Equations we use in Chem 132 - Chemistry

Kinetics equations1) Rate laws and integrated rate laws: describe how concentrations of reactants andproducts vary with timeAs an example, consider a first order reaction X → products.Then d[X(t)]/dt = - k[X(t)] is the rate law.An equivalent statement is the “integrated” rate law, [X(t)] = [X(0)] exp(-kt).In these equations, k is the rate constant and has units sec -1 for the first order reaction.These formulae would be different for reactions with different orders.Use the rate law to determine the concentration of a reaction at a given time. You canalso use it to confirm or rule out hypothetical reaction mechanisms.2) Arrhenius Law: describes the temperature dependence of reaction rate constantsk = A exp(-Ea/RT)Ea is the activation energy (height of the mountain we must climb to reach the transitionor intermediate state). Its units are KJ/mol. R is 8.31 J/mol-K and T is temperature andMUST be expressed in Kelvin.A is the prefactor which includes attempt frequency (how many times we try to go overthe hill) and the “luck factor”. Rigorously, the latter has to do with the relative number ofmicrostates in the transition state versus reactant state but a good example of this latterpart is the fraction of collisions – attempts to go over the hill - where the orientation isfavorable for reaction. Units of A are those of the rate constant and depend on the orderof the reaction.Recall that this equation came from calculating the fraction of molecules in a Boltzmannenergy distribution with enough energy to get over the hill. Use this equation todetermine reaction rates with temperature.

Electrochemistry equations:1) Relationship of Gibbs Free energy to electrical work∆G = -nFEElectrical work is QE where Q is the charge moved and E the potential drop it movesthrough. Here, Q is expressed in units of Coulombs as n, the number of moles ofelectrons transferred in a redox reaction, times the Faraday constant F, which is thenumber of Coulombs in a mole of electrons (F = 96,485 coul/mole). E is in Volts.∆G is the usual Gibbs free energy change for the redox reaction in units of KJ/mol.Note that 1 J = 1 coul x 1 Volt.We use this equation to relate what we learned in thermodynamics where ∆G was used todetermine spontaneity and equilibrium constants to electrochemistry where potential isused to describe the driving force in redox reactions.Note that since n and F are always positive numbers, the criterion for spontaneity ∆G < 0translates to E > 0 for redox reactions.Use this equation when you want to translate potentials for redox reactions into ∆G.2) Nernst equation: enables one to compute redox potential under non-standardconditions (temperature not 298 K and/or concentration not 1 M).E = E 0 – (0.0592/n) log 10 QE and E 0 are the cell potentials under non-standard and standard conditions respectivelyand are expressed in Volts.n is the number of moles of electrons transferred in the redox reaction (deduced from thestoichiometry of the reaction).Q is the reaction quotient.This equation comes directly from the equation ∆G = ∆G 0 + RT ln Q we used inthermodynamics for non-standard conditions. It uses the relationship of ∆G to potential Eabove to convert this to an equation for potential under non-standard conditions. Theconstant 0.0592 is just (RT/F ln 10) where it has been assumed that T = 298 K and wehave converted to base 10 logarithms. Thus, if you wanted to know cell potentials atother temperatures, you could use the more complicated form E = E 0 – (RT/n F) ln Q.The Nernst equation is useful to calculate voltage from concentration (or vice versa) ingalvanic cells or sometimes voltage versus time as concentration changes.

