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MCAT Physics Equations MUST MEMORIZE > General Chemistry Hidden Gems > Flashcards

General Chemistry Hidden Gems Flashcards

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1
Q

Elements

A

-Protons and neutrons have the same mass

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2
Q

Mass Number

A

A-Protons and neutrons (top)

-Good approximation of the mass of an atom

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3
Q

Atomic Number

A

Z-protons (bottom)

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4
Q

Atomic weight/Molar Mass

A

1 amu=1 g/mol, 6.022x10^23 amu= 1 gram

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5
Q

Empirical Formula

A

Whole number ratio of elements to each other

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6
Q

Molecular Formula

A

Exact number of elemental atoms in a molecule

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7
Q

Energy is always required to

A

bread a bond. Energy is NOT released during bond breaking

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8
Q

Periodic Table

A
  • Period: horizontal row
  • Group/Family-vertical column, elements in the same family have similar chemical properties such as the same number of bonds or similarly charged ions
  • Alkali metals, Alkaline earth metals, Halogens, Nobel Gases
  • Representative/main group elements and transition metals
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9
Q

Metals

A
  • large atoms that tend to lose electrons to form cations or positive oxidation states
  • Ductility, malleability, thermal and electrical conductivity, luster
  • All metals except mercury exist as solids at room temperature
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10
Q

Non-metals

A
  • form anions

- Lower melting points than metals

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11
Q

Group 1A

A

(alkali metals)

  • low densities, low melting points
  • Highly reactive, especially with nonmetals to form ionic ions
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12
Q

Group 2A

A

(alkali earth metals)

-Harder/ denser, melt at higher temps than Group 1A (alkali metals)

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13
Q

Group 5A

A
  • Can form 3 covalent bonds
  • Except for nitrogen all other groups 5A can form 5 covalent bonds using d orbitals
  • Nitrogen forms strong pi bonds to make double and triple bonds
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14
Q

Group 6A

A

(Chalcogens)

  • Oxygen is the 2nd most electronegative element
  • Oxygen is divalent and can form strong pi bonds
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15
Q

Group 7A

A

Halogens
-Highly reactive
-Fluorine always has an oxidation state of -1 meaning it can’t make more than one bond
Fluorine,Chlorine: Diatomic gases at room temp
Bromine: Diatomic liquid at room temp
Iodine: Diatomic solid at room temp

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16
Q

Small atoms have less room to stabilize charge by spreading it out making them bond more strongly to

A

water resulting in grater heats of hydration

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17
Q

The p orbitals on atoms that are too big don’t

A

overlap well so they don’t bond (2nd row is fine, 3rd is not)

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18
Q

Transition metals

A
  • Lose electrons from their S subshell first and then from their d subshell
  • Try to even out their d orbitals so they each have the same number of electrons
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19
Q

Atomic Size

A
  • Cations are smaller than neutral counterparts because the loss of an electron decreases shielding
  • Anions are larger than neutral counterparts
  • Elements with the same number of electrons tend to get smaller as there are more protons
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20
Q

Z-effective nuclear charge

A
  • Nuclear charge Z minus the average number of electrons between the nucleus and the electron in question
  • Increase left to right and top to bottom (bottom right has largest z-effective nuclear charge)
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21
Q

Atomic radius increases

A

right to left, increases top to bottom. Bottom left has biggest atomic radius

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22
Q

Ionization energy

A

Increases left to right and bottom to top. Top right highest ionization energy
-Second ionization energy>first ionization energy

Ionization energy refers to the amount of energy needed to remove an electron from an atom.

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23
Q

Electronegativity

A
  • Tendency of an atom to attract an electron in a bond with another atom
  • Increases left to right and bottom to top
  • Undefined for noble gases (same with electron affinity)
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24
Q

Electron Affinity

A
  • Willingness of an atom to accept an additional electron

- Increases from left to right and bottom to top

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25
Q

Metallic Character

A

-Increases from right to left and top to bottom

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26
Q

Types of reactions

A
  1. Combination: A+B—> C
  2. Decomposition: C—->A+B
  3. Single Displacement: A+BC—>AB+ C
  4. Double Displacement: AB+CD—> AC+B
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27
Q

Crystals

A
  • Ionic crystals-consist of oppositely charged ions held together by electrostatic forces
  • Molecular crystals-composed of individual molecules held together by intermolecular bonds
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28
Q

Ionic Compounds Nomenclature

A
  • Add -ic to end of cation with greater positive charge and -ous to ion with less positive charge
  • Monoatomic anions and simple polyatomic anions are given the suffix -ide
  • Polyatomic anions with multiple oxygens end with the suffix -ite (less oxygens) or -ate (more oxygens)
  • To name an ionic compound put the cation name in front of the anion name
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29
Q

Acids Nomenclature

A
  • If the name of the anion ends in -ide the acid starts with hydro- and ends with -ic
  • If the acid is an oxyacid the end -ic is used for the species with more oxygens and -ous for the species with less oxygens
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30
Q

Binary Molecular Compounds Nomenclature

A

-The name begins with the name of the element that is farthest to the left and lowest in the periodic table

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31
Q

Principle Quantum number

A

(n)

