Marjy's Inorganic Chemistry Formulas

Front 1

Planck's Quantum Theory

Back 1

∆E = hƒ

ƒ is frequency
h is planck's constant (6.6E-34 J*s). Light energy in discreet quanta

Hint 1

Me is energy of light.

Front 2

Universal gas constant?

Back 2

R = .08206 L*atm/K*mol
R = 8.314 J/K*mol

Front 3

Ideal gas law

Back 3

PV = nRT
P: pressure
V: volume
n: number of moles
R: universal gas constant = 8.314 J/Kmol
T: temperature in kelvin

Front 4

Standard molar volume at STP

Back 4

22.4 Liters

At STP, 1 mol of ANY ideal gas will occupy 22.4 L.

Front 5

Mole fraction (X)

Back 5

number of mole of gas "a" divided by total number of moles of gas in sample

Front 6

Partial pressure

Back 6

P(a) = (Xa) (Ptotal)

P(a): partial pressure of gas "a"
Xa: mole fraction of gas "a" (Mole fraction is # of moles of gas "a"/ total # of moles of gas in sample)
Ptotal: total pressure of gaseous mixture

Front 7

Average translational kinetic energy

Back 7

KE = (3/2) RT
KE: kinetic energy, found from RMS velocity

This is the avg kinetic energy for one mol of gas molecules at a given moment

Front 8

Graham's Law

Back 8

Ratio of RMS velocities of 2 gases in homogeneous mixture

V1/V2 = square root (m2)/square root (m1)
V: RMS velocities
m: mass of gas molecules

Graham's law describes effusion and diffusion

Front 9

Effusion

Back 9

Spreading of gas from high pressure to low pressure through a pinhole

effusion rate 1/effusion rate 2 = √(M2) / √(M1)
M: molecular weights

Front 10

Diffusion

Back 10

Spreading of one gas into another gas, or into empty space

diffusion rate 1/diffusion rate 2 = √(M2) / √(M1)

Front 11

Heat current (I)

Back 11

I = Q/t

The rate of heat flow

In a steady state system the rate of heat flow is constant across any number of slabs between two heat reservoirs

Front 12

Energy flow

Back 12

heat flow:
∆T = IR
T: temperature
I: heat current (Q/t)
R: resistance to heat flow

electron flow:
∆V = IR
V: voltage
I: electrical current
R: resistance to electrical flow

Fluid flow:
∆P = QR
P: pressure
Q: heat
R: resistance to flow

thicker conduits = greater flow
longer conduits = less flow

Front 13

PV work

Back 13

at constant pressure, work is equal to pressure multiplied by change in volume:

W = P(∆V)
[at constant pressure]

No PV work is done if volume is constant!

a system at rest can still do PV work.

PV work takes place when a gas expands against a force regardless of whether or not the pressure is constant

Work is a PATH FUNCTION: a different pathway results in a different amount of work

Front 14

1st Law of Thermodynamics equation

Back 14

any energy change to a system must equal heat flow into system + work done by system

∆E = Q + W

Work done on the system is positive and work done by the system is negative

Front 15

The average kinetic energy of a single molecule in any fluid

Back 15

K.E. (avg per mole) = 3/2 RT

Front 16

Enthalpy (H)

Back 16

∆H = ∆U + P(∆V)

(constant pressure!)
H: enthalpy (joules)
U: internal energy
PV: Pressure x Volume

Front 17

Change in enthalpy at constant pressure with PV work only?

Back 17

∆H = Q
constant pressure, closed system at rest, PV work only
H: enthalpy
Q: heat

Front 18

Heat of reaction equation

Back 18

the change in enthalpy from reactants to products

∆Hreaction = (∆Hf products) – (∆Hf reactants)

Front 19

Gibbs Free energy (G)

Back 19

The maximum non–PV work available from a reaction. It is an extensive property and a state function

∆G = ∆H – T∆S
G: gibbs free energy
H: enthalpy
T: temperature (constant)
S: entropy

All variables refer to the system and not surroundings

only good for constant T & P reactions

–∆G = spontaneous reactions
∆G = 0 reaction is in equilibrium
+∆G = non–spontaneous reaction

G of the universe is not conserved, an isolated system can change its gibbs free energy

A higher temperature will favor the direction favored by entropy.

Front 20

Molarity (M)

Back 20

moles of a compound/solute divided by volume of solution

mol/L

M = moles solute/volume solution

Front 21

Molality (m)

Back 21

mol/kg

m = moles solute/kg solvent

Front 22

mole fraction (X)

Back 22

no units, since it is a ratio

X = moles solute/total moles of all solutes and solvents

Front 23

mass percentage

Back 23

mass % = (mass solute/total mass solution) x 100

Front 24

Part per million (ppm)

Back 24

ppm = (mass solute)/(total mass solution) x 10^6

Front 25

heat of solution (∆H(sol))

Back 25

the overall change in energy of the reaction in the solution (enthalpy)

∆H(sol) = ∆H1 + ∆H2 + ∆H3

–∆H(sol) = stronger, more stable bonds
+∆H(sol) = weaker, less stable bonds

Front 26

Total pressure of solution?

