| Name | Formula | Comments |
| n = c /l | Used to interconvert wavelength and frequency of
electromagnetic radiation. The speed of light is c = 3.00 x 108 m s-1. n has units of time-1, usually s-1. l has units of length, often nm if in the UV, visible or IR portions of the spectrum. Convert to m if using m s-1 for the units of c. | |
| Planck's equation | E = h n | E is the photon energy equivalent of an
electromagnetic wave having a frequency of n
(s-1). h is Planck's constant, 6.63 x 10-34 J s. The units of E are J (per photon). |
| de Broglie's equation | l = h/(mv) | Yields the equivalent wavelength of an object of mass m with speed v. |
| Balmer-Rydberg equation | 1/l = R[(1/m2 - 1/n2)] | Yields the wavelength (l) of light resulting from an electronic transition. n is the higher energy level, m is the lower energy level of the electron. R is the Rydberg constant, 0.01097 nm-1 |
| Bond Order | Bond Order = (Number of Bonding Electrons - Number of Antibonding electrons)/2 | Used to calculate the bond order after putting together the Molecular Orbital diagram |
| Enthalpy Change | DH = Hfinal -
Hinitial
DH = [sum of DHof of products] - [sum of DHof of reactants]
|
The enthalpy change of a chemical reaction is
positive if heat flows into the reaction, and negative if the reaction
generates heat. Enthalpy is a state function, meaning that the value DH will be the same, regardless of the path taken from
reactants to products, and regardless of the rate of the reaction.
DHof of a substance is the standard enthalpy of formation of the substance. We say "standard" because the quantity refers to the state of the substance (s, l, g or 1 M solution) at standard temperature and pressure (298 K, 1 atm). We say "formation", because the quantity refers to the heat absorbed or evolved when one mole of the substance is formed in its standard state from the elements in their standard states.
|
| Hess' Law | For any chemical change made in several steps, the net DH is equal to the sum of the DH values of the separate steps | The important point here is that the separate steps must add up to the correct reaction before you add up the DH values. If you must reverse a reaction, change the sign of DH. If you multiply it by an integer, multiply its DH by the same integer. |
| Specific heat capacity | c = q/(mDT) | The quantity c (expressed in J g-1oC-1) is the specific heat capacity, which is a property of the substance being heated or cooled. q is the quantity of heat (in J) which raises the temperature of a mass m (in g) by an amount DT (in oC). The specific heat capacity of water is 4.18 J g-1oC-1. |
| Boltzmann equation | S = k ln(W) | The entropy (S) of a system equals the Boltzmann constant k (1.38 x 10-23 J K-1) times the ln of W, the number of ways that that particular arrangement of particles in the system can be achieved. More probable arrangements have higher entropy. |
| Entropy change | DS = Sfinal -
Sinitial
DS = [sum of So of products] - [sum
of So of reactants] |
The entropy change of a chemical reaction is
positive if entropy increases, i.e. if the products have more entropy thsn
the reactants. This would be the case for an increase in the number of
moles, for meltinf or vaporization, for heating a substance, and so on.
Entropy is a state function, meaning that the value of DS will be the same, regardless of the path taken from
reactants to products, and regardless of the rate of the reaction.
So of a substance is the standard absolute entropy of the substance. |
| Free energy change | DG = Gfinal -
Ginitial
DGo = [sum of DGof of products] - [sum of DGof of reactants]
|
The free energy change of a chemical reaction is
negative if the reaction as written occurs spontaneously. Free energy is a
state function, meaning that the value DG will be
the same, regardless of the path taken from reactants to products, and
regardless of the rate of the reaction.
DGof of a substance is the standard efree energy of formation of the substance.
