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CHEM 162: Final Exam Study Guide
Chapter 16: Chemical Equilibrium
equilibrium: state where the forward and reverse
reactions or processes occur at the same rate
– Know that concentrations are not changing at
equilibrium, but they need not be equal to one
another.
– Be able to indicate when equilibrium is achieved
given concentration vs. time plots
Use the law of mass action to write equilibrium
expressions for Kc or Kp for homogeneous and
heterogeneous reactions.
– Include only gases; omit pure liquids and solids.
For general reaction, j A + k B
K
c
=
[C]l [D]m
[A] j [B]k
or
Kp =
lC +mD
PCl PDm
PAj PBk
Extent of reaction
– For large values of Kc or Kp (>103), the reaction
essentially goes to completion.
→ The equilibrium mixture consists mostly of
products (product favored)
→ The equilibrium lies to the right.
– For small values of Kc or Kp (<10-3), the reaction
does not occur to any significant degree.
→ The equilibrium mixture consists mostly of
reactants (reactant favored)
→ The equilibrium lies to the left.
– For intermediate values (10-3 < Kc or Kp < 103),
the equilibrium mixture contains appreciable
amounts of both reactants and products.
→ For Kc or Kp > 1, equilibrium lies to the right.
Equilibrium positions: Set of equilibrium
concentrations or partial pressures of reactants and
products for a system.
– While Kc or Kp for a reaction are constant for a
given temperature, various equilibrium positions
are possible for that reaction depending on the
initial concentrations of reactants and products.
Relating Kp and Kc: Kp=Kc(RT)Δn
– Be able to solve for Δn for any homogeneous or
heterogeneous reaction.
– Know that Kp=Kc for Δn=0.
Know Kc or Kp are unitless!
CHEM 162 Final Exam Review
Reaction Quotient (Q): instant state of system, not
necessarily at equilibrium.
– Also determined using law of mass action.
– Q < K: too many reactants
→ System shifts right to make more products.
– Q > K: too many products
→ System shifts left to make more reactants.
– Q = K: system at equilibrium
Determination of Kp or Kc
– Be able to solve for either using a variety of
experimental data.
– Given equilibrium concentrations or partial P’s.
– Given initial & changes in conc. or partial P’s.
– Given total pressure at equilibrium.
Equilibrium Problems Solving for Kp or Kc.
1. Get balanced chemical equation
2. Write equilibrium expression
3. Set up equilibrium ICE table.
– Let x=change in conc. or partial pressure.
4. Substitute equilibrium conc. or pressures into
equilibrium expression for Kp or Kc.
5. Solve for x, using quadratic method if necessary.
6. Substitute value for x into equilibrium conc. or
pressures to solve for Kp or Kc.
Le Chatelier's Principle
– A system at equilibrium will shift (if possible) to
minimize any stress (change in concentration,
pressure, volume, or temperature)
– Predict shifts in equilibrium given specific changes.
– Only changes in temperature (T) affect Kp or Kc.
– For endothermic reactions, Kp and Kc ↑ as T ↑.
– For exothermic reactions, Kp and Kc ↑ as T ↓.
Characteristics of the Equilibrium Expression
– For reverse reaction, equilibrium expression is
reciprocal of that for forward reaction
– Multiplying coefficients by factor, n, raises equilibrium
constant to nth power
– Multiple Equilibria
– When a reaction can be expressed as a sum of
two or more reactions, the equilibrium constant
for the overall reaction is simply the product of
the equilibrium constants for the individual
reactions.
– Be able to manipulate a series of reactions to
determine the equilibrium constant for an overall
reaction.
page 1 of 8
Chapter 17: Equilibrium in Aqueous Phase
•
Know the properties of acids and bases.
• Know Arrhenius definitions for acids and bases.
– Know the general form of an Arrhenius acid-base
neutralization reaction:
– acid + base → water + salt
•
Know terms monoprotic, polyprotic, etc.
• Know Brønsted-Lowry definitions for acids and bases.
– Recognize conjugate acid-base pairs.
•
• Know the common strong bases: LiOH, NaOH, KOH,
Ca(OH)2, Sr(OH)2, Ba(OH)2.
Recognize strong acids and strong bases dissociate or
ionize (break up) completely.
→ Equilibrium lies far to the right.