Back 26

total pressure of solution is the sum of partial pressures

Pv = XaPa + XbPb
Pv: total vapor pressure
XP: partial pressure of respective solvent

Front 27

Raoult's law

Back 27

vapor pressure of a solution is proportional to mole fraction of liquid and vapor pressure of pure liquid

Pv = (Xa)(Pa)
Pv: vapor pressure of solution
Xa: mole fraction of liquid a
Pa: vapor pressure of pure liquid a

Front 28

Solubility product (Ksp)

Back 28

the equilibrium constant of a solvation reaction. It is a constant found in a book. Unitless.

pure solids and liquids are EXCLUDED from equilibrium expression because have mole fraction of 1

Ksp = [products]^coefficient/ [reactants]^coefficient

changes ONLY with temperature

Front 29

Henry's Law

Back 29

demonstrates that solubility of gas is proportional to its vapor partial pressure

as pressure decreases, solubility of gas decreases (ex: opening can of soda)

C = ka1Pv or Pv = Xaka2

C: solubility of gas a (mol/L)
ka1: henry's law constant
Pv: vapor partial pressure of gas a above the solution

Front 30

Heat Capacity (C)

Back 30

measure of energy change needed to change temperature of substance

C = Q/(∆T), Q = C(∆T)

always positive, temperature will always increase when added to a substance at constant volume or pressure

heat capacity does not change with temperature

Front 31

Specific heat capacity (c)

Back 31

heat capacity per unit mass

Q = mc∆T
m: mass
c: specific heat capacity
T: temperature

c(water) = 1 cal/g˚C (definition of 1 calorie)

Front 32

Van't Hoff Factor (i)

Back 32

number of particles into which a single solute particle will dissociate when added to solution

Front 33

freezing point depression

Back 33

equation for ideally dilute solution

impurities (solute) interrupt crystal lattice and lower freezing point

only nonvolatile solutes

∆T = kfmi
T: temperature
kf: specific constant substance being frozen
m: molality (mol/L)
i: van't hoff factor (# particles dissociated)

Front 34

osmotic pressure

Back 34

osmotic pressure = iMRT
M: molarity
R: resistance
T: temperature
i: van't hoff factor (# dissociated particles)

measure of tendency of water (or some solvent) to move into solution via osmosis

only relevant when comparing 1 solution with another

pressure pulling into a solution

Front 35

pH

Back 35

measure of H+ ion concentration (mol/L)

pH = –log[H+]

scale runs from 0 to 14, each point on pH scale corresponds to 10X difference in H ion concentration

acid at pH 2 produces 10X as many H ions as acid at pH 3 and 100X as many H ions as acid at pH 4

at 25˚C: pH of 7 is neutral, lower pH is acidic and higher pH is basic

Front 36

autoionization of water

Back 36

H2O + H2O ––> H3O+ + OH–

Kw = equilibrium constant for this reaction

Kw = [H+][OH–]

at 25 degrees: Kw = 10^–14 (lies far to left)

[H+] = [OH–] = 10^–7 mol/L = pH 7

pH + pOH = pKw = 14

Front 37

acid dissociation constant (Ka)

Back 37

acid has its own equilibrium constant in water

HA + H2O ––> H3O+ + A–
Ka = [H+][A–]/[HA]

Front 38

base dissociation constant (Kb)

Back 38

for every Ka there is a Kb

equilibrium constant for reaction of conjugate base (A–) with water

A– + H2O ––> OH– + HA
not reverse reaction of Ka

Kb = [OH–][HA]/[A–]

Front 39

product of Ka & Kb

Back 39

Ka X Kb = [H+][OH–] = Kw

KaKb = Kw

pKa + pKb = pKw = 14

larger the Ka, smaller the pKa and stronger the acid

Ka greater than 1 or pKa less than 0 indicates a strong acid

Front 40

Henderson–Hasselbalch equation

Back 40

Predicts where at the half equivalence point the pH of the solution is equal to the pKa of the acid

pH = pKa + log([A–]/[HA])

form of equilibrium expression for Ka

when log(1) = 0, [A–] = [HA], pH = pKa which is at the half–equivalence point

Front 41

Finding pH at half–equivalence point

Back 41

Ka = [H+][A–]/[HA] = [H+] [A–]/[HA]

–log(Ka) = –log[H+] + –log([A–]/[HA])

pKa = pH – log[A–]/[HA]

pH = pKa + log[A–]/[HA]

Front 42

Finding pH at equivalence point

Back 42

Kb = Kw/Ka

Kb = [OH–][HA]/[A–]

solve for OH– concentration

Find pOH

Subtract pOH from 14 to find pH

Front 43

Reactions that do not occur at standard state

Back 43

∆G = ∆G˚ + RT[ln(Q)]
∆G: Gibbs free energy
∆G˚: Gibbs free energy (standard conditions)
T: temperature
Q: reaction quotient

Front 44

reactions that are at equilibrium conditions

Back 44

at equilibrium, there is no available free energy with which to do work; ∆G = 0

∆G˚= –RT[ln(K)]