  • Transition metals-lag one shell behind the period
  • Representative elements-given by the period in the periodic table
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32
Q

Azimuthal Quantum Number (L)

A

s,p,d, and f

L=n-1

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33
Q

Magnetic Quantum Number

A

(m)
- From -L to L
- Gives the precise orbital of the subshell

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34
Q

Spin Quantum Number

A

-1/2 to 1/2

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35
Q

wavelength=

A

h/mv

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36
Q

Aufbau Principle

A

Each new proton added a new electron is added

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37
Q

Hund’s Rule

A

Electrons will not fit any orbital in the same subshell until all orbitals in that subshell contain at least one electron

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38
Q

Order of subshell filling

A

1S, 2S, 2P, 3S, 3P, 4S, 3D, 4P, 5S, 4D, 5P, 4F, 5D, 5F

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39
Q

STP Conditions

A

0 degrees celsius and 1atm, 22.4 L/mol

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40
Q

Mean Free Path

A

Distance travelled by a gas between collisions

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41
Q

Kinetic deals with rate of reaction as it moves towards equilibrium, and thermodynamics deals with

A

balance of reactants and products after they have achieved equilibrium

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42
Q

Kinetic Molecular Theory

A
  • Gas molecules exert no forces other than repulsive forces due to collisions
  • Gas molecules make completely elastic collisions
  • Average kinetic energy of gas molecules is directly proportional to temperature of gas
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43
Q

PV=nRT is a

A

state equation. When gases do work by expanding, the pressure drops to compensate for the volume increase but it also drops to compensate for the kinetic energy loss since the nRT side drops as well

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44
Q

K.E.=

A

(3/2)RT

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45
Q

v1/v2=

A

squareroot (m2/m1)

Find K.E. from temperature, plug into this formula to find relative velocities from relative mass

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46
Q

Effusion

A

Spreading of gas from very high pressure to very low pressure through a “pinhole”
-Effusion rate 1/Effusion rate 2=sqrt(M2/M1)

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47
Q

Diffusion

A

is the spreading of one gas into another gas or into empty space

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48
Q

PV=nRT adjustments

A
  • Real gases deviate from ideal behavior when their molecules are close together
  • High pressure pushes gas molecules together
  • Low temperature causes gas molecules to settle near each other
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49
Q

P+ a(n/V)^2=nRT

A
  • b is a measure of volume actually occupied
  • a is strength of intermolecular attractions
  • A and B generally increase with molecular mass and complexity of gas
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50
Q

Vreal>Videal because

A

gases have volume

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51
Q

Preal

A

gases attract each other

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52
Q

k=zpe^Ea/RT

A

P=steric factor
Z=fraction of collisions having the effective spatial orientation p
-Increasing the rate is NOT a statement about the equilibrium

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53
Q

Rateforward= k f [A]^a[B]^b

what is overall order?

A

a+b is overall order
-If looking at a table…when concentration is doubled reaction rate quadruples making the order=2
[Reactants]/xt=[Products]/yt where x and y are constants

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54
Q

Elementary reactions

A

If the MCAT tells you a reaction is elementary then you can use the coefficients as the order.. otherwise. No

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55
Q

Two step reactions

A

-If second reaction is rate limiting the [reactants] for the second step depends on the rate constant of the first step
1. NO+Br2—>NOBr2. Fast step
2. NOBR2+NO—->2NOBr2 Slow step
k1[NO][Br2]=k-1[NOBr2]
…k1/k-1[NO][Br2]=[NOBr2]
-Since the rate is set by the slow step k2 [NOBr2][NO]={NOBr]
Making k2k1/k-1 [NO]^2[Br2]

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56
Q

Catalysts

A
  • A catalyst may lower the activation energy or increase the steric factor “p”
  • The binding of a catalyst is almost always exothermic and the rate of catalysis depends upon the strength of the bond between the reactant and the catalyst
  • too weak and not enough absorption
  • Too strong and too much energy is required to remove the reactant
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57
Q

Heterogenous Catalyst

A

-in a different phase than the reactants and products

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58
Q

Effects of solvent on Rate

A
  • the solvent bonds may stabilize an intermediate
  • Degree of solvation affects k
  • Rate constant is a function of solvent and temperature
  • Solvent bonds must be broken before
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59
Q

Equilibrium

A
  • the point where the forward rate is equal to the reverse rate
  • Equilibrium is the point of greatest entropy
  • Equilibrium is a dynamic process
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60
Q

Law of Mass Action

A
  • The law of mass action is good for all chemical equations, including non-elementary equations
  • This means use the coefficients as the exponents regardless of molecularity
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61
Q

The equilibrium constant K=

A

[Products^coefficients/ [Reactants]^coefficients
-If you look closely this means K=kf/kr BUT only for elementary equations
K depends only upon temperature
-Equilibrium constant for a series of reactions is equal to the product of the equilibrium constants for each step
-Do not confuse equilibrium constant with equilibrium
-Don’t include solids or pure liquids

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62
Q

Partial Pressure Equilibrium Constant

A

Kp=K(RT)^delta n

  • K is concentration equilibrium constant
  • Kp is partial pressure equilibrium constant, for gas reactions you can use this equation to find K
  • n is the sum of the coefficients of the products minus the sum of the coefficients of the reactants
  • Any two or more single reactions or series of reactions resulting in the same products from identical reactants must have the same equilibrium constant for a given temperature
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63
Q

