Chemistry 17 General Chemistry II
Contributor(s): Allison Arabelo and Tanya Olegario Edited by: Tanya Olegario and Christian Loer Llemit Last Updated: 1617A
Table of Contents First Long Exam I.
Thermochemistry
1
II.
Spontaneous Ch Change
4
III.
Chemical Ki Kinetics
8
Second Long Exam I.
Chemical Equilibrium
13
II.
Acids and Bases
18
III.
Solubility
31
IV.
Experiment 5 Post La Lab
33
V.
Experiment 6 Post Lab
34
VI.
Experiment 7 Post La Lab
35
VII. VII. Expe Experrimen imentt 8 Pos Postt Lab Lab
36
Third Long Exam I.
Complex Ions and Coordination Compounds
39
II. II.
Bond Bo ndin ing g in in Com Compl plex ex Ions Ions:: Cry Cryst stal al Fiel Field d The Theor ory y
41
III.
Electrochemistry
43
IV.
Experiment 9 Post La Lab
48
V.
Chemical Change and Electric Work
49
First Long Exam Thermochemistry System and Surroundings System o Part of the universe being studied Types 1. Open system – system that freely exchanges energy exchanges energy and matter with with its surroundings 2. Closed system – system that exchanges only energy only energy and not matter with with its surroundings The bomb calorimeter is an example of a closed system 3. Isolated system – does not exchange energy or matter with its surroundings o Surroundings – the rest of the universe that is not part of the system Energy, U o The capacity to do work Energy can change from potential to kinetic energy and vice versa o o Kinetic energy, e K the energy of motion
= 12
Where;
= kinetic energy in Joules (J) = mass of moving object in kilograms (kg) = velocity of moving object in meters per second (m/s)
Potential energy, V Stored energy that has the potential to do work Energy due to condition, position or composition Associated with forces of attraction or repulsion between objects Heat and Heat Capacity o Thermal Energy Kinetic energy associated with random mo lecular motion In general, proportional to temperature An intensive property Physical property independent of the amount of material in the system o Heat (q) Heat flows from high to low temperatures which could lead to: Temperature change Phase change process (no change in temperature) An isothermal process (no Units Joule (J) – SI unit for heat Calorie (cal) – Heat required to change the temperature of one gram of water by one degree Celsius Conversion: 1 cal = 4.184 J Heat Capacity o Quantity of heat required to change the temperature of the system by one degree Mass of the system multiplied by the specific heat capacity, c. Molar Heat Capacity If the system is one mole of substance Specific Heat Capacity, c If the system is one gram of substance o
1
Heats of Reaction and Calorimetry Heat of the Reaction, q rxn o Amount of heat exchanged between a system and its surroundings when a chemical reaction occurs within a system at constant temperature Reactions which cause energy differences may be classified into: o Exothermic Occurs when the temperature of the system increases due to the evolution of heat Heat is released into the surroundings Has an overall negative q rxn Example: combustion Endothermic Occurs when the temperature of an isolated system decreases Heat is absorbed by the system Has an overall positive qrxn Calorimetry o Process of measuring the heat of chemical reactions or physical changes as well as heat capacity. Heat released by by the system must be absorbed by the surroundings and vice versa. Types 1. Bomb Calorimetry Constant volume 2. Coffee Cup Calorimetry Constant pressure Well-insulated and isolated Measures temperature change Pertinent equations are:
= = ∆
Work, w o Force acting through a distance
= × = × ×
Where;
o
is the work in Joules (J) is the force in Newton (N) is the mass in kilograms (kg) is the acceleration in meters per second squared (m/s ) is the displacement in meters (m) 2
Chemical reactions may also do work via pressure-volume via pressure-volume work. Gas formed pushes against the atmosphere causing the volume to change. In this case:
=∆
Where P is is the pressure and
∆ is the change in volume
First Law of Thermodynamics o Also known as the Law of Conservation of Energy States that: o Energy can neither be created nor destroyed; rather, it transforms from one form to another The total energy of an isolated system is constant 2
o
An isolated system is unable to exchange either heat or work with its surroundings Mathematically:
= 0
Heats of Reaction: ∆U and ∆H o Internal Energy, U Total energy (potential and kinetic) in the system Includes translational kinetic energy, molecular vibrations, bond vibrations, intermolecular attractions, chemical bonds, and energy associated in electrons and atoms Note: Note: A system contains only internal internal energy. Q and w are are not contained in a system; they are only energy transfer and occur during a change in the system, such that:
= ∆ = o
o
At constant volume:
∆ = 0 = = At constant pressure: ∆ = Since, = ∆ and ∆ = Therefore,
= ∆ ∆ ∆
=
Sign Conventions Work, w Heat, q
o
and
work is done on the system work is done by the system heat is absorbed by the system heat is released by the system
Enthalpy, H
= ∆ = ∆ ∆ ∆ ∆ At constant pressure and temperature:
∆=∆∆=
∆H in an exothermic reaction is negative while ∆H in an endothermic reaction is positive is positive Enthalpy changes accompanying a change of state Enthalpy of vaporization Liquid to gas Enthalpy of fusion Solid to liquid Standard Enthalpy of Reaction, ∆H ° The enthalpy change of a reaction in which all reactants and their products are in their standard their standard states The standard state refers to the pure element or compound at a pressure of 1 bar (1.01325 atm) and at the temperature of interest Heats of reaction may be computed from standard enthalpies of formation using the formula: 3
Δ° = ΣνΔproducts ΣνΔreactants
Standard Enthalpies of Formation, ∆H f ° The enthalpy change that occurs in the formation of one mole of a substance in the standard state from the reference forms of the elements in their standard states The standard enthalpy of formation of a pure element in its reference state is 0 o Functions of State Any property that has a unique value for a specified state of the system Does not depend on the path but rather on the initial and final state Also called state function Notes: U is a function of state; ∆U has a unique value between two states ∆U is easily measured while U is not o Path-dependent Functions Changes in heat and work are not functions of state Hess’s Law o An indirect determination of ∆H o ∆H is an extensive property It is directly proportional to the amount of substance in the system Example: N2(g) + O2(g) → 2 NO(g) ΔH° = +180.50 kJ
N2(g) + O2(g) → NO(g)
o
ΔH° = +90.25 kJ
∆H changes sign when a process is reversed Example:
NO(g) → N2(g) + O2(g) o
ΔH° = -90.25 kJ
Hess’s Law of Constant Heat Summation If a process occurs in stages or steps (even hypothetically), the enthalpy change for the overall process is the sum of the enthalpy changes for the individual steps Example:
N2(g) + O2(g) → NO(g) + O2(g) NO(g) + O2(g) → NO2(g) N2(g) + O2(g) → NO2(g)
ΔH° = +90.25 kJ ΔH° = -57.07 kJ ΔH° = +33.18 kJ
Spontaneous Change Spontaneity o Spontaneous Occurs without intervention. Some spontaneous processes occur slowly and others rapidly Spontaneous does not mean fast enthalpy change alone cannot predict spontaneity Occurs in the direction in which the: Potential energy decreases For chemical systems, the internal energy U is equivalent to potential energy Enthalpy of the system decreases This is mainly true because mostly exothermic reactions are spontaneous but there are exceptions since some endothermic reactions are spontaneous 4
Nonspontaneous Requires the system to be acted on by an external agent o If a process is spontaneous, the reverse process is non-spontaneous. Entropy o A thermodynamic quantity related to the number of ways the energy of a s ystem can be dispersed among the available microscopic energy l evels. o The Boltzmann Equation for Entropy o
=
Where;
S = states; the microscopic energy levels available in the system W = microstates; the particular way in which particles are distributed amongst the states k = the Boltzmann constant; effectively the gas constant per molecule = R/N The greater the number of arrangements (microstates) of the microscopic particles (atoms, molecules, ions) among the energy levels in a particular state of the system, the greater the entropy of the system Four situations generally produce an increase in entropy: 1. Pure liquids or liquid solutions are formed from solids 2. Gases are formed from either solids or liquids 3. The number of molecules of gas increases as a result of a chemical reaction 4. The temperature of a substance increases For changes occurring at a constant temperature:
o
o
∆ =
Evaluating S and ∆S o Phase Transitions ∆S is given by:
∆ = ∆
o
Exchange of heat can be carried out reversibly Trouton’s Rule
o
Raoult’s Law
o
∆ = ∆ ≈ 87 −− =
Third law of thermodynamics States that he entropy of a pure, perfect crystal at 0K is zero. Standard Molar Entropy, S° Check table of values A function of temperature Used to compute for ∆S using the following formula:
Δ= [ Σνp products Σνr reactants] °
°
The Second Law of Thermodynamics o States that all spontaneous processes occur in the direction of increasing entropy of the universe. ∆Stotal = ∆Suniverse = ∆Ssystem + ∆Ssurroundings > 0 5
o
Gibb’s Energy and Gibb’s Energy Change At constant pressure and temperature, there is only pressure-volume work such that:
= =Δ
Enthalpy change is reversible and the entropy can be calculated under the following conditions: Large surroundings Infinitesimal change in temperature For the universe:
∆ =∆ ∆ =∆ ∆ ∆ = ∆ ∆ For the system: = ∆=∆∆ ∆ =∆
Standard Free Energy, ∆G° The Standard Free Energy of Formation, ∆G f ° o The change in free energy for a reaction in which a substance in its standard state is formed from its elements in reference forms in their standard states. o The Standard Free Energy of Reaction, ∆G°
Δ°= ∆°∆°
Δ = Σνp Δf Σνr Δf °
°
°
Gibb’s Energy Change and Equilibrium Gibb’s energy change is a function of temperature o o Relationship between S and S °
∆= = = = ∆ = ° ° =° 1 =°
o
↔ 2NH = ° = ° = ° ∆ = 2 × (° ) (° ) 3×° ∆ =2° ° 3° ∆ =∆° Example: N2(g) + 3H2(g)
3(g)
Relationship of ∆G° to ∆G for Non-standard Conditions
