial pressures when working with a gas phase reaction.
  • Kais used for the dissociation of weak acids in water.
  • Kbis used for the dissociation of weak bases in water.
  • Kwis the equilibrium expression for the dissociation of water into its ions.
  • Kspis used for the dissociation into ions of sparingly soluble solids in water.
practice writing the equilibrium expression
Practice Writing the Equilibrium Expression

4NH3(g) + 7O2(g) 4NO2(g) + 6H2O(g)

First write the equilibrium expression using no concentration values.

What is the value for K if the concentrations are as follows

NH3 1.0 M

O2 1.0 M

NO2 1.4 M

H2O 1.8 M

equilibrium with gases
Equilibrium with Gases
  • Equilibria involving only gases can be described using pressures or concentrations
  • If using pressures,
    • use pA not [A]
    • KP not KC
    • be sure all pressure are in the same units

N2(g) + 3H2(g) 2NH3(g)

calculating k c from k p
Calculating Kc from Kp
  • where
    • Δn is the difference in moles of gas on either side of the equation (np – nr)
    • R is the gas law constant: 0.08206
    • T is Kelvin temperature

For: N2(g) + 3H2(g) 2NH3(g)

    • Δ n = (2) – (1+3) = -2
    • h
practice
Practice
  • Setup the expression for KP in terms of KC, R and T

2NO(g) + Cl2(g) 2NOCl(g)

what the equilibrium constant tells us
What the equilibrium constant tells us…
  • if we know the value of K, we can predict:
    • tendency of a reaction to occur
    • if a set of concentrations could be at equilibrium
    • equilibrium position, given initial concentrations
  • If you start a reaction with only reactants:
    • concentration of reactants will decrease by a certain amount
    • concentration of products will increase by a same amount
using this we can make ice charts
Using this we can make ICE charts

The following reaction has a K of 16. You are starting reaction with 9 O3 molecules and 12 CO molecules.

Find the amount of each species at equilibrium.

slide33

Practice 1 ICE Charts

  • Consider the following reaction at 600ºC
  • 2SO2(g) + O2(g) 2SO3(g)
  • In a certain experiment 2.00 mol of SO2, 1.50 mol of O2 and 3.00 mol of SO3 were placed in a 1.00 L flask. At equilibrium 3.50 mol were found to be present. Calculate
  • The equilibrium concentrations of O2 and SO2, K and KP
slide34

Practice Problem 2 ICE Charts

  • Consider the same reaction at 600ºC
  • In a different experiment .500 mol SO2 and .350 mol SO3 were placed in a 1.000 L container. When the system reaches equilibrium 0.045 mol of O2 are present.
  • Calculate the final concentrations of SO2 and SO3 and K
the reaction quotient q
The Reaction Quotient (Q)
  • 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
what q tells us
What Q tells us

IF THEN

Q = K reaction is at equilibrium

Q > K too much products,

left shift

Q < K too much reactants,

right shift

example 1 reaction quotient
Example 1 Reaction Quotient

For the synthesis of ammonia at 500°C, the equilibrium constant is 6.0 x 10-2. Predict the direction the system will shift to reach equilibrium in the following case:

example 2
Example 2

In the gas phase, dinitrogen tetroxide

decomposes to gaseous nitrogen dioxide:

Consider an experiment in which gaseous N2O4 was placed in a flask and allowed to reach equilibrium at a T where KP = 0.133. At equilibrium, the pressure of N2O4 was found to

be 2.71 atm.

Calculate the

equilibrium

pressure of NO2.

example 3
Example 3

At a certain temperature a 1.00 L flask initially contained 0.298 mol PCl3(g) and 8.70x10-3 mol PCl5(g). After the system had reached equilibrium, 2.00x10-3 mol Cl2(g) was found in the flask.