Reaction Quotient

A

Q=products^coefficients/ Reactants^coefficients

  • Since products always move towards the equilibrium, Q will always move towards K
  • If Q=K reaction is at equilibrium
  • If Q>K then the [products] is greater than it should be and the reaction will shift leftward
  • If Q
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64
Q

Le Chatelier’s Principle

A
  • When a system at equilibrium is stressed the system will shift in a direction to reduce that stress
  • Three types of stress
  • Addition or removal of product or reactant
  • Changing Pressure
  • Heating or cooling the system: if a product of a reaction is heat, adding heat will shift the equilibrium to more reactants… not counterintuitive once you realize that heat is produced when bonds are made… and bonds are broken by adding heat
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65
Q

Molarity Effect

A

-If you concentrate a solution the equilibrium will shift to the side with less moles

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66
Q

Open versus closed systems

A

system deltaenergy deltamass
Open. Yes. Yes
Closed Yes No
Isolated No No

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67
Q

Extensive Properties

A

-Change with amount

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68
Q

Intensive Properties

A

do not change with amount

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69
Q

State functions

A

-state properties describe the state of a system. The change in a state property going from one state to another is the same regardless of the process via which the system changed

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70
Q

Path functions

A
  • opposite of state functions

- ie work, heat

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71
Q

Work

A

W=PdeltaV under constant pressure

  • If force is constant pressure is constant by P=F/A
  • Work can be different going from one state to another… if pressure changes too then there is a different amount of work done
  • If we have a piston on its side and add heat so that it expands (but keep the gas the same temperature) it does work on the surroundings
  • The temperature of the gas stays the same meaning pressure decreases as volume expands. This means that the pressure is decreasing so the force is not constant.
  • Force can not be constant if temperature is constant
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72
Q

Second Law of Thermodynamics

A

-Heat can not be completely changed into work in a cyclical process
-Efficiency=1-Tc/Th
K.E.avg=(3/2)kT

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73
Q

Enthalpy

A

deltaH=deltaU+PdeltaV

  • Two systems may have the same amount of internal energy but if they are at different pressures they have a different capacity to perform work…hence enthalpy was invented
  • Is a state function
  • An ideal gas enthalpy depends only on temperature
  • Is an extensive property
  • At constant pressure change in enthalpy is equal to heat
  • An element at 25 C and 1 atm is arbitrarily assigned an enthalpy value of 0J/mol
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74
Q

Standard Enthalpy of Formation

A

deltaHknotf
-Change in enthalpy that creates one mole of that product from its raw elements
-For a reaction involving no change in pressure deltaH=q
-deltaHreaction=deltaHf products-deltaHf reactants
-deltaHreaction>0 the reaction is endothermic
deltaHreaction<0 the reaction is exothermic
-Hess’s law says when you add reactions you can add their enthalpies

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75
Q

Entropy

A

(Joules/Kelvin)
-is a state function
-is an extensive property (increases with number, volume, and temperature)
-a reaction can be unfavorable in enthalpy and proceed…but it can’t be unfavorable with entropy and succeed
-reactions at equilibrium have achieved maximum universal entropy
-if a reaction increases the number of gaseous molecules then the reaction has positive entropy (nature likes more molecules)
delta S=dqrev/T (defined by change in heat per kelvin in a reversible process)

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76
Q

Gibbs Free Energy (G)

A

DeltaG= deltaH-TdeltaS

  • must minimize delta G to achieve equilibrium
  • A negative delta G value corresponds to a spontaneous process
  • This equation is only good for constant temperature reactions
  • Is an extensive property
  • Is a state function
  • Represents the maximum Non-PV work available from a reaction
  • An isolated system can change its gibbs free energy
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77
Q

There are three ways to transfer energy in a system:

A

Heat(q) and Work

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78
Q

Heat

A

Movement of energy from hot to cold

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79
Q

Conduction

A

-thermal energy transfer via molecular collisions
Q/t=kA (Th-Tc)/L
-k is thermal conductivity
-The rate of heat flow (Q/t) would be the same in all slabs even if they each had different lengths, thicknesses, and different thermal conductivities
-the order of the slabs wouldn’t matter either
-A higher conductivity results in a lower temperature difference across any slab of a given length

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80
Q

Convection

A
  • thermal energy transfer via fluid movement

- Differences in pressure or density drive warm fluid in the direction of cooler fluid

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81
Q

Radiation

A
  • Rate at which an object radiates is its Power P
  • P=sigmaE (Te^4-To^4)
  • E -emissivity of surface (between 0 and 1)
82
Q

Solution

A

-homogenous mixture of two or more compounds in a single phase

83
Q

Solvent

A

compound which there is more of

84
Q

Solute

A

Compound there is less of

85
Q

Types of solutions

A

Ideal solutions-made from compounds with similar properties; compound can be interchanged within the solution without changing the spatial arrangement of the molecules or the intermolecular attractions

  • Ideally dilute solution-solute molecules are separated so that they have no interaction with each other; mole fraction of solvent is 1
  • Nonideal solutions
86
Q