Δ=ΔΔ °=∆°∆° For ideal gases: ∆= ∆° Δ=Δ°Δ Δ=Δ°lnQ
Example: (same given as previous example) 6
∆ =∆°
Δ=Δ°Δ
Δ=Δ°∆° Δ=Δ° o
Relationship of ∆G to the Equilibrium Constant, K eq
Δ=Δ°lnQ If the reaction is in equilibrium, then: Δ=Δ°lnK = 0 Δ°=lnK
o
Note: Every chemical reaction consists of both a forward and a reverse reaction. The direction of spontaneous change is the direction at which the free energy decreases. Positive ∆G - formation of reactants (reverse reaction) is favored Negative ∆G - formation of products (forward reaction) is favored Summary of criteria for spontaneous change: ∆H
o
o
∆S
∆G
Results Spontaneous at all T Spontaneous at low T Nonspontaneous at high T Nonspontaneous at low T Spontaneous at high T Nonspontaneous at all T
Activities For pure solids and liquids, a = 1 For ideal gases, a = P/Po The reference state is the gas at 1 bar at the temperature of interest For ideal solutes in aqueous solutions, a = c/co The reference state is a 1 M solution Thermodynamic Equilibrium Constant, K eq A dimensionless equilibrium constant expressed in terms of activities Note: Often Keq is equal to Kc or Kp but not always Used to determine ∆G using the following formula”
Δ=Δ°lnK
∆G° and K as Functions of Temperature o Recall:
Δ°=lnK °=∆°∆°
Therefore:
∆° ∆° = ∆° =
o
Plotting lnKeq against 1/T yield a increasing linear graph with slope -∆H°/R Demonstrates the temperature dependence of the equilibrium constant Assume that ∆H° and ∆S° do not vary significantly with temperature. 7
o
Van’t Hoff Equation:
∆° 1 1 = Coupled Reactions A non-spontaneous reaction may occur by: o Changing the conditions of the reaction: Coupling two reactions, one with a positive ∆G and one with a negative ∆G This results to an overall spontaneous process Example: Smelting Copper Ore Chemical Kinetics The Rate of a Chemical Reaction o The rate of change of concentration with time General Rate of Reaction o The rate of appearance of the products and the negative of the rate of disappearance of the reactants Given the reaction: aA + bB gG + hH
1 ∆ = 1 ∆ = 1 ∆ = 1 ∆ = ∆ ∆ ∆ ℎ ∆
Measuring Reaction Rates o In the plot of concentration versus time, the instantaneous rate at time t is given by the slope of the line tangent to the plot at time t. It follows that the initial rate is given by the slope of the line tangent to the plot at time = 0. Another way to determine the instantaneous rate is by evaluating - ∆A/∆t, with a short ∆t interval. o The Rate Law o Given the reaction: aA + bB gG + hH
= [ ][]
Where;
k = rate constant m = order of the reaction with respect to A n = order of the reaction with respect to B
= o
o
Method of Initial Rates Used to determine the order of the reaction and the rate constant, k. Requires determination of the initial rates at different concentrations. The initial concentration of one reactant is varied while the other concentrations are held constant. Effect of doubling the initial concentration of one reactant: Zero order in the reactant – no effect on the initial rate First order in the reactant – the initial rate doubles Second order in the reactant – the initial rate quadruples Third order in the reactant – the initial rate increases eightfold Method of Integrated Rate Laws Used to determine the order of the reaction and the rate constant, k. The data may be substituted into integrated rate laws to find the rate law that gives a consistent value of k The goal is to find the graph that yields a straight line. After determining k , the integrated rate laws may be used to determine the concentration at a given time and vice versa. 8
Zero-Order Reactions k is in mol L-1 s-1 Integrated Rate Law: Plotting [A] against t yields a linear plot (shown below) with slope –k
[] = []
First-Order Reactions k is in s -1 Integrated Rate Law: Plotting ln[A] against t yields a linear plot with slope –k Half-life, t 1/2 The time taken for half of the reactant to be consumed. For a first order reaction:
ln[] =ln[]
0.693 / = 2 = May be used to determine k for a first order reaction Second-Order Reactions k is in L mol -1 s-1 Integrated Rate Law:
1 = 1 [] []
Plotting 1/[A] against t yields a linear plot with slope k Half-life, t 1/2 The time taken for half of the reactant to be consumed. For a second order reaction:
1 / = []
Pseudo-first Order Reactions Used for reactions with more than one reactant Simplify the kinetics of complex reactions by: Using a low concentration for the reactant under study Using high concentrations for the rest of the reactants, making them approximately constant. By doing so, the reaction appears to be a first order reaction, hence the term pseudo first order . 9
Theoretical Models for Chemical Kinetics o Collision Theory Collision Frequency Refers to the number of collisions during the reaction Can be determined using the Kinetic Molecular Theory Only a fraction of this will yield a reaction. Orientation of molecules is important Collisions must have sufficient kinetic energy Activation Energy The minimum energy above the average kinetic energy that molecules must bring to their molecules for a chemical reaction to occur If the activation energy is high: Few molecules will have sufficient kinetic energy Reaction is slower Increasing the temperature will yield: More molecules with sufficient kinetic energy Increased reaction rate o Transition State Theory Discusses the existence of a transition state called the activated complex. Activated Complex Has a very short life and cannot be isolated With structure and molecular weight
Effect of Temperature on Reaction Rates Arrhenius Equation o
=−/ = 1 = 1 1
Reaction Mechanisms o Step-by-step description of a reaction o Elementary Steps Any molecular event that significantly alters a molecule’s energy of geometry or produces a new molecule Reversible Exponents for concentration terms are the same as the stoichiometric factors for the elementary process Intermediates Produced in one elementary step and consumed in another Rate Determining Step Slowest elementary step in a reaction mechanism Molecularity Number of reactant particles in the step Types: 1. Unimolecular – one reactant particle 2. Bimolecular – two reactant particles 10
o
o
o
Reaction mechanism must be consistent with: Stoichiometry for the overall reaction The experimentally determined rate law Types 1. A slow, first step followed by a fast step 2. A fast, reversible, first step followed by a slow step The Steady State Approximation Method used to derive a rate law Based on the assumption that one intermediate in the reaction mechanism is consumed as quickly as it is generated Its concentration remains the same at some stage of the reaction The system is said to have reached a steady state Example: 2 N2O5 4 NO2 + O2 The reaction mechanism is given by:
↔
N2O5 NO2 + NO3 NO3 + NO2 NO + NO2 + O2 NO3 + NO 2NO2
(Let k f and k b be forward and backward rate constants) (rate constant k 2) (rate constant k 3)
Use the steady-state approximation to derive the rate law. In these steps, NO and NO3 are intermediates. You have: production rate of NO = k 2 [NO3] [NO2] consumption rate of NO = k 3 [NO3] [NO] A steady-state approach makes use of the assumption that the rate of production of an intermediate is equal to the rate of its consumption. Thus, we have: k 2 [NO3] [NO2] = k 3 [NO3] [NO] and solving for [NO] gives the result, [NO] = k 2 [NO3] [NO2] / (k 3 [NO3])
(1)
For the other intermediate NO 3, production rate of NO3 = k f [N2O5] consumption rate of NO3 = k 2[NO3] [NO2] + k 3[NO3] [NO] + k b[NO3] [NO2] Applying the steady-state assumption gives: k f [N2O5] = k 2[NO3] [NO2] + k 3[NO3] [NO] + k b[NO3] [NO2] Thus,
] [3] = [] + [[] + []
(2)
Let's review the three equations (steps) in the mechanism: Step one is in equilibrium and thus can not give a rate expression. Step two leads to the production of some products, and the active species NO causes further reaction in step three. This consideration led to a rate expression from step two as:
[] = [ ] [ ]
(3)
Substituting (1) in (2) and then in (3) gives:
[] = [5] =[5] 2
11
where k = k f k 2/(k b + 2 k 2).
Catalysis Alternative reaction pathway of lower energy o o Types 1. Homogenous All the species in the reaction are in s olution 2. Heterogenous The catalyst is in the solid state Reactants from gas or solution phase are adsorbed Active sites on the catalytic surface are important
12
Second Long Exam Chemical Equilibrium Exists between two opposing reactions occurring simultaneously with the same rate Dynamic process Forward rate = reverse rate Physical Equilibria o Vapor Pressure of a liquid o Saturation Rate of dissolution = rate of recrystallization of solute Distribution coefficient, which represents partitioning of a solute between two immiscible solvents Reactants are never completely consumed [Reactants]+[Products]=constant o Equilibrium Constant and Activities cC( ) + dD( ) (1); For the general reaction ( )+ ( ) • • • •
• •
• •
•
= = ; (
) (
)
(
) (
)
⇌
aq
aq
= 1; ∴ = [] At low concentrations (dilute, ideal solutions), h
=
[
]
°
:
•
1 =
=
°
[ ]
[ ]
[ ]
[ ]
•
Reverse the reaction, inverse K Multiply the equation by a number, raise by that number Divide by n, take the nth root Combine two equations, multiply their
•
o • • • •
Large K; most reactants are converted to products Small K; small amounts of reactants are converted to products : equilibrium constant in terms of partial pressures Activity of gases are defined by partial pressure relative to a reference pressure ( °) = °× Temperature dependent For pure solids and liquids, a=1; the reference state is the pure solid or liquid For gases with ideal gas behavior assumed, activity is replaced by the numerical value of the gas pressure in bars The reference state is the gas at 1 bar at the temperature of interest The activity of gas at 0.50 bar pressure is a=0.50 For solutes in aqueous solution with ideal behavior assumed (no in terionic attractions), activity is replaced by the numerical value of the molarity Reference state is a 1M solution Activity of the solute in a 0.25M solution is 0.25 When eq. expression is written in terms of activities, the eq. constant is called the thermodynamic equilibrium constant (K) May be equal to
• • •
•
•
•
13
•
o
o o o
( ) For predicting the direction of the reaction
[ ] [ ]
=[
] [ ]
Δ −2 =
•
System is in equilibrium.
