system.surroundings.universe.thermodynamic state of a system.

– The number of moles and identity of each substance.– The physical states of each substance.– The temperature of the system.– The pressure of the system.

Thermochemical standard state conditions– The thermochemical standard T = 298.15 K.– The thermochemical standard P = 1.0000 atm.

• Be careful not to confuse these values with STP.Thermochemical standard states of matter

– For pure substances in their liquid or solid phase the standard state is the pure liquid or solid.– For gases the standard state is the gas at 1.00 atm of pressure. For gaseous mixtures the partial pressure must be 1.00 atm.– For aqueous solutions the standard state is 1.00 M concentration.

State Functions are independent of pathway:– T (temperature), P (pressure), V (volume), E (change in energy), H (change in enthalpy – the transfer of heat), and S (entropy)

Examples of non-state functions are:– n (moles), q (heat), w (work)

oreactants

n

oproducts

n

o298 SnSnS o

reactantsn

oproducts

n

o298 HnHnH

oreactants

n

oproducts

n

o298 GnGnG

– The relationship describes the spontaneity of a system.• The relationship is a new state function, G, the Gibbs Free

Energy.• Sign conventions for G.

G > 0 reaction is nonspontaneous G = 0 system is at equilibrium G < 0 reaction is spontaneous

G = H-TS at constant T and P

heat transfer outheat transfer out(exothermic), -q(exothermic), -q

heat transfer inheat transfer in(endothermic), +q(endothermic), +q

SYSTEMSYSTEM∆E = q + ww transfer inw transfer in

(+w)(+w)Compression of systemCompression of system

w transfer outw transfer out(-w)(-w)Expansion of systemExpansion of system

• Free energy has the relationship G = H -TS.

• Because 0 ≤ H ≥ 0 and 0 ≤ S ≥ 0, • there are four possibilities for G.

Forward reaction H S G spontaneity< 0 > 0 < 0 at all T’s.< 0 < 0 T dependent at low T’s.> 0 > 0 T dependent at high T’s.> 0 < 0 > 0 Nonspontaneous at all T’s.

%100solution of masssolute of mass = w/w%

solution of Literssolute of moles = M

• Molality is a concentration unit based on the number of moles of solute per kilogram of solvent. solvent of kg

solute of molesm

B of moles ofnumber +A of moles ofnumber A of moles ofnumber

AXMole fraction

• Colligative properties are properties of solutions that depend solely on the number of particles dissolved in the solution.– Colligative properties do not depend on the kinds of particles dissolved.

• Colligative properties are a physical property of solutions. There are four common types of colligative properties:

1. Vapor pressure lowering2. Freezing point depression3. Boiling point elevation4. Osmotic pressure

Vapor pressure lowering is the key to all four of the colligative properties.

• Addition of a nonvolatile solute to a solution lowers the vapor pressure of the solution.– The effect is simply due to fewer solvent molecules at the solution’s surface.– The solute molecules occupy some of the spaces that would normally be occupied by

solvent.• Raoult’s Law models this effect in ideal solutions.

Boiling Point ElevationFractional Distillation

Freezing Point Depression

mimi bbff K T KT

M

M

RTwhere: = osmotic pressure in atm

= molar concentration of solution

R = 0.0821 L atmmol K

T = absolute temperature

Osmotic Pressure

KineticsFour factors that affect the rate of reaction

nature of reactantconcentrationtemperaturepresence of a catalyst

Average rate

Rate of reactionRate constantOrder of reactantOverall order of reaction

Integrated rate laws ([ ] with respect to time)If zero order [A]0 - [A] = ak t

first order

second order

t dD+

t cC+or

t bB-

t aA-= Rate

k t aAAln 0

k t aA1

A1

0

General rate expression

Differential rate laws ([ ]with respect to rate)Rate = k [A]m[B]n[C]p

If zero order Rate = k[A]0 = k first order Rate = k[A]1 = k[A] second order Rate = k[A]2

•Change in rate is independent of the change in concentration of a zero order reactant•Change is rate is directly proportional to the change in concentration of a 1st order reactant•Change in rate is directly proportional to the square of the change in concentration of a 2nd order reactant

0 order[A] vs t

[A]

t

ln [A] vs t1st order

ln [A]

t

1/[A] vs t2nd order

1/[A]

t

Initial rateInstantaneous rate

EquilibriumEquilibrium is a macroscopic event, regardless of the direction of approach the forward rate and the reverse rate are the same and changes in concentration cannot be observed.

Products favored 103 > K > 10-3 Reactants favored

Kc =[C]c[D]d

[A]a[B]b

Mathematical RelationshipsReverse the reaction equation K’ = 1/KScale the reaction equation K’’ = Kn

Add equations Krxn = K1 K∙ 2 K∙ 3 ...Relationship to Kp = K(RT)n

Heterogeneous SolutionsFor mixtures the pure solids, pure liquids, and components that represent large amounts of solvents can be removed from the equilibrium expression and rolled into the equilibrium constant.

To solve for concentrations using K you will need a balanced reaction equation and an ICE table.