Colloid

A

a colloid is a solution except the solute forms into larger particles
ie, aerosol, foam, emulsion
-can be separated with a semipermeable membrane using dialysis

87
Q

London Dispersion Forces

A

Nonpolar molecules are held together by weak bonds caused by instantaneous dipole moments
-the bonds of a nonpolar solute are too strong to be broken by the weak forces of a nonpolar solvent

88
Q

Electrolyte

A

a compound which forms ions in aqueous solution

-Strong electrolytes create solutions which conduct electricity well and contain many ions

89
Q

Nitrite

A

NO2-

90
Q

Nitrate

A

NO3-

91
Q

Sulfite

A

SO3^2-

92
Q

Sulfate

A

SO4^2-

93
Q

Hypochlorite

A

ClO-

94
Q

Chlorite

A

ClO2-

95
Q

Chlorate

A

ClO3-

96
Q

Perchlorate

A

ClO4-

97
Q

Carbonate

A

CO3^2-

98
Q

Bicarbonate

A

HCO3-

99
Q

Phosphate

A

PO4^3-

100
Q

Molarity

A

(mol/L) moles of the compound divided by the volume of solution

101
Q

Molality

A

(mol/kg) moles of solute divided by kilograms of solvent

102
Q

Normality

A
  • measures the “number of equivalents” per liter of solution
  • Depends on the type of reaction taking place in the solution
  • An acid-base reaction with H2SO4 would be a 2 normal because two equivalents of protons
103
Q

Solution Formation

A

3 steps to making a solution

  1. Break bonds between solute molecules
  2. break bonds between solvent molecules
  3. Formation of bonds between solvent and solute
    - At constant pressure the enthalpy change of a reaction equals the heat
    - For condensed phases not at high pressure the enthalpy change approximately equals the internal energy change (delta H~=delta U)
104
Q

Heat of Hydration

A
  • the combined enthalpy of steps 2 and 3. Breaking bonds then forming bonds
  • A solution with negative enthalpy will give off heat when it forms creating a stronger bond
  • A solution with a positive enthalpy will take in heat when it forms
  • Positional entropy always increases in the formation of a solution so solution formation ALWAYS has a positive entropy
105
Q

Vapor Pressure

A
  • vapor pressure increases with temperature
  • Vaporization is an endothermic process
  • When vapor pressure equals local atmospheric pressure a compound boils
106
Q

Clausius-Clapeyron equation

A

ln(Pv)= -(deltaHvap/R)(1/T)+C

  • deltaHvap is the heat of hydration
  • Pv is vapor pressure
  • R is 8.314 J/molK
  • Vapor pressure increases with temperature
107
Q

Melting Point

A
  • temperature at which the vapor pressures of the solid is equal to the vapor pressure of the liquid
  • Above the melting point the liquid vapor pressure is greater than that of the solid
  • Below the melting point the liquid vapor pressure is less than that of the solid
108
Q

Nonvolatile Solute

A

-when a solute with no vapor pressure is added to a solution it reduces that solutions vapor pressure by taking up room at the surface of the liquid reducing the area in which the liquid interacts with the vapor

109
Q

Raoult’s Law

A

Pv=XaPa

  • For nonvolatile solutes- if 97% of the solution is solvent, the vapor pressure will be 97% of the vapor pressure of the pure solvent
  • For volatile solutes-if 97% of the solution is solvent, then the vapor pressure will be 97% of the pure solvent PLUS 3% of the vapor pressure, of the pure solute
110
Q

Vapor pressure if a nonideal solution will deviate from Raoult’s law

A
  • Negative heats of solution form stronger bonds so less molecules can break free so there is lower vapor pressure and there will be a negative deviation
  • Positive heats of solution form weaker bonds so more molecules can break free so there is increased vapor pressure and there will be a positive deviation
  • A positive heat of solution leads to an increase in vapor pressure while a negative heat of solution leads to a decrease in vapor pressure
111
Q

Saturated

A

when the rate of dissolution and precipitation are equal

112
Q

Solubility

A
  • number of moles of solute per liter of a solution that can be dissolved in a given solvent
  • depends on the common ions in the solution
  • solubility is found from the solubility product
  • solubility product is not the same thing as solubility, you look up the SP in a table and it is independent of common ions
113
Q

For most salts, crystallization is

A

exothermic

114
Q

Solubility Product (Ksp)

A
  • Equilibrium of a solvation reaction has its own equilibrium constant called the Ksp
  • Solids and pure liquids have a mole fraction of approximately 1 and are excluded
  • Set Ksp equal to products over reactants raise to the power of their coefficients!!!
  • Changes ONLY with temperature, solubility depends on temperature and common ions in solution
115
Q

Example of solubility product (Ksp)

A

example:
BaF2(s)Ba2+(aq)+2F-(aq)
Then Ksp=[Ba^2+][F-]^2
If “x” moles per liter of BaF2 dissolve then there will be x mol/L of Ba^2+ and twice as many (2x) mol/L of F-. If Ksp is 2.4x10^-5 then…. 2.4x10^-5=(x)(2x)^2
-Solve for “x” and that is the solubility of BaF2 in mol/L
What if we had added 1 mol of NaF to the solution?
-It would completely dissociate forming 1 mole of F- and 1 mole of Na+
-Remember, the order in which you mix doesn’t make a difference so choose the most convenient
-2.4x10^-5=(x)(2x+1)^2
-X~=2.4x10^-5. We can check our assumption by comparing with the original equation. Since 2x=4.8x10^-5 and this is «1 assumption=valid