>
•
System goes towards the reactants
<
•
System goes towards products
•
o
Same form as Concentration used in Q are not equilibrium values
= °× is the equilibrium constant in terms of partial pressures
•
o
(
=
•
•
) ; difference in stoichiometric coefficient of gaseous pdts and reactants
= 8.314 × 10
•
o
⋅
Gas phase reactions and reactions in aq. So lutions are homogeneous Occur within a single phase •
o
Equilibrium constant expressions do not contain concentration terms for solid or liquid phases of a single phase (pure solids and liquids) An equilibrium constant expression includes terms only for reactants and products whose concentrations or partial pressures can change during a chemical reaction The concentration of the single component within a pure solid or liquid phase cannot change Activities (a) of pure solids and liquids are set to be = 1. Thus, the effect is the same if not included. K relates to thermodynamic stability A reaction is considered as going to completion in the forward direction if > 10 or not occurring in the forward direction if < 10 •
•
•
o o
•
o
Δ Δ
Recall : •
Δ
−10
< 0 for spontaneous and > 0 for nonspontaneous processes = 0; system is at equilibirium At this point, infinitesimal change will cause a net change to occur If system is left undisturbed, no net change occurs with time
14
10
o o
Δ Δ
•
o
•
o
o
Δ Δ Δ Δ Δ
As the temperature increases, the magnitude of decreases. At low T, the magnitude of exceed the magnitude of and is (+) Nonspontaneous Higher T, the magnitude of exceed the magnitude of and is (-) Spontaneous [Theoretical Standpoint] All chemical reactions reach equilibrium and no chemical reaction goes totally to completion ° ( ) = ° = °+ ln
ΔΔ ΔΔ Δ
At only one T are the reactants in their standard states in equilibrium with products in their standard states, at only one T does °=0 Therefore; it is more convenient to use Equilibrium is described in nonstandard conditions using Q is the reaction quotient
•
Δ Δ
•
o •
Δ Δ Δ
° At equilibrium; Therefore; °=
•
A.
Δ
( )
=0
ln
When ° is small a. The equilibrium mixture lies about midway between the two extremes of pure products or reactants in their standard states
b. The effect of nonstandard conditions can be deduced from the slope of the curve Mixtures with Q>K are to the right of the equilibrium point and undergo spontaneous change in the direction of lower Gibbs energy Eventually comes to equilibrium Mixtures with Q
•
•
•
Δ Δ
Δ
•
15
•
o
Minimum is the equilibrium point of the reaction
° ln
ln
• • • • •
<0
•
°<0
•
•
=0
•
°=0
•
•
Δ Δ
>0
•
°>0
•
•
Δ Δ =
°
°
Δ Δ Δ 1 1 Δ2 1 ⋅−1−1
•
Δ Δ
− +
If ° and ° are constant, it describes straight line with a slope of and a y-intercept of ° /
•
Δ Δ
=
°
•
=
Δ °/
°
Temperatures are in Kelvin are equilibrium constant at respective T ° is the enthalpy of reaction in = 8.3145 ( °& )
⋅−1
Reaction or process is spontaneous in the forward direction for the stated conditions Forward reaction is spontaneous when reactants and products are in their standard states > 1 whatever the initial concentrations or pressures of reactants and products
Reaction is at equilibrium under stated conditions Reaction is at equilibrium when reactants and products are in their standard states = 1; which can only occur at one T
Reaction or process is nonspontaneous in the forward direction under the stated conditions Forward reaction is nonspontaneous when reactants and products are in their standard states < 1 whatever the initial concentrations or pressures of reactants and products
Only when all the reactants and products are in their standard states Otherwise; = °+ ln •
Δ Δ 16
Δ
•
o
When an equilibrium system is subjected to a change in T, P or a [reactants], the system responds by attaining a new equilibrium that partially offsets the impact of the change
o •
2g
•
2 g ⇌
g
c
2
2SO ( ) K = 2.8 × 10 at 1000K For 2SO ( ) + O ( ) If initially in equilibrium, If +1 mol SO into the 10.0 L flask Favors the reverse reaction SO becomes SO & O Q >K For H + I 2HI K = 49 at 450°C Removing HI Goes forward, towards products Removing I Goes backward, towards reactants Adding H Forward Adding HI, remove I Backward Add HI, remove H Need to calculate Q to know equilibrium direction
2
c
2 2 c
2 ⇌c
•
2
•
•
2
2
•
2
•
o •
Adding/removing a gaseous reactant/product Towards the formation of removed species Adding inert gases to a constant volume reaction mixture Increase total pressure but partial pressures of reacting species are unchanged Adding an inert gas to a constant volume equilibrium mixture has no effect on equilibrium condition Change pressure by changing volume of the system Decreasing volume increases pressure Moving a piston downward Increase pressure, decrease volume Shift towards products Lowers the number of molecules Increases product Moving a piston upward Decrease pressure, increase volume Shift towards reactants Greater # of molecules 2SO 2SO ( ); For ( )+O ( ) An increase in external pressure causes a decrease in the reaction vol ume and a shift in equilibrium to the right
•
•
•
•
•
2g
2 g ⇌
g
17
•
Note
When the volume of an equilibrium mixture of gases is reduced, a net change occurs in the direction that produces fewer moles of gas When the volume is increased, a net change occurs in the direction that produces more moles of gas To decrease the volume of the mixture, increase external P. To increase the volume; Lower external Pressure Transfer to a larger container Add inert gas at constant P Volume of the mixture must increase to make room for the gas added • • • •
o • • • • •
Addition of heat; favors reaction in which heat is absorbed (endothermic) Removal of heat; favors reaction in which heat is evolved (exothermic) Increase in T shifts equilibrium in the direction of endothermic reaction Decrease in T shifts equilibrium in the direction of the exothermic reaction Principle effect: Changing the equilibrium constant K
o • • • • •
Speeds both forward and reverse reactions Equilibrium is achieved more rapidly Equilibrium amounts are unchanged Does not change the equilibrium constant An equilibrium condition is independent of the reaction mechanism because equilibrium is a thermodynamic quantity
•
o •
Have H+
o
Have OHStrong vs Weak Electrolytes Strong electrolytes exist as ions •
o
•
18
Weak electrolytes exist as partly ions and molecules Neutralization reaction involves a combination of H + ions and OH- to form H2O + •
o
•
o • • •
+ − → 2 4
, ( ) not proven to exist Solvents aren't recognized Amphiphrotic substances
•
o • • •
Proton Donor (H+) All Arrhenius acids are B-L acids Ex. NH + H O NH + OH H O functions as acid Conjugate pairs: NH (b ) & NH (a ) H O (a )& OH (b ) ( ) Greater , the farther the equilibrium lies in the forward direction H O is the strongest possible acid in water Water has a leveling effect on HCl and HClO (HClO > HCl)
2 2 • •
•
• •
+
→
1 2 2
4+
−
4−+ 1 2
4
4
o
Proton acceptor All Arrhenius bases are B-L bases ( ) Greater , the farther the equilibrium lies in the forward direction Accounts for amphiphrotic substances Can function as acids or bases In an acid-base reaction, the favored direction of the reaction is from the stronger to the weaker member of the conjugate pair The stronger the acid, the weaker the conjugate base • • •
o
•
o
•
•
o o
Does not always have conjugates Forms adducts (B-A); coordinate bond formation Addition compound + Accounts for transition metals and Boron containing compounds • •
o
→
o • • •
Electron pair acceptor Has ionizable Species with vacant orbitals that can accommodate electron pairs Incomplete valence shells HCl is not a Lewis acid
+
•
o • •
Electron pair donor Have lone-pair electrons available for sharing
19
•
o
Water contains low concentration of detectable ions Ions form because amphiphrotic For each H2O acting as an acid, another acts as a base 0 & ions formed Reversible + + Equilibrium is displaced far to the left Assume a=1 for H 2O molecules ][ =[ ] In pure H2O, 0 = •
o o o o
+ − 2 2 ⇌ + − + +− − ≠ ++ −− +− + − + − + − •
o
• •
o
•
is the ionic product of H 2O Equilibrium condition for the self-ionization of water ][ =[ ] = 1.0 × 10 at 25°C [ ][ ] applies to all equilibrium solutions
−14
•
•
= log[ = log[ = log([
o
] ]
][
]) = 14.00
] & [ ] to be very small Expect [ Usually < 1.0 M Use activities for accuracy; pH and pOH are only approximations ]=[ ] Neutral when [ Pure water: = 7.00 (25° ) •
o o
•
o
• •
+
[H O ] from autoionization should be considered ex. pH of 1 × 10 M HCl? Reactions: 2H O H O + OH H C l + H O H O + Cl Concentrations: [H O ] = [OH ] = x [H O ] = [Cl ] = 1.0 × 10 M Satisfy K ; [H O ][OH ] = (x + 1 × 1 0 )x = 1.0 × 10 x = 9.5 × 10 M [H O ] = (9.5 × 10 ) + (1 × 1 0 ) = 1.05 × 10 pH = 6.98
−8
2
• •
• •
•
∴
++ w+ + •
•
⇌ →+ + − 2 − −
−
−8 −8
−
−8 −8 −8
−14 −7
o •
When salts of strong bases and strong acids do not hydrolyze; pH = 7 i.e. NaCl When salts of strong bases and weak acids hydrolyze; pH > 7 i.e. NaCH COO The anion acts as a base When salts of weak bases and strong acids hydrolyze;
•
•
20
pH < 7 i.e. NH Cl The cation acts as an acid When salts of weak bases and weak acids hydrolyze; Cations are acids, anions are bases; whether the solution is acidic or basic depends on the relative values of & for the ions