PCl5(g) PCl3(g) + Cl2(g)

Calculate the equilibrium concentrations of all the species and the value of K.

approximations
Approximations
  • If K is very small, we can assume that the change (x) is going to be negligible compared to the initial concentration of the substances
  • can be used to cancel out when adding or subtracting from a “normal” sized number to simplify algebra
example 4
Example 4

At 35°C, K=1.6x10-5 for the reaction 2NOCl(g) ⇄ 2NO(g) + Cl2(g)

Calculate the concentration of all species at equilibrium for the following mixtures

2.0 mol NOCl in 2.0 L flask

1.0 mol NOCl and 1.0 mol NO in 1.0 L flask

2.0 mol NO and 1.0 mol Cl2 in 1.0 L flask

le chatelier s principle
Le Chatelier’s Principle
  • can predict how certain changes or stresses put on a reaction will affect the position of equilibrium
  • helps us determine which direction the reaction will progress in to achieve equilibrium again
  • system will shift away from the added component or towards a removed component
change concentration
Change Concentration
  • equilibrium position can change but not K
  • system will shift away from the added component or towards a removed component

Ex: N2 + 3H2 2NH3

  • if more N2 is added, then equilibrium position shifts to right (creates more products)
  • if some NH3 is removed, then equilibrium position shifts to right (creates more products)
adding gas
Adding Gas
  • adding or removing gaseous reactant or product is same as changing concentration
  • adding inert or uninvolved gas
  • increase the total pressure
  • doesn’t effect the equilibrium position
change the pressure by changing the volume
Change the Pressure by changing the Volume
  • only important in gaseous reactions
  • decrease V
    • requires a decrease in # gas molecules
    • shifts towards the side of the reaction with less gas molecules
  • increase V
    • requires an increase in # of gas molecules
    • shifts towards the side of the reaction with more gas molecules
change in temperature
Change in Temperature
  • all other changes alter the concentrations at equilibrium but don’t actually change value of K
  • value of K does change with temperature
  • if energy is added, the reaction will shift in direction that consumes energy
  • treat energy as a
    • reactant: for endothermic reactions
    • product: for exothermic reactions
catalyst
Catalyst
  • The use of a catalyst may speed up a reaction, but it speeds it up in both the forward and reverse direction, therefore a catalyst has no effect on the equilibrium state of the system.
slide55
energy + N2(g) + O2(g) ⇄ 2NO(g)
  • endo or exo?
    • endothermic
  • increase temp
    • to right
  • remove O2
    • to left
  • increase volume
    • no shift
  • add N2
    • to right
le chatlier s simulation
Le Chatlier’s Simulation
  • http://www.learnerstv.com/animation/animation.php?ani=120&cat=chemistry
another dynamic equilibrium
Another Dynamic Equilibrium
  • Equilibrium occurs when the solution is saturated
slide59
Ksp
  • The solubility product constant (Ksp) is similar to the Keq.
  • When a solid is added to water, some dissolves (and splits into ions) while some remains a solid.

Ex: NaCl(s) Na+(aq) + Cl-(aq)

  • The mass action expression for this is:

Ksp = [Na+][Cl-]

(remember… solids are 1)

common ion effect
Common Ion Effect
  • The solubility of a solid is lowered if the solution already contains ions common to the solid
    • Dissolving silver chloride in a solution containing silver ions
    • Dissolving silver chloride in a solution containing chloride ions
precipitation
Precipitation
  • Opposite of dissolution
  • Can predict whether precipitation or dissolution will occur
  • Use Q: ion product
    • Equals Ksp expression but doesn’t have to be at equilibrium
    • Q > K: more reactant will form, precipitation until equilibrium reached
    • Q < K: more product will form, dissolution until equilibrium reached
qualitative analysis
Qualitative Analysis

Process used to separate a solution containing different ions using solubilities

Example Problem

A solution of 1.0x10-4 M Cu+ and 2.0x10-3 M Pb2+. If I- is gradually added, which will precipitate out first, CuI or PbI2?

example
Example
  • The molar solubility for MgCl2 is 0.0056M. Calculate Ksp
salt simulator
Salt Simulator
  • In groups of two use the simulator to discover the Ksp and Le Chatlier’s Principle
solutions of acids or bases containing a common ion
Solutions of Acids or Bases Containing a Common Ion
  • Common Ion
    • Ion provided in solution by an aqueous acid (or base) as well as a salt

a. HF(aq) and NaF (F- in common)

b. HF(aq) H+(aq) + F-(aq) Excess F- added by NaF

    • Equilibrium shifts away from added component. Fewer H+ ions present.
    • pH is higher than expected.