116
Q

Nearly all ionic compounds containing nitrate (NO3-), ammonium (NH4+), and alkali metals (Li+, Na+, K+…) are

A

soluble

117
Q

Ionic compounds containing halogens (Cl-, Br-, I-) are

A

soluble, EXCEPT for silver, mercury and lead compounds (Ag+, Hg2^2+, Pb2+)

118
Q

Sulfate compounds (SO4^2-) are

A

soluble, EXCEPT for mercury, lead, and the heavier alkaline metals (Hg2^2+, Pb^2+, Ca^2+, Sr^2+, Ba^2+)

119
Q

Compounds containing the heavier alkaline metals (Ca^2+, Sr^2+, Ba^2+) are

A

soluble when paired with sulfides (S^2-) and hydroxides (OH-)

120
Q

Carbonates, phosphates, sulfides, and hydroxides (CO3^2-, PO4^3-, S^2-, OH-) are generally

A

insoluble other than in the cases already mentioned

121
Q

Solubility factors

A

-pressure and temperature affect solubilities. Pressure on liquids and solids has little effect, but pressure on a gas increases its solubility. For an ideally dilute solution, the increase in pressure of a gas a over a solution is directly proportional to the solubility of a gas a, if the gas does not react with, or dissociate in, the solvent. This relationship is given by Henry’s Law

122
Q

Henry’s Law

A

C=ka1PV

  • C is the solubility of the gas a(in mol/L)
  • ka1-is Henrys law constant which varies with each solute-solvent pair
  • Pv is the vapor partial pressure of the gas a above the solution
  • DOES NOT agree with Raoult’s Law because both are approximations
123
Q

When to use Rauolt’s or Henry’s

A
  • The partial vapor pressure of a solution component is always proportional to its mole fraction. If the component predominates as the solvent the Pv is proportional to the pure vapor pressure. If the component represents a tiny amount of solution Hendry’s law says that the vapor partial pressure is proportional to a Henry’s law constant.
  • Raoult’s law is more accurate when looking at the vapor partial pressure of a solvent with high concentration
  • Henry’s law is more accurate when looking at the vapor partial pressure of a volatile solute where the solute has a low concentration
  • In an ideally dilute solution, the solvent obeys Raoult’s Law and the solute obeys Henry’s Law
124
Q

As temperature increases the solubility of salts

A

generally increases. This is NOT true for gases. The solubility of gases in liquid decreases with increased temp.
-Can remember this by remembering why hot waste water is hazardous to aquatic life. It contains less oxygen and also tends to float on cold water limiting its gas exchange with the air.

125
Q

Heavier, larger gases tend to be

A

more soluble

126
Q

Gases that react with a solvent have

A

greater solubility

127
Q

Heat Capacity

A

C, in units of J/K
measure of the energy change needed to change the temperature of a substance
-C=q/delta T
-There is a constant volume and constant pressure heat capacity
-When pressure is held constant some energy escapes as PV work so at constant pressure a substance can absorb energy with less change in temperature by expelling some of the energy to the surrounding as work
-Thus, Cp>Cv

128
Q

Specific heat capacity

A

(usually c, J/kg*K

  • heat capacity divided by unit mass (not molality!)
  • you might get a question with specific heat capacity per volume, per mole, etc.
  • Each phase of a substance has its own specific heat
129
Q

Coffee Cup Calorimeter

A
  • An example of a constant pressure calorimeter because it measures energy change at atmospheric pressure
  • Used to measure heats of reaction
  • Specific heat capacity ‘c’ of water is 1cal/g*C
  • you measure the change in temperature because at constant pressure q=delta H
130
Q

Bomb Calorimeter

A
  • An example of a constant volume calorimeter because it measure energy change in a confined volume
  • A small blast chamber is inside of a larger chamber with the inner chamber surrounded by water
  • Heat is transferred to water outside the blast chamber and the change in temperature is recorded so the heat of the reaction can be figured out using the equation q=CdeltaT
131
Q

Heat of fusion

A

enthalpy change associated with melting

132
Q

Heat of vaporization

A

enthalpy change associated with boiling

133
Q

Each phase of a substance has its own specific heat

A

-the slope of the heating curve (heat(x-axis) v. temperature(y-axis)) is the reciprocal of specific heat

134
Q

Enthalpy and entropy are positive for

A

melting and vaporizing

135
Q

Melting and boiling are

A

endothermic

136
Q

Enthalpy and entropy are negative for

A

condensing and freezing

137
Q

Condensing and freezing are

A

exothermic

138
Q

Triple Point (phase diagram)

A

the one point where a substance can exist in equilibrium as a solid, liquid, and gas

139
Q

Critical Temperature (Phase diagram)

A

The temperature above which a substance cannot be liquefied regardless of pressure applied

140
Q

Critical Pressure (phase diagram)

A

The pressure required to produce liquefaction while the substance is at the critical temperature

141
Q

Critical Point (phase diagram)

A

-defined by the critical temp and pressure

142
Q

Supercritical fluid

A

fluid beyond the critical point has characteristics of both gas and liquid
-During the phase change temp goes towards breaking bonds not increasing temperature

143
Q

Colligative Properties

A

-depend only upon the number of molecules, not the type.