4
•
•
+ =
o •
• •
log
] contribution due to the self-ionization of water can be ignored unless The [ extremely dilute (resp. bases) Dilution of SA or SB is exothermic Ex. HCl dissolution (SA) is assumed to go into completion Strong Acids
4 2 4
HClO HClO H SO HNO HI HBr HCl
Strong Bases • • • • • • • • •
LiOH NaOH KOH RbOH CsOH Mg(OH) Ca(OH) Sr(OH) Ba(OH)
2 2 2 2
o •
Go into equilibrium rather than completely dissociating Reversible Usually have -COOH Carboxylic acids Any acid/base not in above table The conjugate of weak is weak
• • • •
o •
The stronger the H-X bond, the weaker the acid is Consider bond length and polarity Stronger bonds are characterized by short BL and high bond dissociation energies Lower bond length, higher bond energy Acid strength increases as the heterolytic bond dissociation energy decreases Lower energy required to convert H-X into ions, greater acid & strength When comparing binary acids of elements in the same row, acid strength increases as the polarity of the bond increases When comparing binary acids of elements in the same group, acid strength increases as the length of the bond inc reases HF < HCl < HBr < HI •
+ −
•
21
•
Hydrogenated form of anion oxides , , , Focus on the attraction of electrons from the OH bond toward the central atom Factors Promoting Electron withdrawal from O-H bonds High EN of the central atom Ex. HOCl vs HOBr HOCl is more acidic; Cl is more EN < < < > Large # of terminal oxygen atoms in the acid molecule Ex. H2SO4 & H2SO3 Highly EN terminal O atom tends to withdraw electrons from OH bonds, weakening bonds and increasing acidity H2SO4 has 2 terminal O atoms, therefore H 2SO is stronger Oxidation # changes
4 24 •
•
•
• • • •
•
24 24 4 4 4 4
•
•
•
•
•
Compounds or ions with more resonance structures are more stable Ex. AcOH vs etOH AcOH more acidic because highly EN terminal O atom Has resonance; CH3COO- negative charge is spread out, therefore weaker base, therefore stronger acid • •
Group
Examples
EDG: Strongly Activating EDG: Moderately Activating EDG: Weakly Activating EWG EWG: Strongly deactivating EWG: Moderately deactivating
2 65 2 + 2 ,
,
,
,
,
,
,
,
,
,
,
,
,
•
Fundamental factor affecting the strength of an amine as a base concerns the ability of the lone pair of electrons on N to bind a proton taken from an acid Lone pair electrons cannot bind a proton as strongly, and the base is weaker Ex. BrNH2 is a weaker base than NH 3 Hydrocarbon chains have no electron withdrawing ability If attached to an amine group, pK values are lower than for ammonia due to electron donating ability of CH & CH CH •
•
b
•
22
2
•
o
Ex.
4
Three ionizable H atoms Triprotic acid Ionizes in three steps 1. H PO + H O H O + H PO • •
4 2 ⇌ P+ 2 −4 Ka1 = = 7.1 × 10− P 2. H2 PO− O+ + HPO2− 4 + H2O ⇌HP 4 Ka 2 = = 6.3 × 10−8 P 3. HPO2− 4 + H2O ⇌ +HO+ −+ PO− 4 [H O ][PO4 ] Ka = = 4.2 × 10−1 2− [HPO4 ] 1 is so much larger than +2 & Essentially all the is produced in the first ionization step So little of the 2 4− forming in the first ionization step ionizes any further, such that we can assume [2 4− ] = [ + ] in the solution [42− ] ≈ , regardless of the molarity of the acid ex. 2 4 Strong acid in first ionization and weak in the second Ionization is complete in the first step, means that in most 2 4 solutions, [2 4 ] ≈ 0 We can treat a 0.50 2 4 solution as if it were 0.50 + & 0.50 4− initially, then determine extent of ionization (production of additional + & 42− •
[
][
[
[
]
]
][
[
]
]
•
•
•
o
• •
(
)
o
Behavior of amphiphrotic system depends on its dissociation constant
2 1
1
=
=
=
= =
•
o o
Ions may act as acids or bases Equilibriums in reactions between ions (or ion + water) is described by means of acid ionization constant or base ionization constant & are related The product of the ionization constants of an acid and its conjugate base equals the ion product of water ( ) ( )× ( )= ( )× ( )= • •
o
ex. • •
4+ + 2 ⇌ + 4
+ + acts as an acid giving up a proton to water Described by
23
o o
=
�
[
][
]
The stronger the acid, the weaker its conjugate base The weaker the acid, the stronger its conjugate base Does not mean that the conjugate base of a weak acid a SB When we compare the values of i.e. & ; it is still a weak base The conjugate base of a weak acid is a weak base The conjugate of weak is weak •
−
• •
•
o
Acid interacts with solvent CB also interacts with solvent = when is the solvent Solvolysis would then be called hydrolysis •
o
2
o • •
A reaction between an ion and water In pure water at 25°C, [H O ] = [OH ] = 1.0 × 10 M; pH = 7.00 Pure water is pH neutral When NaCl dissolves in water at 25°C, complete dissociation into Na & Cl ions occur, and pH remains at 7.00 Na + Cl + H O NR A reaction producing H O did not occur Cl did not hydrolyze When NH Cl is added to water, pH falls bellow 7 because [H O ] > [OH ] Produced additional [H O ] NH + H O NH + H O Ammonium ion hydrolyzes ] Greater , greater extent of hydrolysis and will give a solution with greater [
+
•
+
•
• •
−
−
4 4+ 2
→
−7
+
+
+
++
⇌
+ →
−
−
−
Metal ions i.e. are acidic Accept electron pairs as water Transition metals act as an acid Greater charge smaller radius more acidic •
2
−
•
→
o
Solving Write the equations separately Indicate expected concentrations of molecules and ions in solu tion Ex. 0.1 M CH COOH + 0.1 M HCl solution CH COOH + H O H O + CH COO H C l + H O H O + Cl ] = 0.1 M + x Common ion: [ If adding a SA to a WA [anion] greatly reduced Le Chatelier's; increasing [product] ([common ion]) shifts equilibrium to the reverse direction Not limited to WB & WA Can also be applied to buffers Suppression of weak electrolytes by addition of common ion •
•
• •
•
•
2
2 ⇌+ −+ → +
• •
o
• •
24
−
o
The salt of a weak acid is a strong electrolyte Ions become completely dissociated from one another in aqueous solutions The anion is an ion common to the ionization eq. of the weak acid Presence of anion suppresses the ionization of the weak acid In solving common-ion problems, assume that the ionization of the WA or B does not begin until both the WA or WB and its salt have been placed in solution Consider that ionization occurs until equilibrium is reached •
• •
•
o
•
+ −
a b
When [WA] = [CB], K = [H O ] pH = pK When [WB] = [CA], K = [OH ] pOH = pK
o
a
•
o
b
•
o
Suppression of ionization of WB by the common cation
• •
%
=
× 100%
Extent of ionization of a WA or WB The % ionization of a WA or WB increases as the solution becomes more dilute Ionization constant and volume are indirectly proportional
o o
•
•
pH changes very slightly on the addition of small Has 2 components Acid neutralizer Base neutralizer Components must not neutralize each other Rules out SA & SB Possible components Weak acid and conjugate base Weak base and conjugate acid
o o
• •
o
•
o
• •
WAorWB ≈ 1 (very slightly > 1 or < 1) conjugate
o o
[
]
[
]
•
≈ ≈
) ( + ) If solution is to be an effective buffer, ( will always be valid Equilibrium concentrations of buffer components will be very nearly the same as stoichiometric concentrations •
o
=
+ log
=
+ log
•
(
+
)
+
(
Derived from
)
=
[
][
[
]
]
25
=
+ log
=
+ log
+
+
•
Avoids the need for an ICE table •
•
o •
∴
•
< 10 Molarity of each buffer ≫ by ≥ 100x Has limitations; 0.10 < [
Valid only when
[
]
]
≈
is valid
Options 1. Find WA/WB with desired pH, prepare solution with equal molarities if acid and salt (impractical) 2. Use WA/WB, establish appropriate [A - ]/[AH] (or [B + ]/[BOH] ) to obtain a. Select WA with pK close to desired pH b. Calculate necessary [CB]/[acid] to give desired pH c. Calculate necessary concentration of CB and acid 3. Six methods for preparing buffer solutions a. Depending on the pH range required and the type of experiment the buffer is to be used for
a
o •
•
•
•
First use stoichiometric principles to establish how much one before component is consumed and how much of the other component is produced New concentrations of the WA (or WB) and its salt can be used to calculate the pH of the buffer Assume that the neutralization reaction proceeds to completion and determine new stoichiometric concentrations New stoichiometric concentrations are substituted into the equilibrium constant expression and the expression is solved for [ ], which is converted to pH
+
26
•
Dilute and concentrated buffers will have the same pH Only the buffer capacity changes
o •
•
Amount of acid or base that a buffer can neutralize before its pH changes appreciably Maximum buffer capacity exists when the concentrations of a weak acid and its conjugate base are kept large and approximately equal to each other
o •
• •
pH range over which a buffer effectively neutralizes added acids and bases and maintains a fairly constant pH o When the ratio falls to 0.10, the pH decreases by 1 pH unit from When the ratio increases to a value of 10, pH increases by 1 unit Therefore, the range of 2 pH units is the maximum range to which a buffer solution should be exposed
•
=
o o o o
+ log
[
]
[
]
