formulas !! HIGH YIELD

Planck relation (frequency)

E = hƒ

h = 6.626×10^-34 m²*kg/s

angular momentum on an electron (bohr model)

L = nh/2π

energy of an electron (bohr model)

E = - R_H/n²

Planck relation (wavelength)

E = hc/λ

Energy of electron transition (Bohr model)

E = R_H[1/n_i² - 1/n_f²]

maximum number of electrons within a shell

2n²

max number of electrons within a subshell

4l+ 2

l = n-1 (between 0 and n-1)

Dipole moment

p = qd

Formal charge

V- N_nonbonding - 1/2 N_bonding

Valence electrons -dots - sticks

moles from mass

Moles = mass of a sample/ Molar mass

gram equivalent weight

GEW = molar mass / n

(where n = number of equivalent ions of desired)

equivalents from mass

equivalents = Mass of compound/ gram equivalent weight

molarity from normality

Molarity = normality / n

percent composition

% composition = mass of element in formula/ molar mass x 100%

percent yield

% yield = actual yield/ theoretical yield x 100 %

avogadro’s number

6.022 × 10^(23) mol ^-1

normality

N = equivalents/ L

equivalents = Mass of compounds (g)/ gram equivalent weight (g)

1 mole of any ideal gas at STP =

22.4 L

Collision theory

rate = Z x ƒ

Z = total number of collision per second

f = fraction of collisions that are effective

rate = rate of rxn

Arrhenius equation

k = Ae^^(-E_a/RT)

where k = reaction rate constant

A = Frequency factor

Ea = activation energy (J/mol)

R = universal gas constant = 8.314 J/mol *K

T = Abs temp (Kelvin)

(e): Euler's number, a mathematical constant approximately equal to 2.718.

universal gas constant

R = 8.314 J/mol *K

definition of rate

rate = - ∆ [A] / a∆t = - ∆[B]/b∆t = ∆C/c∆t = ∆D/d∆t

when aA+bB → cC + dD

rate law

rate = k[A]^x[B]^y

when aA+bB → cC + dD

Radioactive decay

[A]_t = [A]₀e^^-kt

([A]t): The concentration of reactant (A) at a specific time ((t))

([A]0): The initial concentration of reactant (A) (at time zero, before the reaction starts).

(e): Euler's number, a mathematical constant approximately equal to 2.718.

k = The first-order rate constant. For a first-order reaction, its unit is always s^-1(per second) or another unit of reciprocal time

t = elapsed time

Q < K_eq; ∆G < 0

reaction proceeds in forward direction

Q = K_eq ; ∆G = 0

rxn is in dynamic equilibrium

Q > K_eq ∆G > 0

reaction proceeds in reverse direction

K_eq

ratio of products to reactants at equilibrium

Q reaction quotient

calculated value that relates the reactant and product concentrations at any givnen time during a reaction

K_eq >1

products are present in greater concentration at equilibrium

Keq ≈ 1

products and reactants are both present at equilibrium at reasonably similar levels

K_eq <1

reactants are present in greater concentrations at equilibrium

K_eq« 1

amount of reactants that have been converted to products can be considered negligible in comparison to the initial concentration of reactans

3 types of stress to a system

pressure and volume, temperature, and changes in concentration

increasing concentration of reactants or decreasing concentration of products will shift the rxn to

right

increasing the concentration of products or decreasing concentration of reactants will shift rxn

left

increasing pressure on a gaseous system (decreasing its volume) will shift the reaction toward side with

fewer moles of gas

decreasing pressure on a gaseous system, increasing volume will shift rxn toward

side with more moles of gas

increasing temp of an endothermic rxn or decreasing the temp of an exothermic rxn will shift rxn to

right

decreasing temp of an endothermic rxn or increasing the temp of an exothermic rxn will shift rxn to

left

kinetic product

higher in free energy, form at lower temps; fast products

thermodynamic productws

lower in free energy than kinetic products, more stable, proceed more slowly, more spontaneous

K_eq for aA+bB → cC + dD

[C]^c[D]^d/ [A]^a[B]^b

reaction quotient Q_c for aA+bB → cC + dD

Q_c = [C]^c[D]^d/ [A]^a[B]^b

First law of thermodynamics

∆U = Q - W

where

∆U = change in internal energy

Q = heat energy

W = work done

+Q vs -Q

(+Q) = Heat is absorbed by the system (endothermic).

(-Q) = Heat is released by the system (exothermic)

+W vs -W

(+W) = Work is done by the system on the surroundings (e.g., a gas expanding). Energy leaves the system, which is why it is subtracted.