HNO3(aq) → H (aq) +
+
NO3−(aq)
Ca(OH)2(aq) → Ca+2(aq) + 2 OH−(aq)
and H2SO4: H2SO4(aq) → H (aq) +
+
•
H+(aq) + F−(aq)
Write balanced equations and equilibrium expressions
for the dissociation of any acid.
– Omit pure liquids and solids.
H+(aq) + A−(aq)
HA(aq)
K
•
HSO4−(aq)
Recognize weak acids dissociate or ionize (break up)
only to a small degree.
→ Equilibrium lies far to the left.
HF(aq)
•
Recognize how structure, bond strength and bond
polarity influence the properties of an acid.
Be able to explain:
– why HF is weak while HCl, HBr, and HI are strong
– why some ternary oxyacids are strong while others
are weak (e.g. HNO3 vs HNO2).
– Ka values for different acids based on structure.
• Autoionization of water:
a
=
[H+ ] [A - ]
[HA]
Recognize the strength of an acid is inversely
related to the strength of its conjugate base.
→ Strong acids have conjugate bases that are weaker
than H2O.
→ Weak acids have conjugate bases that are stronger
than H2O.
→ Be able to write the net ionic equation for the
conjugate base reacting with H2O to form the
conjugate acid and OH−.
• Know Lewis definitions for acids and bases.
– Identify the Lewis acid or base in a given reaction.
– Know highly charged cations (Al3+, Cu2+, etc.) can
act as Lewis acids and why.
CHEM 162 Final Exam Review
H3O+(aq) + OH−(aq)
2 H2O(l)
Kw = [H+][OH–]=1.0x10–14
Recognize hydronium ion, H3O+ = H+ + H2O
• Know the strong acids: HCl, HBr, HI, HNO3, HClO4,
H2SO4.
•
•
Kw = water’s ion-product or dissociation constant
–
–
–
–
When [H+] = [OH–], the solution is neutral.
When [H+] > [OH–], the solution is acidic.
When [H+] < [OH–], the solution is basic.
Solve for [H+] or [OH–] given the other then classify
the substance as acidic, basic, neutral.
pH scale: pH = 7: neutral and pH < 7: acidic
and pH > 7: basic (or alkaline)
[H+] = 10−pH
pH [OH−] = 10−pOH
pH = − log [H+]
pOH = − log [OH−]
pH + pOH = 14.00
Know # of sig figs in [H+] or [OH−] determines the #
of decimal places in pH and pOH.
Calculate pH of Strong Acids and Strong Bases
– Strong acids and bases ionize completely.
– Accounting for all the H+ and OH– ions,
→ [H+] = original molarity of strong acid
→ [OH–] = molarity of base × (# of OH– in base)
pH Calculations for a Weak Acids
– Weak acids remain mostly undissociated.
– Write the equation for the dissociation of acid.
– Set up ICE table with [HA] given or calculated.
– Calculate pH of weak acid solution.
– Use percent ionization of solution to get Ka.
– Use [HX] – x ≈ [HX] approx. for:
x
<5%
[HA]
– Use Quadratic Method or method of successive
approximations for:
Percent ionization =
x
≥0.05%
[HA]
[H + ] at equilibrium
[HA]
×100%
page 2 of 8
Chapter 17: Equilibrium in Aqueous Phase (Continued)
Convert Ka and Kb: Kw = Ka · Kb = 1.0×10–14
Acid-Base Properties of Salts
– salt = ionic compound
– Soluble salts dissociate into ions in water.
– Classify a given salt as acidic, basic, neutral.
Ions that produce acidic solutions
– NH4+
– Highly charged metal ions: Al3+, Zn2+, etc.,
except cations of strong bases
Ions that produce basic solutions
– anions that are conj. bases of weak acids.
– all anions except anions of strong acids
– Note: SO42− + H2O → HSO4− + OH−
Ions that produce neutral solutions
– do not react with H2O to make H+ or OH−
Salts with an acidic cation and basic anion
– Classify a given salt as acidic, basic, neutral.
– If Ka>Kb, salt is acidic.
– If Kb>Ka, salt is basic.
– Account for different ion concentrations by multiplying
initial concentration with Ka or Kb.
pH Calculations for a Weak Bases
– Write equation for reaction of base with H2O.
– Write equilibrium expression for the weak base.
– The weaker the base, smaller the Kb.
– Set up ICE table with [A–] given or calculated.
– Solve for x to calculate [OH–] and pOH of weak
base solution.
Polyprotic Acids
– Write the stepwise dissociation of any
polyprotic acid.