Substance whose color depends on the pH of the solution it is added to Chosen based on how acidic or basic the solution is Does not affect the pH of the solution if added in small amounts ) It is the indicator that gets affected by the ions in solution ( If 90% or more of the indicator is in the form of , the solution will take on the acid color If 90% or more is in the form of , the solution will take on the base (anion) color ] [ ], the indicator is in the process of changing from one form to the If [ other and has an intermediate color Complete change occurs over a range of ~ 2 pH units = at the middle of the range Exists as: Weak acid ( ) •
• •
o
o
•
≈ −
−
+
27
One color ) Conjugate base ( Having a different color
•
−
•
o
Equivalence point of a neutralization reaction The point at which both acid and base have been consumed Neither acid nor base is in excess End point of the indicator Point in a titration at which the indicator changes color If the endpoint of the indicator is near the equivalence point of the neutralization, the color change marked by that endpoint will signal the attainment of eq. Point • •
o
•
o
•
o o
Graph of pH versus For equal volumes of acid solutions of the same molarity, the volume of base required to titrate to the equivalence point is independent of the strength of the acid
o
• • •
pH has a low value at the beginning of the titration At the equivalence point pH rises very sharply by approx. 6 units Beyond the equivalence point, pH rises slowly
o
o
Similar corresponding properties with titration of a SA with a SB
28
o
•
Neutralization of a WA involves the direct transfer from the acid molecule to the ions Recall: Neutralization of a SA, protons transferred from ions Ex. + + + + Calculations involve a stoichiometric part and an equilibrium part to take into account the partial ionization of the WA Similar to calculations for addition of a SB to a buffer solution Four regions of interest Initial pH Is higher as compared to titration of a SA WA only partially ionized Initial sharp increase in pH at the s tart of the titration Anion produced by the neutralization of the WA is a common ion that reduces the extend of ionization of the acid Over a long section of the curve preceding the equivalence point, pH changes only gradually Solutions corresponding the this portion of the curve are buffer solutions At the point of half neutralization, = [ ]=[ ] At equivalence point, >7 CB of a WA hydrolyzes and produces Beyond the equivalence point, the titration curve is identical to that of a SA with a SB pH is established entirely by the concentration of unreacted Steep portion of the titration curve at the equivalence point occurs at a relatively short pH range (7-10) Indicators that change color below pH 7 cannot be used
−
•
− → 2 − + − → 2 2
+
• •
•
•
•
•
•
−
−
−
•
•
o • •
Expect 3 equivalence points First two equivalence points come at equal intervals on the volume axis Third required more base/stronger base ex. titrated with 1 mol NaOH required to reach first equivalence point First equivalence point should come in a somewhat acidic solution and the second in a mildly basic solution Third equivalence point can only be reached by a strongly basic solution
•
4
29
First equivalence point Solution is essentially Acidic solution > For , the reaction that produces predominates over the one that produces + + = 6.3 × 10 + + = 1.4 × 10 1 mol NaOH required to convert 1 mol to 1 mol 2nd equivalence point in the titration of Basic solution >
24
• •
2− 24 − + −8 2−4− 2 ⇌ + −42− 24 2 ⇌ 4 − 2− −12 4 2 4 4 −2 4− 2 ⇌ 42− − 4 4− •
•
•
•
+
=
= 2.4 × 10
When can ionize (hydrolyze) only as a base pH of third equivalence point is easy to calculate For reasonable concentrated solutions (>0.10 M) pH values are independent of solution concentration )) = ( log( • •
+
12
30
•
o
o
o
( ) Constant for the equilibrium established between a solid solute and its ions in a saturated solution In general, the dissolution of a slightly soluble compound and its solubility product expression are represented as: + ( ) ] [ ] =[ Varies with temperature
o
⇌+ +− − →
The solubility product constant can be used to calculate the solubility of a compound at 25°C When comparing the solubility of different salts, get s first. No information from . Molar solubility Molarity in saturated aqueous solution •
•
o
•
•
o o o
o
Ions in saturated solutions come from a single source, the pure solid solute If we add another source of a common ion, solubility changes The solubility of a slightly soluble ionic compound is lowered in the presence of a second solute that furnishes a common ion : ICE table Write the equilibrium of the dissolution of the solute •
•
o o o
o
Initial conc. of the SOLVENT
In x M
Solubility of solute
In s
Equilibrium concentration
In s+x
expressions most effective for slightly soluble solutes Cannot be used for highly soluble ionic compounds Ionic solutions with moderate to high concentrations, activities and concentration are far from equal Limited to slightly soluble or essentially insoluble solutes
o • •
• •
Increase rather than decrease solubility As total ionic concentration increases, interionic attractions become more important Activities become smaller than stoichiometric concentration For ions involved in the solution process; higher concentration must appear in solution before equilibrium is established Solubility increases
o •
Assumption for calculations are that all dissolved solute appears in s olution as separated cations and anions Not valid Solute might not be 100% ionic
31
Some of the solute might enter in molecular form Some ions might join together in ion pairs i.e. Two oppositely charged ions held together by electrostatic attraction Additional solute must be present for the product of ion concentrations to equal solute, making the true solubility of the solute greater than the expected Degree of ion-pair formation increases as mutual electrostatic attraction of the anion and cation increases Ion-pair formation more likely if charge is higher
• •
•
•
•
+
o •
Reversible reaction between a solid solute and its ions in aqueous so lution is never the sole process occurring Self-ionization of water occurs Can generally be ignored Reactions between solute ions and other solution species Acid base reactions Complex ion formation Calculations based on are erroneous if other equilibrium processes are not taken into account
•
• •
•
•
o •
•
+ −
Reaction quotient applied to the solubility equilibria ] ] =[ ×[ Any possible dilutions must be considered before calculating for
o
> Precipitation should occur if Supersaturated < Precipitation cannot occur if Unsaturated = A solution is saturated if Precipitation is complete only if amount of solute remaining is very small If 99.9% or more of a particular has precipitated Less than 0.1% of the ion in solution •
•
•
o
• •
•
o o
o o
Selective precipitation Technique in which two or more ions in solution, each capable of being precipitated by the same reagent, are separated by the proper use of that reagent One ion is precipitated while the other remains in solution Primary condition: Significant difference in so lubilities of the substances being separated Usually means significant difference in their values Slow addition of a concentrated solution of the precipitating reagent to the solution from which precipitation is to occur Higher solubility, last to precipitate If the solution has it is pH dependent
•
o
o o
−
32
•
+ − ⇌ 2+ [
Reaction of + o Complexation o o
•
o o o o
� + → =[
If
[
]
gets used up but gets again, so use
Emits light from UV to visible 300 nm - 700 nm Color of the solution is the color complement of the light absorbed h h = =
0 →h → → → → 0 • •
o
]
is LR
•
•
]
Transparent side for incident light Frosted site facing you <
=
•
Where; A = absorbance b = path length c = [complex] = molar absorptivity Valid if; Monochromatic Single wavelength Needs monochromator Filter
•
eq
ε
•
•
•
o o o •
o
=
= log =
Fraction of light that passed through sample If transparent, 1 (100%)
•
o o
1
log
= log
Linear form of B-L equation + error Blank solutions negate absorption of others that are not the complex "Taring" Subtract blank absorbance from the observed absorbance HCl used as solvent because ions tend to hydrolyze ) ] in water and upon undergoing hydrolysis produces Exists as [ ( [ ( ] ) ) or rust May further hydrolyze into ( Gives the yellow-brown color May interfere with the color of and give erroneous absorbance read by the spectrophotometer May increase absorbance ] ) HCl shifts equilibrium to the left, preventing the formation of [ ( • •
o
•
+ + 2 6 2 5 2+ 2+
•
2 5 2+
•
33
o
2+
Solution with highest theoretical concentration of (S5) was used to determine ) analytical wavelength ( Absorbance curve flats out at Provides optimum precision Analytical wavelength corresponds to blue Opposite of red-orange, as seen by the naked eye Blank solution removed absorbance read from KSCN and HCl
•
• •
o o
•
2 ≈ 2+ =(
) + = +
• •
1 To determine molar absorptivity ( ) Plot absorbance vs concentration of Plot absorbance vs concentration of
o • •
2+ 2+
Since [ ] is unknown Standard Solutions KSCN in large excess in the standard solutions Reaction pushed to completion with as LR [ ]=[ ] Equilibrium concentration of calculated from [ ] & [ ] We need to use equilibrium concentrations, •
•
2+ ∴ 0 10 0 002 −5 + 02 1 00 10 − 10 •
•
•
•
•
•
2+
[
]
[
]=[
] =
=[
] =
( .
( . )( .
)( .