a. NH4OH and NH4Cl (NH4+ in common)

b. NH3(aq) + H2O(l) NH4 +(aq) + OH-(aq)

    • Equilibrium shifts to the left. pH of the solution
    • decreases due to a decrease in OH- concentration
equilibrium calculations
Equilibrium Calculations
  • Consider initial concentration of ion from salt when calculating values for H+ and OH-
slide75
Acid + Base  Salt + Water
  • Acid + Base Conjugate Base + Conjugate Acid
  • Some solvents are amphiprotic
    • Water can act as an acid and a base!
    • Methanol can act as an acid and a base!
  • Autoprotolysis
    • Some solvents can react with themselves to produce an acid and a base
      • Water is a classical example
  • Weak acids dissociate partially, weak bases undergo partial hydrolysis. Strong acids and bases are strong electrolytes.
kw dissociation of water
Kw (Dissociation of Water)
  • Water is amphiprotic it also undergoes autoprotolysis
  • Kw = 1.0E-14 at about 25 ˚C
  • This is where the pH scale we commonly use originates from!
  • What is the concentration of hydronium and hydroxide ions in neutral solution? What is the pH? What is the pOH?
weak acid weak base equilibria
Weak Acid & Weak Base Equilibria
  • Weak acids produce weak conjugate bases, and weak bases produce weak conjugate acids
  • Ka is a “special” equilibrium constant for the dissociation of a weak acid (found in standard tables)
  • Kb is a “special” equilibrium constant for the hydrolysis (or dissociation” of a weak base.
calculations
Calculations……..
  • What is the pH of a 1.0 M solution of acetic acid (HAc)?
  • What assumption can you make?
    • If [acid] is about 1000 times the Ka value, it’s concentration in solution won’t change much!
    • Use an “I-C-E” table to look at this.
    • The text goes into a more elaborate discussion of approximations. I will allow approximations if the concentrations or pH values do not change in the hundred’s decimal place.
slide83
What is the pH of a 4.0 M solution of phosphate ion?
    • Write reaction
    • Calculate Kb
    • Setup “I-C-E” table
    • Make assumptions
    • Solve algebraically.
buffers
Buffers
  • Buffers resist the change in pH because they have acid to neutralize bases and bases to neutralize acids.
  • Made from a weak acid (HA) and the salt of its conjugate base (A-, where the counter ion is gone for example), or a weak base and the salt of its conjugate acid.
features of buffers
Features of Buffers
  • Buffers work best at maintaining pH near the Ka of the acid component, usually about +/- 1 pH unit. This is their buffer capacity .
  • Buffers resist pH changes due to dilution.
  • All seen when we use the “Buffer Equation”
henderson hasselbalch buffer equation
Henderson-Hasselbalch (Buffer) Equation
  • A modification of the equation for the dissociation of a weak acid.
  • The pH is the pH of the buffered solution, pKa is the pKa of the weak acid.
  • What is the pH of a buffer solution made from 1.0 M acetic acid and 0.9 M sodium acetate?
  • You add .10 moles of sodium hydroxide to the above solution? What is the new pH?
h h equation buffers
H-H Equation & Buffers….
  • If [A-] = [HA] pH = pKa!
    • This is what we see at half-way to the equivalence point in the titration of a weak acid with a strong base!
  • Dilution does not change the ratio of A- to HA, and thus the pH does not change significantly in most cases
how do you prepare buffers
How do you prepare buffers?
  • Select a buffer ‘system’ based on the pH you want to maintain.
  • Use the H-H equation to calculate how much acid and conjugate base you need
  • Take one of three approaches:
    • Mix acid and base forms, measure pH and adjust with strong acid or strong base
    • Start with a solution of weak acid, and add strong base until you reach the desired pH
    • Start with a solution weak base and add strong acid until you reach the desired pH
    • REMEMBER, STRONG Acids or Bases react completely!
  • Dilute as necessary, adjust pH further if needed with strong acid or base.
slide90
You want 1L of a buffer system that has a pH of 3.90?
    • What acid/conjugate base pair would you use?
    • How would you go about figuring out how much of each reagent you might need?
    • How would you prepare and adjust the pH of this solution?