– Recognize the [H+] is determined only by 1st
dissociation step for most polyprotic acids.
– Solve for pH and equilibrium ion concentrations.
Given K 'w =
1
[H + ] [OH − ]
= 1.0×1014,
− H+ (from strong acids) will completely react with any
base, A−.
− OH− (from strong bases) will completely react with
any acid, HA.
− Be able to write neutralization reactions for H+ or
OH− added to any solution.
CHEM 162 Final Exam Review
Acid-Base Titrations
− Distinguish between endpoint (when indicator changes
color) and equivalence point (when equal amounts of
H+ or HA and OH− present)
− Know general pH range for equivalence point of
following:
− strong acid-strong base titration (pH≈7)
− weak acid-strong base titration (pH>7)
− strong acid-weak base titration (pH<7)
− Recognize that pH=pKa at halfway to equivalence
point for a weak acid-strong base titration.
− Given titration data, calculate the pH an any point during
an acid-base titration.
− Account for amount of H+ and OH− neutralized and new
initial [HA] or [A−].
− Set up ICE table and solve for x to get [H+] or [OH−].
− Calculate pH based on [H+] or [OH−].
Acid-Base Indicators
− Know when color changes become apparent
(at pKa±1).
− Know effective range of indicators based on Ka.
− Given the pKa for different acid-base indicators, explain
which indicators would be effective for specific acid-base
titrations.
The Common Ion Effect
− The shift in equilibrium caused by the addition of a salt
with an ion in common with the dissolved substances
− Given a compound added to an acid or base at
equilibrium, predict equilibrium shifts (to the left or right),
if [H+] ↑ or ↓, if [OH−] ↑ or ↓, and if pH ↑ or ↓,
− Calculate the pH of an acid or base solution when a
common ion is added.
buffer: a solution of a weak acid or weak base and its
conjugate that can resist large changes in pH
upon addition of a small amount of strong acid or
strong base.
Determining pH and [H+] for any buffered solution
− Know pH=pKa for [HA]=[A−] (at halfway to the
equivalence point for a titration)
− Calculate the number of moles of strong acid (H+) or
strong base (OH−) added to the buffer.
− Carry out the neutralization reaction, and determine the
number of moles of weak acid/base and its conjugate
present after reaction with the strong acid (H+) or strong
base (OH−).
− Calculate the pH using the Henderson-Hasselbalch
equation:
[A - ]
pH = pKa + log
[HA]
page 3 of 8
Chapter 17: Equilibrium in Aqueous Phase (Continued)
Blood pH Buffer System
− Know the chemical equation for the buffer system for
blood pH.
− Use Le Châtelier’s Principle to explain how the
buffer system maintains the pH of blood.
− Define acidosis and alkalosis and use Le Châtelier’s
Principle to explain how the conditions occur.
Know how to prepare a buffer system to maintain a
given pH.
− Choose a weak acid with pKa near buffer pH range.
− Use Henderson-Hasselbalch to determine the
[A - ]
ratio to get desired pH for a buffer system.
[HA]
− Be able to determine the amount of weak acid
(volume and concentration) and/or the mass of a
salt containing the conjugate base to prepare a
buffer to maintain the desired pH.
Chapter 18: The Colorful Chemistry of Metals
Lewis Acids and Bases
– Lewis acid: electron-pair acceptor
– Lewis base: electron-pair donor
– Recognize that any Brønsted-Lowry acid is also a Lewis acid, and any Brønsted-Lowry base is a Lewis base.
– Recognize that most nitrogen-containing compound is a Lewis base, so like NH3, they can act as weak bases in
water—e.g. CH3NH2 + H2O
CH3NH3+(aq) + OH−(aq).
Chapter 15: Chemical Kinetics
chemical kinetics: study of the factors that influence
reaction rates
reaction rate: a positive quantity expressing the
concentration change with time
Use experimental data given to
– determine a general reaction rate given
concentrations of reactants/products over time
– determine the rate of disappearance/
consumption of a reactant or rate of
appearance/production of a product given the rate
of disappearance/appearance of another
reactant/product in the reaction
Determine reaction rate
– Given experimental data of concentrations and
time.
Know the terms: rate law, rate constant (k)
Reaction Order:
– zero-order, first-order, second-order
– Determine overall order for a reaction given rate
law.
Distinguish between the instantaneous reaction
rate and the average reaction rate.