)
)
=2×10
= 0.02
Unknown Solutions KSCN is not in excess anymore To allow reaction to proceed to its equilibrium state Unlike in Standard solutions, can be determined from the equilibrium concentrations of the products and reactants Another blank was used Given the absorbance calculated (from the standard solutions), [ can be determined using equation of the best fit curve •
• •
•
•
•
•
o
SA: •
o
]
=
−
→ ∞
All dissociate Completely ionizes, i.e. [HCl] 0 in + If polyprotic (3 acidic H) i.e. ( ) = (0.1 ) + ( ) ) ( ) =( + ( ) ) ( ) ( ) =( + For salts, consider components that undergo hydrolysis (can produce WA/WB) + i.e. + + •
o
2+
[
2+
•
+ − → → 4 − + 1 − 4 ⇌ 22−4 + 22−4 ⇌ −4 + 4 ⇌ 4 4+ + − ⇌ + 4 2
34
]
+ 2+ − 2 ⇌ 4 Produces
•
•
o o
if
(resp. bases)
( ) ( ) +2 The molar solubility is obtained via titration Obtain hydroxide concentration, determine molar solubility from =4 • •
=
•
•
Only true for this system If the metal is trivalent,
•
In terms of hydroxide concentration,
•
4 = (3
•
)
=
2
[
]
12 ∑=1 12 2 2+ =
• • • •
Source: 0.1 M Less titrant was used as compared to just water Can be explained using Le Chatelier's principle Addition of ions shift the equilibrium backward Use the for the room T of water does not change unless the temperature changes or solvent Use Van't Hoff equation Relates and temperature Depends on the enthalpy of reaction Not always directly proportional
•
•
4 =
[
]
Lower than s calculated without a common ion
• • • • • •
Source: 0.5 M KCl More HCl was needed to titrate, therefore more hydroxide was present Thus, the molar solubility is higher compared to just water Increased solubility because of interionic interactions ) are positively and negatively charged; thus the ions do not Ions from the ( precipitate because the and solvate and ions
2 + − 2 2
• • •
− 2+
0.1 < < 0.5 (Same temperatures) The addition of ethanol, a solvent less polar than water, decreased the dielectric constant of the solvent and as a consequence, decreased the polarity of the solvent Lower polarity, lower solubility (in this system) Salt is ionic It should be observed that the solubility decreases; less salt needed to s aturate the solution
35
•
Addition of 5 drops 0.10 M BaCl2
•
Ion
Net ionic equation
242− 2− 42− 42− 4− − − − − −
2+ 2+ 2+ 2+ 2+
,
,
,
,
+ + + + +
Color of ppt
242− → 24 2− → 42− → 4 42− → 4 4− → 2 4 2 (
)
(
)
NVR
White White Yellow White White None
o
4+
; won't form phosphate because pKa of CH3COOH is higher than the pKa to reach the phosphate NVR means that it remained; precipitate
Ppt
24 → 2+ − 2 2 4 4 4 2 4 2 → 2+ − 24−
NVR
+2
(
(
+2 )
h
+
NVR NVR
)
(
) +4
3
+4
•
Ion
242− 2− 42− 42− 4−
+
Net ionic equation
Color of ppt
2+ 242− → 24 2+ 2− → +
White
+
White
NVR
2+ 4− → 2 4 2 +
(
)
(
White
)
36
+2
o
+
( ) . Precipitates of relatively strong basic anions will dissolve in CH 3COOH
Ppt
24 → 2+ − 2 2− 4 2 2+ 4− → 2 4 2
NVR
+2
(
)
+
+2
(
)
(
+
+
(
)
)
o
•
•
This is done to determine which ion that precipitates in neutral medium with BaCl 2 will not ppt in acidic medium Recall: the predominant species in acidic medium is dichromate becomes dichromate, and fomr yellow soltution, turns orange
Ion
Reactions involved
42− 42− •
Observation
2+ 42− → 4 42− + → 72− 2 − − − − − − +
2
• •
42−
+2
+
White ppt formed Yellow
→
orange solution
Performed for test solutions that did not precipitate with BaCl 2 . ) For , , ( h . : > >
• •
Based on the values (lower values, higher fraction dissolves if 1:1) AgI is least soluble (Lowest ) If 1:1, can tell you which is more soluble If different ratios, use ) AgCl is most soluble (Highest Will first react The species that gives most silver ions is AgCl, therefore with ammonia, it will react most [ ( ) ] is colorless complex; formed If NH3 is added, AgBr reacts next AgI doesn't dissolve even with concentrated ammonia < < .
•
•
2 +
•
37
• • • • •
Note: done only to CH3COO- and NO3- ions which did not ppt to both BaCl 2 and AgNO3 Presence of the smell of vinegar signals presence of CH3COOGoes from stronger to weaker acid Both strong, therefore NVR Ion
Reaction with H2SO4
− CH_3 COO^H^→CH_3 COOH − 3
NVR
38
Observations Vinegar-smelling gas ---
Third Long Exam Complex Ions and Coordination Compounds Werner's Theory of Coordination Compounds o Coordination Compounds Compounds made up of two simpler compounds The charge on a complex is the sum of its constituent charges or oxidation states Dot formula suggest coordination compounds also means also means Coordinate Covalent Bond Pair of electrons from a donor shared with an acceptor Frequently are formed in Lewis acid-base reactions i.e. NH3 + BF3 Ammine Complexes Dilute NH3 reacts with metal ions to form insoluble metal hydroxides or hydrated oxides
⋅2
⋅ 6 ⋅ 6
+ 2 2 → 2+ + + 3 3 → 3
When in excess, ammine complex containing NH 3 molecules form by bonding to metal ions through coordinate covalent bonds i.e.
+
− 4 ⇌ + 2 − 6 ⇌ + 2
o
Certain metal atoms (primarily transition metals) have two types of valence or bonding capacity Primary Valence Based on the number of electrons the atom loses in forming the metal ion Secondary Valence Responsible for the bonding of other groups called ligands to the central metal ion Complex Any species involving coordination of ligands to a metal center Metal center can be an atom or an ion Complex can be a cationic, anionic, or neutral molecule
o
Complex Cation
Complex Anion
Neutral Complex
363 432 o
o
4 6
Coordination Number Number of points around the metal center at which bon ds to ligands can form Range from 2-12; 6 is most common (followed by 4) CN = 2; limited to most complexes of Cu(I), Ag(I), Au(I) CN > 6; not often found in members of first transition series; more common in 2nd & 3rd series CN = 3, 5; rarely stable Depends on ratio of radius of central atom or ion to the radii of the attached ligands Coordination Sphere Includes the metal atom or ion and the l igands coordinated to it; does not include uncoordinated counter ions Ligands Lewis bases Central atoms act as Lewis Acids
o
Coordination compound
39
Donor Atom Atom in a ligand that donates a lone pair of to form a coordinate covalent bond Monodentate Ligand Ligand that uses one pair of electrons to form on e point of attachment to the central metal atom or ion i.e. monoatomic anions such as halide ions, polyatomic an ions (OH-), simple molecules (NH3; called ammine when it is a ligand), and more complex molecules (methylamine CH3, NH2)
Bidentate Ligand attaches to metal center at two points Polydentate i.e. oxalate (ox, ethylenediamine (en); Can donate two electron pairs; one from each N atom Polydentate Ligands that are capable of donating more than a single e lectron pair from different atoms in the ligand and to different sites in the geometric structure of a complex Lone pairs of electrons of a polydentate ligand must be far enough apart to attach to the metal center at two or m ore points; the donated pairs of electrons must be on different atoms i.e. ethylenediamine (en), oxalato (ox), ethylenediaminetetraaceto (EDTA); hexadentate Chelate Complexes Complexes that have a metal atom or i on and polydentate ligands that form rings
−
Isomerism o Isomers Substances that have the same formula but differ in structure and properties
40
Structural Isomers Differ in basic structure or bond type (diff. Ligands are bonded to the metal center and through which atoms) Ionization 4 3 5 3 5 4 Pentaamminesulfatochromium(III) chloride vs Pentaamminechlorochromium(III) sulfate Differ in which neutralizes the complex ion Coordination The ligands can be distributed differently between two complex ions
;
Hexaamminecobalt(III) hexacyanochromate(III) vs Hexaaminechromium(III) hexacyanocobaltate(III) Linkage Isomerism Ligands attach to CA in different ways ; Pentaamminenitrito-NCobalt(III) ion vs Pentaamminenitrito-O-Cobalt(III) ion Stereoisomers Same number and types of ligands and mode of attachment, but differ in th e space occupied around the metal center Geometric Isomerism Cis and trans Fac and mer Optical Isomerism Dextrorotatory or levorotatory
+ +
Bonding in Complex Ions: Crystal Field Theory Focus on how electrons in the d orbitals of a metal ion are affected when they a re in a complex Crystal Field Theory o Bonding in a complex ion is considered to be an electrostatic attraction between a positively charged nucleus of a central metal ion and el ectrons in the ligands o Repulsions also occur between the ligand electrons and electrons in the central ion o Focuses on the repulsions between ligand electrons and d electrons of the central ion o Recall
All five d orbitals are alike in energy when in a n isolated atom or ion but they are unlike in their spatial orientations ; E raised Maximum repulsion occurs with the Other three ligands approach between lobes of the orbitals and there is a gain in stability over the head-to-head approach Lowered with respect to average d-orbital E Crystal Field Splitting of an Octahedral complex Difference in E between 2 groups of d o rbitals Removal of degeneracy of the d orbitals Pairing depends on the magnitude of and P
& −
o
Δ >
Greater stability if Low spin
Δ
4h − paired with one in the lower level 41
Ligands are strong-field ligands, produce large Strong field complex
Δ <
Greater stability obtained by keeping the electrons unpaired Maximum number of unpaired electrons High spin complex If ligands are weak field ligands, produce only small ) four electrons assigned to lowest energy orbitals; goes into
Δ
(i.e. + ; instead of
Δ
again
Extra stability offset by pairing energy (P) Forces an electron already occupied May be assigned into 2 to avoid pairing energy Requires energy and offsets the extra stability acquired by placing first three electrons in the lower level
o
Note
Δ
Greatest for square-planar complexes Smallest for tetrahedral complexes Splitting is small, almost all tetrahedral complexes are high spin High spin complex has more unpaired electrons than a low spin complex Measuring change in weight of the complex in a magnetic field allows us to determine if high or low spin
Δ
o
−