There are 4 colligative properties: vapor pressure, boiling point, freezing point, osmotic pressure

144
Q

Vapor Pressure

A

-Adding a nonvolatile solute lowers the vapor pressure so the remaining surface area has to have more molecules vaporizing and condensing…this increases the boiling point

145
Q

Boiling Point Elevation

A

-The equation for an ideally dilute solution due to the addition of a nonvolatile solute is:
deltaT=kbmi
-Kb is a substance dependent constant
-m is molality
-I is the van’t Hoff factor- the number of particles into which a single solute particle will dissociate when added to solution
-You can’t use this equation for volatile additions
-Making predictions: An endothermic heat of solution indicates weaker bonds, which leads to higher vapor pressure, so you can predict the boiling point will go down.

146
Q

Freezing Point Depression

A

deltaT=kfmi
-If you add a liquid solute the impurities will initially lower the melting point but as the mole fraction of the “impurity” goes up the solvent becomes the impurity preventing the solute from freezing. This the freezing point of a solution will rise as solute is added.

147
Q

Osmotic Pressure

A
  • Measure of the tendency of water to move into solution via osmosis
  • The pressure pulling a liquid into a solution
  • II=iMRT
  • M is molarity
  • i is number of dissociated particles
  • Let’s say you have a semi-permeable membrane separating two side of a U-shaped tube filled with water. If you have more salt on one side and the membrane is only permeable to water the water will flow more to the salt side. The pressure at the bottom of the solution on the salt side will be greater because more liquid is above it. The difference in pressure is the osmotic pressure
  • Extra pressure on solution side is osmotic pressure
148
Q

Osmotic Potential

A
  • Partial measure of a system’s free energy
  • When a solute is added osmotic potential becomes negative
  • Water flows from higher osmotic potential to lower osmotic potential
149
Q

Arrhenius Acids and Bases

A

Acid-Anything that produces hydrogen ions in aqueous solution
Base-anything that produces hydroxide ions in aqueous solution

150
Q

Bronsted and Lowry

A

Acid-anything that donates a proton

Base-anything that accepts a proton

151
Q

Lewis

A

Acid-anything that accepts an electron pair
-all simple cations except alkali and heavier alkaline earth metal cations
-Smaller the cation and the higher the charge, the stronger the acid strength
Base- Anything that donates an electron pair
-most general, includes all B/L

152
Q

Conjugate Acid/Base

A
  • The reactants are referred to as the acid and base while the products are referred to as the conjugate acid and conjugate base
  • Strong acids have weak conjugate bases
  • However, weak acids may have strong or weak conjugate bases
  • Weak acid/weak conjugate is possible
153
Q

Amphoteric

A

substances that can act as either acids or bases depending on their environment

154
Q

Strong Acids

A 
Hydroiodic Acid (HI)
Hydrobromic Acid (HBr)
Hydrochloric Acid (HCl)
Nitric Acid (HNO3)
Perchloric Acid (HClO4)
Chloric Acid (HClO3)
Sulfuric Acid (H2SO4)
155
Q

Strong Bases

A 
Sodium Hydroxide (NaOH)
Potassium Hydroxide (KOH)
Amide Ion (NH2-)
Hydride Ion ( H-)
Calcium Hydroxide (Ca(OH)2
Sodium Hydroxide (Na2O)
Calcium Hydroxide (CaO)
156
Q

The second proton donated by a polyprotic acid can usually be ignored if the

A

Ka values differ by >10^3

157
Q

Percent dissociation of an acid decreases with the acidity of the solution but more concentrated acids are

A

more acidic

158
Q

Three factors that determine acidity

A
  1. Strength of the bond holding the hydrogen molecule
  2. The polarity of the bond (polar bonds are easier to separate)
  3. Stability of the conjugate base

H-F. H-Cl. H-Br. HI

Increasing Acidity

159
Q

Keeping conjugate stability in mind, the molecules with the most oxygens makes the

A

strongest acid

160
Q

Hydrides

A

-Acidity increases down (period 2 to 3) the periodic table and to the left (from groups 4A to 6A

161
Q

Equilibrium constants for Acid Base reactions

A 
-Autoionization of water
Kw=10-14
pH+pOH=14
-Acid dissociation constant
HA+H2O---> H3O+ +A-
Ka=[H+][A-]/[HA]
-Base dissociation constant 
The equilibrium constant for the reaction of the conjugate base with water 
NOT the reverse of the reaction for Ka
A-+H2O--->OH-+HA

KaKb=Kw
The larger the Ka and the smaller the pKa the stronger the acid
-A Ka>1 or a pKa<0 indicates a strong acid. The same is true for a base

162
Q

Finding the pH of strong acids and bases

A
  • Very strong acids and bases dissociate almost completely
  • Ex. If you have a 0.01M HCl solution it is completely dissociated so you will have 0.01M of H+ ions. Since 0.01=10^-2 the pH is 2
163
Q