(-W) = Work is done on the system by the surroundings (e.g., a gas being compressed). The two minus signs cancel out ((-(-W) = +W)), adding energy to the system.

heat transfer ( no phase change)

q = mc∆T

q = heat energy (joules or cal) +q = absorbed heat -q= released heat

m= mass (g) or kg

c = specific heat capacity (J*(g *°C)

∆T= Tf - T₀ ( Celsius or K)

heat transfer during phase change

q =mL

q = heat transer

m = mass

L = latent heat (J/g)

Generalized enthalpy of reaction

∆H_rxn = H_products - H_reactants

₊ means

-means exothermic

Standard enthalpy of reaction

∆H°_rxn = ∑H°_f,products -∑H°_f,reactants = total energy absorbed - total energy released

entropy

∆S = Q_rev/T ;(J/K)

+∆S = more disordered

-∆S = more ordered

Qrev= reversible heat (J)

T = temp (K)

second law of thermodynamics

∆S_universe = ∆S_system + ∆S_surroundings > 0

Standard entropy of rxn

∆S°_rxn = ∑∆S°_f,products+ ∑∆S°_f,reactants

Gibbs free energy

∆G = ∆H -T∆S

Standard Gibbs Free energy of rxn

∆G°rxn = ∑∆G°_f,products- ∑∆G°_f,reactants

Standard gibbs free energy from equilibrium constant

∆G°rxn = -RT ln K_eq

R = 8.314 J/(mol*K)

T = temp K

Keq = equilibrium constant

gibbs free energy from reaction quotient

∆Grxn = ∆G°rxn + RT ln Q = RT ln Q/ K_eq

if Q/Keq is less than one Q<Keq nat ln will be negative and free energy will be negative

if Q/Kee is greater than one Q>Keq then positive and free energy will be positive

if equal reaction will be ln 1 = 0

standard conditions

298 K, 1 atm, 1 M

fusion

melting

freezing

crystallization or solidification

vaporization

evaporation or boiling

condensation

gas to a liquid

Sublimation

solid to gas

deposition

gas to solid

+∆H + ∆S

spontaneous at Hi T

+∆H - ∆S

nonspontaneous at all T

-∆H + ∆S

spontaneous at all T

-∆H -∆S

spontaneous at low T

ideal gas law

PV = nRT

P = pressure (N/m² Pa)

V= VOLUME (m³)

n = moles (mol)

R = ideal gas (8.314 J/(mol*K) use (0.08206 L*atm/Mol*K if P in atm and V in L)

T = temp (K)

1 atm =

101.3kPA

Density of a gas

ρ=m/V = PM/RT

m = mass

V= volume

P = pressure

M = molarmass

R = ideal gas = 0.08026 L*atm/ (mol*K)

T = temp

density = mass over volume

Combined gas law

P₁V₁/T₁ = P₂V₂/T₂

P = Pressure

V = volume

T = temp

avogadro’s principle

n/V = k (constant)

n₁/V₁ = n₂/V₂

_{n= moles}

V = volume

k = constant so just assume set equal

Boyle’s Law

PV = k or P1V1=P2V2

Pressure

Volume

k = constant

Charles’s law

V/T = k

k = constant

V = volume

T = temp (K)

Gay-Lussac’s law

P/T = k or P1/T1 = P2/T2

pressure

T in Kelvin

Dalton’s law total pressure from partial

P_T = P_A+P_B+P_C ….

DAlton’s law partial pressure from total pressure

P_A = X_AP_T

Henry’s Law

[A] = k_H x P_A or [A]1/P1 = [A]2/P2

concentration of A = constant * pressure

avg kinetic energy of a gas

KE = 1/2mv² =3/2k_BT

kB is boltzman constant = 1.38×10^-23 J/K

T in kelvin

root mean square speed

u_rms = sqrt√(3RT/M)

R = 8.314 J/(mol*K) J=kg*m²/s²

T = kelvin

M = molar mass (kg/mol)

Graham’s law

r1/r2 = sqrt√(M2/M1)

M = molar mas

r = rate of effusion for gas 1 and 2 (V/t volume per time)

van der Waals equation of state

(P + n²a/V²) (V-nb) = nRT

Pressure, Volume, moles, T in K, R= 0.08206 atm*L/(mol*K)

a = intermolecular attraction constant (L²*atm/mol²)

b= moleculr volume constant (L/mol)

% composition by mass

mass of solute/mass of solution * 100%

mole fraction

X_A=moles of A/total moles of all species

molality

moles of solute/kilograms of solvent = m

Dilution formula

M_iV_i = M_fV_f

M= molarity

V = volume

solubility product constant

K_sp = [Aⁿ⁺]^m[B^m-]ⁿ

for A_mB_n ↔mAⁿ⁺(aq) + nB^m- (aq)

ion product

IP = [Aⁿ⁺]^m[B^m-]ⁿ

for A_mB_n ↔mAⁿ⁺(aq) + nB^m- (aq)

if IP< K_sp

solution is unsaturated, if more solute added it will dissolve

IP = Ksp

solution is saturated at equilibrium and no change in concentration

IP > Ksp

solution is super saturated, precipiatate will form

Raoult’s Law (vapor pressure depression)

P_A= X_AP°_A

_Xa= mole fraction

P°= pressure of pure

Pa is = pressure of solvent

boiling point elevation

∆T_b = iK_bm

i = # of particles it will split into vant hoff

m = molality (mol/kg)

Kb = molal boiling point elevation constant

∆Tb = Tboiling solution - Tboiling solvent (how much higher from pure state)

freezing point depression

∆T_f= iK_fm

m = molality (kg/mol)

i = vant hoff factor amount of particles splits into # , if no split = 1

Kf = fixed constant of freezing °C/m

∆Tf= Tfreezing pure - Tfreezing solvent

osmotic pressure

π= iMRT

(atms)

i = vant hott

M= molarity (mol/L)

R= 0.08206 L*atm/mol*K

T = temp in K

Hypochlorite

CLO-

CLO_2-

chlorite