CHEM 162 Final Exam Review
Determine rate law using initial rates method
– Given data of concentrations and rates.
– Cannot be determined given only the balanced
chemical equation.
half-life (t1/2): the time required for the concentration
of a reactant to decrease by half
Determine the reaction order given experimental
data of reactant concentration over time.
– Recognize the plots giving a straight line for zeroorder, first-order, and second-order reactions.
– Recognize that the half-life is only constant for
first-order reactions.
– Compare slopes for the first and last sets of data
to see if slope changes for each reaction order.
Do calculations given integrated rate laws for zeroorder, first-order, and second-order reactions.
– Solve for concentration at a given time given initial
concentration and rate constant.
– Carry out natural log (ln) calculations for 1st-order
reactions—review your algebra!
– Solve for the time required for the concentration to
decrease to a given amount.
– Solve for half-life given rate constant, k, or vice
versa.
page 4 of 8
Chapter 15: Chemical Kinetics (Continued)
COLLISION MODEL: reactant molecules must
collide to react
Activation Energy (Ea): minimum energy needed
for chemical reaction
Reaction Rate and Temperature
– As T ↑, reaction rate ↑ since molecules move
faster and more molecules have activation
energy.
Three Factors Affecting Reaction Rate
1. Concentration
2. Orientation of molecules
3. Temperature and Kinetic Energy (KE)
– Molecules move faster at higher temps.
– Molecules must have the activation energy
(i.e., suuficient KE to break and make
bonds, so they react)
TRANSITION STATE MODEL
Reaction Energy Profiles
– Indicate transition state, activation energy for
reactants and products, ΔH for a reaction,
effect of a catalyst.
– Distinguish between the activated complex and
the transition state.
– Determine if a reaction is endothermic or
exothermic.
– Identify the number of steps in the mechanism
for a catalyzed reaction and the ratedetermining step.
Catalyst: substance added to a system that
lowers the activation energy of a reaction.
– Know catalysts provide an alternative pathway
that eases the collision geometry requirement
– Recognize that catalysts increase reaction rate
without being consumed in reaction.
– Know homogeneous versus heterogeneous
catalysts.
– Enzymes: catalysts with unique active sites
that speed up biochemical reactions
Arrhenius Equation: k = A e–Ea/RT
where A=frequency factor, Ea=activation energy,
R=8.3145 J/mol·K, T=temperature in K
– Be able to solve for any unknown given other
variables or graphical data and variables.
Two-Point Equation of Arrhenius Equation:
– Solve for the rate constants or activation energy at
two different temperatures
⎛ k 2 ⎞ Ea ⎛ 1
⎞
− 1 ⎟
⎟=
⎜
⎝ k 1 ⎠ R ⎝ T1 T2 ⎠
ln⎜
– Recognize that an Arrhenius plot of ln k versus I/T
at different temps gives a slope equal to Ea/R.
Reaction Mechanisms
– The sequence of steps by which a reaction occurs
at the molecular level
– The slowest step in a mechanism is the ratedetermining step.
– Given the reaction mechanism for a reaction,
determine the rate law.
– If a fast step is followed by a slow step,
determine the rate law only in terms of the
reactants (excluding intermediates).
– Given the experimentally determined rate law,
determine the correct reaction mechanism given
possible mechanisms.
molecularity of a reaction
– unimolecular, bimolecular, termolecular
– Recognize that unimolecular and bimolecular are
common, but termolecular steps are very rare.
– Determine corresponding rate laws for each.
intermediate: species produced in an earlier step
and consumed in later step of mechanism
Distinguish between a catalyst and an intermediate.
– Be able to identify the correct mechanism for a
reaction given information on any catalysts,
intermediates, rate laws, and/or reaction orders.
Chapter 14: Thermodynamics
system: that part of the universe being studied
surroundings: the rest of the universe outside the
system
1st Law of Thermodynamics: Energy is neither
created nor destroyed.
CHEM 162 Final Exam Review
Enthalpy change, ΔH = qreaction at constant
pressure (e.g atmospheric pressure)
– endothermic reaction: ΔH = +
– Ereactants < Eproducts; reaction cools surroundings;
– for physical changes, KEproducts > KEreactants
– exothermic reaction: ΔH = –
– Ereactants > Eproducts; reaction heats surroundings;
– for physical changes, KEproducts < KEreactants
page 5 of 8
Chapter 14: Thermodynamics (Continued)
spontaneous process: occurs without external
intervention or stimulus
nonspontaneous process: only occurs with external
intervention or stimulus
Entropy, S: measure of molecular randomness and
number of arrangements available to a system
– Ssolid < Sliquid < Sgas
– 3rd Law of Thermodynamics: S=0 only for a perfect
crystalline solid at 0K
– S > 0 for all other substances, even naturally
occurring elements; the more complex the molecule
the greater its absolute entropy, S°.