Origin of paramagnetism is the existence of unpaired Pulled into a magnetic field Transition metals vary in paramagnetism; some are diamagnetic Mass is the same If diamagnetic, it is slightly repelled by a magnetic field and weighs less within the field If paramagnetic, it weighs more within the field Strong-field ligands tend to form low -spin, weakly paramagnetic, or diamagnetic complexes Weak-field ligands tend to fo rm high-spin, strongly paramagnetic complexes
o
Additive mixing Primary colors: any three colors that when combined yield white RGB Secondary colors: produced by combining two primary colors C (G+B), M (B+R), Y(R+G) Each secondary is a complement of one the primary colors C: Red, M: Green, Y: Blue Primary + Complementary color = white light Subtractive mixing Some of the wavelength components of W are removed b y absorption; reflected or transmitted light is deficient in some wavelength components Material appears black if completely absorbed Complementary color is transmitted if material absorbs o ne color (Primary or secondary) Subtracting a color from white light yields a complementary color
o
o
Contain species that can absorb photons of visible light and use the energy of photons to promote to higher levels Energies of photons must match energy differences through which the electrons can be promoted
= h =
Shown as colorless (no light absorbed):
42
Noble-gas electron configuration Outer shell of 18 18 in the n-1 shell and two in the n shell (outermost) i.e. CFS of d energy levels produces that accounts for colors of complex ions
−
− +
Δ
o o o
+
To calculate the solubility of AgCl(s) in NH3 (aq), use of Most cations in aqueous solutions exist mostly in hydrated form Successive K values decrease regularly in the displacement process (for displacements involving neutral molecules as ligands) Once degree of substitution of NH3 for H2O (for example) has become large, changes improve for H2O to replace NH3 in a rev. rxn If ligand in a substitution process is a polydentate, displace as many H2O molecules as there are points of attachment Complexes with polydentate ligands have much larger formation constants than do those with monodentate ligands Chelation Effect Additional stability of chelates over complexes with monodentate ligands Partly attributed to the increase in entropy associated with chelation Larger positive value of , the more negative , larger K
o
Δ°
o
Labile
o
Δ°
Complex ions in which ligands can be interchanged rapidly Complex ions of the first transition series except Cr(III) and Co(III)
Inert
Complex ion that exchanges ligands slowly Easier to isolate
o
Recall
o
There is a rxn between Cu(s) and but not Cu(s) and is more readily reduced than Electrode Strip of metal (M) is an electrode Classified according to weather oxidation or reduction takes place there
Ag++aq Zn
Ag+
Zn+aq
o
Anode
Cathode
Where oxidation takes place
Where reduction takes place
Half-cell Electrode immersed in a solution containing ions of the same metal Interactions between metal atoms on the electrode and metal ion in solution Ion is reduced Metal ion from solution collides with electode; gain n electrons from it, converted to M Metal atom (M) is oxidized Metal atom M on the surface may lose n electrons to the electrode and enter the solution as the ion
+
o
+
+ − ⇌
+
43
Forward rxn: oxidation Backward rxn: reduction Electrons remain on the electrode (M(s)) Free electrons are never found in an aqueous solution Measurement of the tendency for to flow from the electrode of one half cell to the electrode of another must be based on a combination of two different half cells Charge is carried through solutions by the migration of ions Wire therefore cannot be used for this connections Solution must either be in direct contact through a porous barrier or joined by a third solution in a salt bridge Electrochemical cell Properly connected combination of two half-cells Anions migrate toward the anode, cations migrate toward the cathode Voltaic (Galvanic cells) Electrolytic cells Cell Voltage Potential difference between the two half-cells or electromotive force (emf) Units: V (Energy per unit charge; J/C) Driving force for electrons Couple Abbreviations for half cells, i.e.
o
−
o
o
E
o
/+
o
Shows the components of a cell
o
Where oxidation occurs; left side of the diagram
Electrode where reduction occurs; right side
o
o
Single vertical line placed between different phases Double vertical line represents boundary between half cell compartments; represents a salt bridge Different species within the same solution are separated by a comma i.e. ||
o o
+ + | | °
o
o
Potentials of individual electrodes cannot be precisely established; therefore we arbitrarily choose a particular half-cell that is assigned an electrode potential of 0 Measures the tendency for a reduction process to occur at an electrode
°/ To determine ° for a standard electron, compare it with a SHE
o
o
SHE is always taken as the anode (left) (SHE) Involves equilibrium established on the surface of an inert metal The equilibrium reaction produces a particular potential on the metal surface, arbitrarily taken to be 0
°
o
Potential difference V of a cell formed from two standard electrodes Intensive property Difference is always taken as Any electrode at which a reduction half-reaction shows a greater tendency to occur than does the reduction of 1M H+ to H2(g, 1 atm) has a positive value for E°
o
o
° = °, °,
, Reaction occurs in a voltaic cell → cell does electrical W
Work of moving electric charges
44
=
−
n is the number of transferred between electrodes(electron # of an electrochemical rxn, AKA charge #; no units) Applies only if the cell operates reversibly
Δ = ° Δ° = Δ < 0 → Property of Redox reactions Δ < 0 → > 0 > 0 → Spontaneous in the forward direction for the stated conditions < 0 → Spontaneous in the reverse direction for the stated conditions = 0 → reaction is at equilibrium for the stated conditions Cell rxn reversed → changes sign ° ° Δ° = ln = . ° ° = ln ⇒ = ln ° ln = ° . log = = 0.0592
o
o
o o
Where pH is that of the unknown solution Consists of two half-cells with identical electrodes but different ion concentrations Identical electrodes, therefore are equal, but there is a potential difference between two half cells Spontaneous change in concentration cell always occurs such that the concentrated solution becomes more dilute and the dilute solution becomes more concentrated
°
o
o o o
Difference in concentration of ions between half cells cause Provides basis for determining values for sparingly soluble ionic co mpounds
° = 0 since species are the same in the half cells
Produces electricity through chemical reactions Stores chemical energy for later release as electricity May consist of a single voltaic cell with two el ectrodes and appropriate electrolytes
o
Cell reaction is not reversible When the reactants have been mostly converted to products, no more electricity produced Positive and negative electrodes are known as the cathode (where reduction Leclanche (Dry) Cell
+ 2 2 − →
Drawback: Zn metal dissolves because of the acidic medium (NH4Cl) NH3 builds up on the electrodes, decreases voltage
Alkaline Cell Superior form of Leclanche Cell Uses NaOH or KOH in place of NH4Cl as the electrolyte
o
− → 2 − 2
Cell reaction is reversible by passing electricity through the cell (charging)
45
Lead-Acid (Storage) battery Oxidation of Pb(s) to Pb2+(aq), followed by precipitation of PbSO4 at each electrode
2+ 2− → 2 2
Alternator powered by the engine constantly recharges the battery; non spontaneous Short circuit when the anode and cathode come into contact with each other Silver-Zinc Cell Button battery Zn(s), ZnO(s) | KOH (satd) | Ag2O(s), Ag(s)
→ 2 Discharging = 1.8
Nickel-Cadmium Cell Rechargeable battery Anode is cadmium metal, cathode is NiO(OH) supported on Ni metal
2 2 → 2 [Discharging] = 1.4
Lithium-Ion Battery Rechargeable Positive electrode: LiCoO2 Negative electrode: Highly crystallized graphite Electrolyte: Organic solvent and ions i.e. LiPF6
o
→ − + −
Materials pass through battery, converts chem. energy to electric energy Run indefinitely as long as they are supplied electrolytes The theoretical maximum energy available as electric energy in a ny electrochemical cell is for the cell reaction : Maximum energy release when a fuel is burned Efficiency Value
Δ° Δ°
° = H°
ε
o o o
Unwanted voltaic cells Significant in acidic medium Strained metal gets more corroded; strained metal is more active (more anodic) than unstrained metal i.e. Preferential rusting of a dented automobile fender Corrosion preferentially occurs at the head and tip of nail Some metals corrode such that the products adhere tightly to the underlying metal and prevent further corrosion (aluminum) Unlike Iron (rust); flakes off, exposes fresh iron surface and corrodes further
o o
o
Cover with oil or polymers (paint) Plate with a second metal (through electroplating) Corrosion begins when coating is cracked (dented) More active element undergoes oxidation red Reduction half reaction occurs on the plating If iron is coated with zinc (galvanizing); even if coating is broken, iron is still protected because zinc is more active than iron (zinc gets oxidized instead) and the corrosion products of zinc protect zinc from further corrosion Cathodic protection Active metal is a sacrificial anode
46
Oxidation occurs at the active metal; cathode supports a reduction half-reaction Cathode is protected as long as active metal remains
o
The process in which a nonspontaneous rxn is driven by the application of electric energy
o
Uses electricity to produce a nonspontaneous reaction ° is negative Reversing the direction of the electric flow changes the voltaic cell into an electrolytic cell
Where oxidation occurs (also for galvanic cells) For an electrolytic cell, anode is positive Electrons withdrawn from anode, therefore (+)
Where reduction occurs Cathode is negative Electrons are forced into cathode, therefore (-)
Sign of each electrode in an electrolytic cell is the same as the sign of the battery electrode to which it is attached. For both cell types, cathode is the electrode at which electrons enter and anode is the electrode from which electrons exit
→ + − →
→ + − →
Anode is (-) Cathode is (+)
+ → + Δ < 0
Releases energy System does W on surr.