Predicting reaction direction from non-equilibrium conditions – Q

The equilibrium expression for Q is the same as for K. Compare the values of Q and K (the ratios of products to reactants)Q=K the system is at equilibriumQ>K favors reactantsQ<K favors products

Factors that affect Equilibrium – [ ], P&V, TFor [ ] check Q or use the teeter totter, For gases – P&V, [ ] is to 1/V which is to P, ↑P, ↓V, or ↑[ ] will push to fewer moles of gas.For T check H is q product or reactant

AcidHCl

BaseNaOH

Arrhenius Produces H+ Produces OH-

Brönsted-Lowery

Donates H+ Accepts H+

Lewis Accepts e- pair Donates e- pair

pH = -log[H+] pOH = -log[OH-] pKa = -logKa etc.Kw = 1E-14 = [H+][OH-] = 1E-7∙1E-7

pKw = pH + POH = 14 = 7 + 7

Henderson-Hasselbach equationpH = pKa + log([salt]) – log([acid])pOH = pKb + log([salt]) – log([base])

Determination of strength – or who wins the tug of war for the proton

Strong acids Ka > 1Weak acids Ka < 1

Strong bases Kb > 1Weak bases Kb < 1

BufferCommon Ion EffectTitrant Equivalence PointPrimary Standard End PointSecondary Standard TitrationSolubilityMolar solubilityKsp - Solubility Product ConstantKa – Acid dissociation constantKb – Base dissociation constantKw – water dissociation constantLe Chatelier’s PrincipleLewis AcidLewis BaseBronsted-Lowry AcidBronsted-Lowry BaseArrenhius AcidArrhenius BaseHenderson-Hasselbach equationConjugate acidConjugate Base

TitrationTitration 1. Add solution from the buret.1. Add solution from the buret.2. Reagent (base) reacts with 2. Reagent (base) reacts with

compound (acid) in solution compound (acid) in solution in the flask.in the flask.

3. Indicator shows when exact 3. Indicator shows when exact stoichiometric reaction has stoichiometric reaction has occurred.occurred.

4. Net ionic equation4. Net ionic equation HH++ + OH + OH-- --> H --> H22OO5. At equivalence point 5. At equivalence point moles Hmoles H++ = moles OH = moles OH--

http://www.chem1.com/acad/webtext/abcon/abcon-2.html

In General How to Solve Acid/Base Equilibria and Buffer Problems1. Are the compounds involved strong or weak electrolytes

a. If strong then there is a 100% dissociation and no way back use a single headed arrow and when adding the simultaneous equations together only and add the products.

i.e. NaOH → Na+ + OH-

b. if a weak electrolyte use a double headed arrow and when adding simultaneous equations together add both reactants and products

i.e. CH3COOH ↔ CH3COO- + H+

2. Determine if reactions are neutralization, dissociation, or equilibrium reactionsI. Neutralization

a. Class I – strong electrolyte/strong electrolyte: produces salt and water. Use ICE table with units of amount (mmol or moles), cancel terms, determine amounts of products. Stop.b. Class II – weak electrolyte/weak electrolyte: Compare Kas or Kbs to determine which predominates products or reactants. Use the equlibrium equation and ICE table with units of concentration (mmol/mL or mol/L) to determine equilibrium concentrations. Stop. c. Class III – strong electrolyte/weak electrolyte: produces salt of the conjugate and water and further actions, Dissociation or Equilibrium reactions must be determined continue to 2 II

II. Dissociate any products and begin again at repeat step 1 then continue to 2 III. Products of the dissociation though by name appear to be the same compounds they are not identical in that they have a different source and are treated separately.

III. Equilibrium – write new stoichiometrically balanced chemical equilibrium equations.

3. Write the equilibrium equation and solve for x and determine [H+] or [OH-]

4. Find pH or pOH5. Determine the pOH from pH or pH from pOH6. Determine the [H+] or [OH-] not found in step 3

Ksp

Write a stoichiometrically balanced equation for the limited dissociation of the solid If provided with the Ksp and looking for molar solubility of products

Write the stoichiometric ratio, i.e. 1:2Multiply the ratio through by x, i.e. x:2xWrite the equilibrium equation, i.e. Ksp = [M2+][Y-]2

Substitute in the stoichiometrically determined x values for the [ ]s, Ksp = (x)(2x)2

Solve for xIf provided a mass determined to be in solution and looking for Ksp

convert mass to molesconvert moles to molarityWrite the equilibrium equation, i.e. Ksp = [M2+][Y-]2

Solve for Ksp

Electrochem• The cathode is the electrode at which reduction occurs.

• The cathode is positive in voltaic cells.• The anode is the electrode at which oxidation occurs.

• The anode negative in voltaic cells.• 1 amp = 1 coulomb/second

coulombs 487,96e 106.022faraday 1 -23 • In all voltaic cells, electrons flow spontaneously from the negative electrode (anode) to the positive electrode (cathode).1. Choose the appropriate half-reactions from a table of standard reduction potentials.2. Write the equation for the half-reaction with the more positive E0 value first, along with its E0 value.3. Write the equation for the other half-reaction by reverse the tabulated reduction half-reaction and change the sign of the tabulated E0 .4. Balance the electron transfer. Do not multiply the potentials.5. Add the reduction and oxidation half-reactions.6. Add their potentials. This creates an E0cell which is positive, which indicates that the forward reaction is spontaneous.

K log RT 303.2G

orlnK -RTG

0

0

The Nernst Equation

The Gibbs Free Energy Relationship

Primary Voltaic Cells• As a voltaic cell discharges, its chemicals are consumed.• Once the chemicals are consumed, further chemical action is impossible.• The electrodes and electrolytes cannot be regenerated by reversing current flow through

cell.– These cells are not rechargable.

Secondary Voltaic Cells• Secondary cells are reversible, rechargeable.• The electrodes in a secondary cell can be regenerated by the addition of electricity.

– These cells can be switched from voltaic to electrolytic cells.• One example of a secondary voltaic cell is the lead storage or car battery.

The Hydrogen-Oxygen Fuel Cell• Fuel cells are batteries that must have their reactants continuously supplied in the

presence of appropriate catalysts.• Hydrogen is oxidized at the anode.• Oxygen is reduced at the cathode.

• Fuel cells are very efficient. – Energy conversion rates of 60-70% are common!