– Recognize ΔS is positive for a reaction that
increases the # of moles of gas particles.
– Recognize if Ssys increases or decreases based on
increased kinetic energy or physical changes.
– Calculate ΔS° given S° data for all the reactants and
products.ΔSuniv = ΔSsys + ΔSsurr
– 2nd Law of Thermodynamics: For any
spontaneous process, the Suniv increases.
– Recognize that the Ssurr=−
ΔH sys
T
Definitions of standard state
1. A gaseous substance with P=1 atm
2. An aqueous solution with a concentration of 1M at a
pressure of 1 atm
3. Pure liquids and solids
4. The most stable form of an element at 1 atm & 25°C
G=Gibbs free energy
– Be able to calculate ΔG° given ΔGf° data
Gibbs’ Equation: ΔG = ΔH - T ΔS
Standard state conditions: ΔG° = ΔH° - T ΔS°
– Be able to calculate ΔG°, ΔH°, T in Kelvins, and/or
ΔS° given the other variables
– Know at what temperatures a reaction is
spontaneous given ΔH and ΔS (e.g. if ΔH and ΔS
are both +ve, then the reaction is only spontaneous
at high T)
Effects of Pressure and Concentration on ΔG:
– Use ΔG = ΔG° + RT ln Q where T is in Kelvins,
R=8.3145 J/mol·K, and Q=reaction quotient
– Know how partial pressure of gases and
concentrations of solutions affect ΔG using Le
Châtelier’s Principle.
At equilibrium, use: ΔG° = – RT ln K where
K=equilibrium constant
– Be able to solve for ΔG° or K for any system at
equilibrium.
– Recognize that the equilibrium point occurs at the
lowest value of free energy available to system.
→ A system will move towards equilibrium to
achieve the lowest free energy.
– Know how K ↑ or ↓ with temperature changes
based on ΔS and ΔH, accounting for bond strength
and available energy states.
Two forms of the van’t Hoff Equation:
– Solve for K1, K2, ΔH°, T1, or T2 using
ln
K2
K1
1 ⎞
⎟ or
⎜⎜
−
R ⎝ T2
T1 ⎟⎠
K
- ΔHo ⎛ T2 − T1 ⎞
⎟
⎜
ln 2 =
K1
R ⎜⎝ T1T2 ⎟⎠
=
- ΔHo ⎛ 1
If ΔG < 0 → a spontaneous reaction
If ΔG > 0 → a nonspontaneous reaction;
reverse reaction is spontaneous.
If ΔG = 0 → reaction is at equilibrium
Chapter 19: Electrochemistry
Electrochemistry: the study of the relationship between electricity and chemical reactions
Be able to determine oxidation numbers for all atoms in an oxidation-reduction (redox) reaction.
– Determine what is oxidized, what is reduced, the oxidizing agent, and the reducing agent.
– Balance the electrons transferred in a redox reaction by balancing atoms gaining/losing electrons
cathode: where reduction occurs
anode: where oxidation occurs
CHEM 162 Final Exam Review
page 6 of 8
Chapter 19 (Continued)
Voltaic (or Galvanic) Cells
Know line notation (cell diagram) for an
electrochemical cell: Zn | Zn2+ || Cu2+ | Cu where
1. anode reaction (oxidation) is shown at left
2. The separation of the two half cells is
indicated by the symbol ||.
3. cathode reaction (reduction) is shown at right
4. single vertical line "|" indicates a phase boundary
(e.g. a solid electrode in an aqueous solution)
For any voltaic cell, be able to identify:
1. The half-reaction at the cathode
2. The half-reaction at the anode
3. The electron flow in the cells and through external
circuit
4. The ion flow in both cells, including the salt bridge
Calculation of Ecell° from Eox° and Ered°:
Eell° = Eox° + Ered°
– The cell's overall voltage or potential (Ecell°) is the
sum of the two half reactions:
General Guidelines for Voltaic Cells
1. Calculated cell voltages, Ecell°, must be positive for
reactions in a voltaic cell.