Finding the pH of weak acids and bases

A

Ex. Supposed you have 0.01M HCN and Ka=6.2x10^-10
Ka=[H+][CN-]/[HCN]=6.2x10^-10
6.2x10^-10=[x][x]/[0.01-x] can be approximated to x^2/[0.01]
-Thus x~2.5x10^-6. This is much smaller than 0.01 so the approximation stands
-For a weak base the process is the same, except that you use Kb and your result is the pOH
-Subtract the pOH from 14 to get pH

164
Q

Titration

A
  • A drop by drop mixing of an acid and base
  • Performed to find the concentration of some unknown by comparing it with the concentration of the titrant
  • The titration curve is sigmoidal when titrating a strong acid with a strong base
  • The midpoint is called the equivalence point or the stoichiometric point
  • This is also the point where there are equal amounts of acid base
  • Titrating a strong acid with a strong base means slowly adding base to an acid…
  • The point where the curve has the greatest slope
  • Thus if it’s the other way around it would be inverted sigmoidal graph
165
Q

Titrating a weak acid with a strong base the graph shoots up…

A

then goes flat, then shoots up again and then goes flat again.

166
Q

If base is stronger than the acid the equivalence point is

A

> 7

167
Q

If the acid is stronger than the. base the equivalence point is

A

<7

168
Q

Half equivalence point

A

point where exactly one half of the acid has been neutralized by the base
pH=pKa
-In other words, concentration where there is the same amount of acid as there is its conjugate base
-Occurs where there is a flat part of the graph…point where you could add the largest amount of acid or base with the least change in pH
-Is the point where solution is most buffered

169
Q

Henderson-Hasselbalch equation

A

pH=pKa+log([A-]/[HA])

170
Q

To make a buffered solution

A
  1. you would start with an acid whose pKa is closest to the pH at which you would want to buffer your solution
  2. Mix equal amounts of that acid with its conjugate base
  3. Would want the concentration of the buffer solution to greatly exceed the concentration of the acid or base that will be added to it.
    - A buffer solution is made from equal and copious amounts of a weak acid and its conjugate base
171
Q

You can’t use the Henderson-Hasselbalch equation to find the

A

pH at the equivalence point
-You must use the Kb of the conjugate base. You can find the Kb from the Ka and Kw. The concentration of the conjugate base at the equivalence point is equal to the number of moles of acid divided by the volume of acid plus the volume of base used to titrate
Kb=Kw/Ka
Kb=[OH-][HA]/[A-]. Solve for [OH-] to find the pOH
-Don’t forget to do
14 -pOH=pH

172
Q

Indicators

A
  • To find the equivalence point use an indicator
  • Usually a weak acid whose conjugate base is a different color
  • Example: We have an indicator HIn where In- represents the conjugate base and we add a small amount of indicator to the acid so as not to disturb the pH. At low pH the HIn form of the indicator predominates. As we titrate and the pH increases the In- form also increases. When the In-concentration reaches 1/10 of the HIn concentration a color change can be detected by the human eye. If we titrate a base with an acid the process works in reverse. Thus, the pH of the color change depends upon the direction of titration. The pH values give the range of the indicator.
173
Q

Range of indicator can be predicted by Henderson Hasselbalch

A

lower range: pH=pKa +log(1/10)—->pH=pKa-1
Upper range: pH=pKa+log(10/1)—>pH=pKa+1

The point where the indicator changes color is called the endpoint

174
Q

Oxidation

A

Loss of electron

175
Q

Reduction

A

Gain of electron

176
Q

Oxidation states

A

possible charge values that an atom may hold within a molecule

177
Q

Oxidation state charts

A

Oxidation state Atom
0 Atoms in elemental form
-1. Fluorine
+1. Hydrogen (Except when bound to a metal, then -1)
-2. Oxygen (except when it is in peroxide like H2O2

General Guidelines below, can't always be true or Redox would never occur
Oxidation State.   Group on PT
\+1.           Alkali Group 1
\+2 alkaline earth metal group2
-3       Nitrogen Family
-2    Oxygen Family
-1.   Halogens
178
Q

Oxidation states must

A

add up to the total charge on the ion

179
Q

Losing electrons

A

increases oxidation state

180
Q

Reducing agent (reductant)

A

Reducing agent gets oxidized because it is giving up electrons

181
Q

Oxidizing Agent (oxidant)

A

Oxidizing agent gets reduced because it is taking electrons

182
Q

Reducing or oxidizing agents are compounds NOT

A

atoms.
Example: Cd(s)+NiO2(s)+2H2O(l)—->Cd(OH)2(s)+Ni(OH)2(s)
-Ni is reduced, NiO2 is the oxidizing agent

183
Q

Oxidation-reduction titrations

A

-To find the molarity of a reducing agent you titrate with an oxidizing agent. You know what concentration you have by balancing the change of electrons. Instead of measuring pH you measure voltage.
There is an electric potential E associated with any redox reaction since in a redox reaction electrons are transferred
-Since the reverse of a reduction half potential is an oxidation half potential you just switch the signs.
-Reduction potential format:2H^+ +2e—->H2, E knot =0.00V

Do NOT multiply the half reaction potential by the number of times it occurs because voltage is an intensive property!!