|
| Free energy change | DGo = DHo - TDSo | The free eergy change can also be calculated if the enthalpy and entropy change for the reaction are known |
| Equilibrium Constant | Keq = exp[-DGo/RT] | The equilibrium constant can be calculated knowing the standard free energy change for the reaction |
| Equilibrium Constant | For a reaction aA + bB ¾ cC + dD, Keq = {[C]c[D]d}/{[A]a[B]b} |
This can only be calculated using EQUILIBRIUM concentrations. At non-equilibrium, the expression yields Q, the reaction quotient instead of the equilibrium constant. |
| Boyle's Law | PV = k | Boyle showed that the product of pressure and volume is a constant. |
| Charles' Law | V/T = k' | Charles showed that the volume of a sample of gas is proportional to the absolute temperature. |
| Avogadro's Law | V = k''n | Avogadro found that the volume of a gas sample was directly proportional to the number of moles of gas. |
| Ideal Gas Law | PV = nRT | P = Pressure (atm), V = Volume (L), n = number of moles of gas, R = the Univeral Gas Constant (0.082057 L atm K -1 mol-1), T = temperature (K) |
| van der Waals equation | P = nRT/(V-nb) - a(n/V)2 | a and b are constants for each gas. The magnitude of b is related to the size of the gas molecules. The magnituse of a is related to the interactions among the gas molecules. |
| Gas Density | r = MW(P /(RT)) | r is the density of a gas in g/L if MW is expressed in g/mol, P is expressed in atm and T is the absolute temperature. Here again, R must bt 0.082057 L atm K -1 mol-1. |
| Pressure equilibrium constant | Kp = Kc(RT)Dn | For gas phase reactions, or at least a reaction involving at least one gas, the pressure based equilibrium constant is related to the concentration based equilibrium constant using the change in the number of moles of gas, Dn. |
| Average Speed of gas molecules | v = [3RT/MW]1/2 | The average speed of a molecule in a gas is related to the molecular weight (MW) of the gas, and to the absolute temperature. |
| Dalton's Law of Partial Pressures | Ptot = S(Pi) | The total pressure is the sum of the partial pressures of the different components in a gas mixture. |
| Graham's Law | rate1/rate2 = (MW2/MW1)1/2 | This works for effusion and for diffusion |
| Cell potential | Eocell = Eocathode + Eoanode | The cell potential is the sum of the anode (oxidation) reaction potential plus the cathode (reduction) potential. Note that the potentials in the tables are listed as reduction reactions. You must reverse the sign of the potential, and the reaction itself, to get the oxidation half-reaction. |
| Free energy of an electrochemical reaction | DG = -nFEcell
DGo = -nFEocell |
The free energy change of the reaction (DG) is related to n, the number of moles of electrons transferred in the balanced REDOX reaction, Faraday's constant, 96487 C/mol e-, and the cell potential, Ecell. This works under standard or non-standard conditions. |
| Nernst equation | Ecell = Eocell - (RT/nF) ln(Q) | Used to find the cell potential under non-standard conditions, i.e. when the temperature is not 298 K and/or the concentrations are not 1 M. The reaction quotient Q is calculated in the same manner as an equilibrium constant. |
| Total charge | q = i t | charge(C) = current (C/s) x time (s)
(Note that 1 Ampere = 1 C/s) |
| Total charge | q = nF | charge = moles electrons x Faraday's
constant (Note that "moles electrons" = the total moles of electrons passed through the external circuit. This is not the same as "n", the number of electrons in the balanced REDOX reaction. |
| Clausius-Clapeyron Equation |
|
The vapor pressure of a liquid (p2 at T2) can be found knowing the value at another temperature (p1 at T1) as long as the heat of vaporization is known (DHvap). Be able to solve this equation for any one of the variables. |
| Henry's Law | solubility = KH ·p | The solubility of a gas in a liquid! KH is the Henry's Law constant for the particular gas in the particular solvent (usually water) p is the partial pressure of the gas above the solution |
| Freezing Point Depression | DTf = Kfm | DTf is the
freezing point depression in oC. Kf is the cryoscopic constant of the solvent (oC kg mol-1) m is the molality of the solute |
| Boiling Point Elevation | DTb = Kbm | DTb is the
boiling point elevation in oC. Kb is the ebbulioscopic constant of the solvent (oC kg mol-1) m is the molality of the solute |
| Raoult's Law | psoln = posolv ½Xsolv | psoln
is the vapor pressure of the solution posolv is the vapor pressure of the pure solvent Xsolv is the mole fraction of the solvent in the solution |
| Osmotic Pressure | P = MRT | P is the osmotic pressure
(atm) M is the molarity of the solute R is the gas constant (8.314 J K-1mol-1) T is the absolute temperature |
| pH | pH = -log10[H3O+] | [H3O+] is the concentration of the hydronium ion expressed in mol/L. [H3O+] = 10-pH. |
| Henderson-Hasselbalch Equation | pH = pKa - log10{[acid]/[base]} |
Here, you need to be able to identify the acid and base in solution. (They must be conjugates of one another.) If Kb for the base is given, you must find Ka for the conjugate acid according to Ka = Kw/Kb |
| Ka | Ka = Kw/Kb | Relates the acid dissociation with the dissociation of its conjugate base. Kw = 10-14 in any aqueous solution |
| Differential rate law for a first order reaction | rate = -d[A]/dt = k[A] | |
| Differential rate law for a second order reaction | rate = -d[A]/dt = -k[A]2 | |
| Integrated rate law for a first order reaction | ln{[A]/[A]0} = kt | [A] is the concentration of the reactant after time t. [A]0 is its concentration at time 0. k is the rate constant, which for a first order reaction will have units of time-1 |
| Integrated rate law for a second order reaction | 1/[A] = 1/[A]0 + kt | As above, but k will have units of concentration-1 time-1 |
| Half life of a first order reaction | t1/2 = ln(2)/k = 0.693/k | This is the time required for the concentration of the reactant to decrease to one half of its original value. |
| Half life of a second order reaction | t1/2 = 1/k[A]0 | Note that half life depends on the initial concentration! |
| Nuclear Decay | rate = kN | k is the decay constant for the particular radioisotope, N is the number of nuclei of that isotope present. Nuclear decay is a first order process! The rate will be expressed in disintegrations per second, if k is expressed in s-1 |
| Nuclear decay half life | t1/2 = ln(2)/k = 0.693/k | |