2. Because cell potentials are intensive properties
(independent of amount), you never multiply by
the coefficients in a half reaction to get the cell
potential.
Strength of Oxidizing and Reducing Agents
– The species reduced = oxidizing agent
– The species oxidized = reducing agent
The strength of an oxidizing agent is directly
related to its Ered°.
→ The more positive Ered°
→ the stronger the oxidizing agent (or tendency
to be reduced)
→ The more negative Ered°
→ the stronger the reducing agent (or tendency
to be oxidized)
– Be able to rank different oxidizing and reducing
agents given Ered° values.
Spontaneity of Redox Reactions
– If Ecell° > 0, the reaction is spontaneous.
– If Ecell° < 0, the reaction is nonspontaneous;
the reverse reaction is spontaneous.
Relations Between Ecell°, ΔG°, and K:
ΔG = – n F Ecell
and
where n=# of moles of electrons transferred,
F=faraday=96,485 J/mol·V
Nernst equation:
Ecell = Ecell° –
At equilibrium at 25°C,
0.025693 V
Ecell° =
ln K
n
0.0592 V
Ecell° =
log K
n
RT
ln Q
nF
or
Not at equilibrium at 25°C,
0.025693 V
Ecell = Ecell° –
ln Q
n
0.0592 V
Ecell = Ecell° –
log Q
n
or
Concentration Cells:
– Recognize that electrons are transferred from the
less concentrated to the more concentrated solution
in an effort to equalize the concentration of ions in
solution.
→ Oxidation occurs in the cell with the less
concentrated solution.
→ Reduction occurs in the cell with the more
concentrated solution.
Electrolysis: the process of using electrical energy to
cause a nonspontaneous redox reaction
Units to know:
faraday (F) = 1 mole of electrons = 96,485 coulombs
1 coulomb = amperes · seconds or 1 C = 1 A·s
1 joule = volts · coulombs or 1 J = 1 V·C
For an electrolytic cell, be able to:
1. identify what is oxidized (reducing agent)
2. identify what is reduced (oxidizing agent)
3. write half reactions
Quantitative Relationships
The mass of product formed (or reactant consumed) at
an electrode is directly proportional to
1. the amount of electricity transferred at the electrode
2. the molar mass of the substance
ΔG° = – n F Ecell°
CHEM 162 Final Exam Review
page 7 of 8
BE PREPARED TO SOLVE PROBLEMS COMBINING
CONCEPTS FROM VARIOUS CHAPTERS.
THE USUAL GENERAL CHEMISTRY PERIODIC TABLE WILL BE PROVIDED,
ALONG WITH THE EQUATIONS ON THE FOLLOWING PAGE.
THE FOLLOWING EQUATIONS WILL ALSO BE PROVIDED:
[A]t = −kt + [A]0
ln [A]t = −kt + ln [A]0
ln
t1/2 =
[A]t
= −kt
[A]0
1
1
−
= kt
[A]t
[A]0
⎛k ⎞ E
ln⎜ 2 ⎟ = a
⎝ k1 ⎠ R
ln
K 2 - ΔHo
=
K1
R
⎛ 1
1 ⎞
⎟⎟
⎜⎜
−
T
T
1⎠
⎝ 2
ΔG = – n F Ecell
Ecell° =
Ecell = Ecell° –
and
t1/2 =
⎛1
⎞
− 1 ⎟
⎜
⎝ T1 T2 ⎠
Ssurr= –
ΔH sys
T
ΔG° = – RT ln K
and
ln
K 2 - ΔHo
=
K1
R
⎛ T1 − T2
⎜⎜
⎝ T1T2
⎞
⎟⎟
⎠
ΔG° = – n F Ecell°
0.0592 V
log K
n
0.025693 V
ln K and
n
Ecell° =
0.025693 V
ln Q
n
Ecell = Ecell° –
and
0.693
k
1
k [A]0
t1/2 =
ΔSuniv = ΔSsys + ΔSsurr
ΔG = ΔG° + RT ln Q
[A]0
2k
0.0592 V
log Q
n
Units to know:
faraday (F) = 1 mole of electrons = 96,485 coulombs
1 coulomb = amperes · seconds or 1 C = 1 A·s
1 joule = volts · coulombs or 1 J = 1 V·C
CHEM 162 Final Exam Review
page 8 of 8