184
Q

Balancing Redox reactions

A
  1. Divide reaction into corresponding half reactions
  2. Balance the elements other than H and O
  3. Add H2O to one side until all O atoms are balances
  4. Add H+ to one side until all the H atoms are balanced
  5. Add e- to one side until the charge is balanced
  6. Multiply each half reaction by an integer so that an equal number of electrons are transferred in each reaction
  7. Add the two half reactions and simplify
185
Q

Galvanic (voltaic) cell

A
  • If two distinct electrically conducting chemical phases are placed in contact and one charged species from one phase can’t freely flow to the other has a tiny amount of charge difference occurs resulting in a potential. By offering an alternative pathway for the electrons to flow you can convert chemical into electrical energy.
  • All phases must conduct electricity but one phase has to be impermeable to electrons
  • The phase that is impermeable to electrons is an ionic conductor carrying the current in the form of ions–usually an electrolyte solution in the form of a salt bridge. Salt bridges aren’t necessary when both anode and cathode are in the same solution.
186
Q

Anode

A

Negative terminal, where ions are oxidized (where they lose electrons)

187
Q

Cathode

A

Positive terminal, where ions are reduced (where they gain electrons)

188
Q

Half cell

A

Strip of metal and solution can be referred to as a half cell

189
Q

Cell potential E /electromotive force (emf)

A
  • the potential difference between the terminals when they are NOT connected
  • Once they are connected the potential difference is reduced due to internal resistance in the galvanic cell
  • the drop in the emf increases as the current increases
190
Q

Current

A

Flows in direction opposite electron flow

191
Q

Electrons in the anode have

A

higher potential energy than those in the cathode so electrons flow anode to cathode

192
Q

Cell potential for a galvanic cell is

A

ALWAYS positive

193
Q

Real cell potential depends upon

A

the half reactions, concentrations of reactants and products, and temperature

194
Q

When a cell has two different solutions a

A

liquid junction has to be used to separate the solutions. Because ions can move across a liquid junction any liquid junction creates an additional small potential difference that affects the potential of the galvanic cell. A salt bridge is a type of liquid junction that minimizes this

  • A good salt bridge provides an ionic conductor between two solutions without creating a strong extra potential within the galvanic cell
  • As electrons leave one half of a galvanic cell and flow to the other, a difference in charge is established. If no salt bridge was used, this charge difference would prevent further flow of electrons. A salt bridge allows the flow of ions to maintain a balance in charge between the oxidation and reduction vessels while keeping the contents of each separate. With the charge difference balanced, electrons can flow once again, and the reduction and oxidation reactions can proceed.
195
Q

A positive cell potential indicates

A

a spontaneous reaction delta G=-nFEmax

196
Q

When all the conditions are standard deltaG knot =

A

-nFEknot

197
Q

For non-standard conditions delta G=

A

=deltaGknot+RTln(Q)

Q is the reaction quotient

198
Q

Delta G knot is specifically described delta G with

A

standard state conditions (1M products and reactants, 1 atm pressure)

199
Q

At equilibrium delta G =0 so delta G knot =

A

-RTln(K)
-K is the equilibrium makeup
-If K=1 then deltaG knot=0
-If K>1 then delta G knot <0
If K<1 then delta G knot >0
This does not mean that id a reaction have K>1 that then the reaction is spontaneous because the spontaneity of a reaction depends on the starting concentrations of products and reactants. It does say that the reaction is spontaneous at standard conditions though

200
Q

How to find potential when concentrations aren’t 1M?

A

E=Eknot-(RT/nF)ln(Q) Nernst Equation

201
Q

Concentration Cells

A
  • A type of galvanic cells
  • Never at standard conditions so you need the Nernst equation to solve for the potentials
  • A limited form of a galvanic cell with a reduction half reaction taking place in one half cell and the reverse of that half reaction taking place in the other half cell
  • The more concentrated side will become less concentrates and electrons will flow accordingly
  • Example: 0.01M Fe2+ on one side and 0.1M Fe2+ at the other side
  • Electrons will flow to 0.1M Fe2+ side to lower the amount of Fe2+ on that side
202
Q

Electrolytic cells

A
  • If we hook up a power source across the resistance of a galvanic cell and force it to run backwards we get an electrolytic cell.
  • ANY electrolytic cell has a negative emf
  • The cathode is marked negative and the anode is marked positive. RED CAT AN OX still applies
  • Often used for metal plating and purifying metals.
  • Example: Piece of copper at cathode end. Electrons flow to cathode where there is a metal spoon. These ions are attracted to the spoon that has an excess of electrons so they plate onto the spoon and form a Cu surface at the cathode.

-A reaction will not run in an aqueous solution if water has a less negative reduction potential than the element trying to be purified.
-Example: Collecting pure sodium through electrolysis
Na+ + e- —-> Na. Eknot=-2.71V
2Cl- —>2e-+Cl2 Eknot=-1.36V
H2O(l)+2e- —>H2(g)+2OH-(aq) Eknot =-0.83 V
Sodium can’t be reduced because water will take the electrons

MCAT Physics Equations MUST MEMORIZE (6 decks)

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