Anode is (+) Cathode is (-)
+ → + Δ > 0
Absorbs energy Surr. does W on system
o
Voltage significantly in excess of the calculated value May be necessary to cause a particular electrode rxn to occur Needed to overcome interactions at the electrode surface More common when gases are involved High overpotential, less products
May have more than one possible reduction/oxidation half reactions occurring at each electrode (No competing reaction) Molten NaCl (With competing reaction (water)) Aqueous NaCl
Affects the reaction favored
(Inert Electrode) Provides a surface on which an electrolysis half-reaction occur; reactants come from electrolyte solution (Active electrode) Participates in the oxidation or reduction half-reaction
47
o
1 mol e− = 96485 C Charge = Q = It # of mol e− = F
Experiment 9: Coordination Chemistry Coordination Compounds o Formed when a metal atom or ion and a molecule with one or more unshared electron pairs (ligand ) interact o Metal center acts as Lewis Acid (electron pair a cceptor) Ligands bonded to metal center Lewis base Electron pair donor Bonds present are electrons coming from the ligands o Ligands introduce magnetic field in the metal center Can have different orientations Can destabilize the compound Changes color and magnetic properties and spin Metal center part of the d block 5 orbitals Clover shaped
o
o
Δ = Δ
In a metal ion surrounded by ligands, the d orbitals are at a different energy level compared to when they are in an isolated metal ion Effect of the crystal field on the d orbitals vary depending on which direction the ligands approach the metal ion i.e. Coordination compound Copper center Ligands donate electron pair vs. H2O Normal covalent bond; shared electrons; from two differe nt electrons Covalent bond comes from ammonia only (lone pairs of ammonia used up) Therefore called coordinate covalent bond AKA Dative i.e. Octahedral Blood red complex 5 H2O ligands Can affect d orbitals in an octahedral manner Can destabilize the orbitals 2 destabilized, 3 others are stable orbitals are higher in energy If spherical ligand effect, will affect everything
+
o
+
&
Crystal Field Splitting energy o The difference in energy of orbitals Octahedral o Higher CFSE, compound absorbs at higher energy (lower wavelength) o Smaller CFSE, absorb at lower energy (higher wavelength) o If compound absorbs at higher energy, color would be observed to be O & Y (observed at V & B; complements) o Weak field, absorbs at higher wavelength o If separation is too large, fill orbitals with lower energy first only
48
o
o
Property: diamagnetic If separation is small, fill lower energy orbitals before going up Property: paramagnetic Small dOct; high spin complex
Barycentre o Sum of the energies of orbitals Crystal Field Theory (CFT) o Explains the various colors of coordination complex
Electrochemistry: Chemical Change and Electrical Work # of electrons gained = # of ele ctrons lost Oxidizing agent is reduced, reducing agent is oxidized Oxidation (electron loss) is always a ccompanied by reduction (electron gain) General Characteristics of Voltaic and Electrolytic Cell o Electron flow is from anode to cathode Anode is always the site of oxidation Cathode is the site of reduction Differ in polarity Voltaic: - to + Electrolytic: + to Chosen cathode is usually more positive E half cell Higher chance to be reduced Most active; strongest reducing property Most negative E cell Highest E cell; most oxidizing Reducing agent gets oxidized Lithium metal gets easily oxidized In reduction reactions, left side contains oxidizing agents Partner of highest oxidation agent is the lowest reducing agent o Voltaic Cell Energy is released from spontaneous redox reaction System does work surroundings Overall cell reaction
+ → + ;Δ < 0 Getting the potential; ° is the overall cell reaction In the anode beaker; oxidation occurs As → ∞ , Zn gets oxidized and disappeared, copper atoms increase
Zinc gets oxidized Becomes copper metal
Salt bridge Closes the circuit Provides electron neutrality or charge neutrality Contains sodium sulfate Sulfate goes into the anode beaker to compensate for the positive charge Sodium goes into the cathode beaker to neutralize the negative charge U-tube; cylindrical glass tub Tip has semi-permeable membrane that allows ion diffusion Sizes and relative charges of the ions used are considered
49
Notation Anode Compartment
Anode compartment (oxidation half-cell)
Cathode Compartment (reduction half - cell)
Phase of lower oxidation state | Phase of higher oxidation state
Phase of higher oxidation state | phase of lower oxidation state
Phase boundary between half-cell
|++||+−| + − → − 2 + − 2 + → h| || , O | h
o
Electrolytic Cell Energy is absorbed to drive the nonspontaneous redox reaction i.e. use battery Surroundings (power supply) does work on system (cell) Overall cell reaction
− + → ;Δ > 0 Relate Gibbs Free energy to electrochemical cell potential Δ° = ° Standard conditions = 96 485 1 V = 1 J/C Δ° = ln ° = ln To make Δ° negative, ° ° = ° °
o
o o o
o o o
o
Not a state function Not additive
h = . = ° log + → +
° ln < 0, > ° Q=1; [reactant]=[pdt], ln = 0, = ° Q<1; [reactant]>[pdt],
Q > 1;reactant < pdt, ln > 0, <
+ →. + →−.
Potential = -0.036 V Left to right; reduction Highest to lowest oxidation numbers Reduction potentials indicated in volts From 3+ 0
→
++ − → + Δ° = 10.771 − 2 → Δ° = 20.440 ___________________________________________________________ + 3− → Δ = 0.109 ° = . − ⇒ 0.036
o
Plot of nE vs oxidation number
50
o o o o
Lower in the diagram, more stable Oxidation process is from left to right Negative slope; from less stable to more stable; more favored Strongest reducing agent Mn is the only reducing agent, only one that undergoes oxidation Negative slope to the right Oxidizing agents are to the right of the positive slope To the right of the steepest slope; strongest (H3MnO4) Recall:
o
o
o
= Δ 0.90 1.28 2.9 0.95 1.5 1.18 i.e. − → − → → → + → + →
Oxidation #
Species
+ + − −
0 +2 +3 +4 +5 +6 +7
o
o o
0 21.180 = 2.36 2.36 11,5 = 0.86 0.86 10.95 = 0.09 0.09 12.9 = 2.99 2.991.28 = 4.27 4.270.90 = 5.17
+ → + + Δ < 0 + + → →++− 2− + 2 →
_____________________________________
2+ → 3+
o
o
Shortcut; draw a line from alternate oxidation states Species between is under; comproportionation Species between is over; disproportionation
− → − { } − −
→ : −− → − → Balancing… − − + − + → − − 2 →
____________________________________________________
2− + → −
51
o
+ dH2O
Green
Red
Hexaaquanickel(II) ion
+en
Blue
Orange
Tris(ethylenediamine)nickel(II) ion
+ NH3
Purple
Yellow
Hexaamminenickel(II) ion
+ +
Trend: Water
Excess CH3COOH dissolves to form the complex, dissolves
+ + + + +
Yellow
Tetraoxochromate(VI)
Yellow to orange solution
Hexanitrocobaltate(III)
Deep blue solution
Tetramminecopper(II) ion
Blood red solution
Pentaaquathiocyanatoiron(III) ion
Cherry red precipitate or solution
Bis(dimethylgyoxime)nickel(II)
+
+ 3 10− ⇌ 2− 8 2 + + 6− → − − − 2+ → [ ]− + + h ⇌
−
52
Redox Rxn
Results
Brown ppt due to MnO2
Colorless sol'n due to
Green sol'n due to − , Brown ppt due to MnO2
4. 5. 6.
2− 2− − 2− − + − → + 2+ → + 2+ 2−
Blood red solution Decrease in intensity of the blood red solution, dissolution of Zn
2. 3.
2− 3− → 2 2− 3− 6+ 2− 5− → 2+ 3 5−
+
+
1.
Equations
2 → 2 ; MnO2 acts as catalyst
Bubbles due to O2 gas
Brown precipitate due to the formation of MnO 2 solid a. Balanced in basic medium Formation of colorless solution due to Mn 2+ Formation of green solution due to MnO4 2- (discoloration of the solution and production of brown MnO 2 ppt may be observed if solution stands) Formation of blood red solution If you add a small amount of zinc, complex is blood red a. Add excess zinc, solution becomes colorless; Zinc does not form complexes with it p Disproportionation reaction; MnO 2 acts as catalyst a. Theoretical frost diagram i. Blue: H2O2 ii. Green: O2 iii. Red: H2O iv. Broken lines; catalyzed
Electrochemistry In the expt: o
Use reduction potentials At 298K o
+ 0.1 | |+0.1 || Cell reaction: + → + ( ) = ( ) . log
o
|+ 0.1 ||+ 0.1 |
o
1.103
| +0.10|| +0.10, +0.10| 0.4606 | +0.10|| , −0.10| Get x concentration, dependent | +0.10|| , −0.10| Depends | +0.10|||+ 0.1 | 0.0296 |0.533, + 0.033|| + 0.1 |
o
o
o
o
53