3) Equilibrium constants of redox reactions.log 10 K eq = n E 0 /0.0592This is just the Nernst equation expressed at equilibrium (where Q = K eq by definitionand ∆G = -n F E = 0). Again, n is the number of moles of electrons transferred in theredox reaction and E 0 is the standard cell potential in Volts.This equation is used to determine concentrations in galvanic cells when equilibrium isreached (i.e. when the redox reaction stops so the battery is dead).Note that another way to write this (see above where we have not assumed that T = 298K) is K eq = exp (n F E 0 /RT). This means that when n F E 0 is a few times RT, theredox reaction will essentially go to completion. Galvanic cells with potentials muchlarger than 0.0592 V will tend to end up completely reacted.Quantum mechanics equations:1) Relationship between photon energy, frequency and wavelength:E = hν = hc/λLight behaves as discrete particles called “photons” in many circumstances (e.g.photoelectric effect). The properties of photons are expressed by this equation.E is the photon energy (standard units of Joules), h is Planck’s constant (6.63x10 -34 J-sec), ν is the frequency of the light in sec -1 (Hz) and λ its wavelength (in meters). Thespeed of light is given by c = 3x10 8 m/sec.2) DeBroglie wavelength:λ = h/mvIn this case, λ refers to the DeBroglie wavelength of a particle. Particles sometimesbehave as waves as for example in electron diffraction or electrons in atoms where theyexhibit the wave-like property of interference. The particle mass (in kg) is represented bym and its velocity (m/s) by v. The significance of this equation is that it tells you whenwe need to think of particles as waves (i.e. when should we use the Schrödinger equationinstead of Newton’s Laws to describe their behavior). When the wavelength of theparticle is comparable to or larger than the size of the system under study (for Chemistry,this means atomic dimensions), you need to use a quantum mechanical description andyou can expect particles to have wavelike properties. This is not an either/or situation –quantum mechanics always works and degenerates to Newton’s Laws for macroscopicobjects but it is not necessary to invoke its complexity when the DeBroglie wavelengthsare small compared to the system of interest.

3) Heisenberg uncertainty principle:∆p∆x ≥ h/4πThis equation states that a particle’s momentum p and position x cannot be knownsimultaneously to better than a certain precision. ∆p stands for the uncertainty inmomentum (p = mv) and ∆x for the uncertainty in position. Once again, h is Planck’sconstant. You can think of this in two ways. One interpretation is that the act ofmeasuring one of these quantities changes the other so that it cannot be measured withgreater than a certain precision. Another interpretation is that these quantities really donot have precise meaning simultaneously so that asking about a particle’s exact positionand velocity is just not a meaningful question.4) Schrödinger wave equationHψ = Eψ :(-h 2 /8π 2 m)d 2 ψ/dx 2 + V(x)ψ = EψHere H is a mathematical operator called the Hamiltonian. It has units of energy and ismeant to describe the total energy (kinetic plus potential) of a particle or set of particles.Again, h stands for Planck’s constant, m is the particle mass and x the particle position(this would be r in three dimensions). V(x) is the potential energy that the particle feels.For us in Chem 132, the most important potential is the Coulomb potential (see below)describing the attraction of electrons to nuclei and interelectron repulsion. We have alsoalluded to the harmonic potential used for springs (Hooke’s Law) describing bondvibrations of molecules. Typically, the input to the Schrodinger equation is V(x) and wesolve for the allowed energies E that the particle can have and the correspondingwavefunctions ψ(x) that the particle can have. In general, ψ(x) is a complex number andits squared modulus |ψ(x)| 2 is proportional to the probability of finding the particle at thelocation x. I would like to explain to you why the Schrödinger equation works and why|ψ(x)| 2 describes its location in a probabilistic way at a more fundamental level that isintuitive but I can’t. I suggest treating it as a brute fact that it works and explainsexperimental data exquisitely well.5) Bohr formulaE n = -Z 2 me 4 /(8n 2 h 2 ε 0 2 ) = 2.178 x 10 -18 (Z 2 /n 2 ) (in Joules)When we solve the Schrödinger equation with the Coulomb potential V = -Ze 2 /4πε 0 r, theallowed energies for an electron bound to a proton that we get are described by the aboveformula. We were able to “derive” this formula by using Bohr’s idea that electrons travelin circular orbits and that the radii are selected by the criterion that the DeBrogliewavelength of the electron folds back onto itself. That assumption plus some classicalmechanics gave us the same result. This formula applies onto to one electron atoms likeH, He + , Li 2+ , etc. Z stands for the atomic number (number of protons in the nucleus) andn is an integer (1,2,3, …) we call the principal quantum number. Remember that the case

where n = ∞ corresponds to a free electron (E ≡ 0) so that E 1 is also the ionization energyfor the hydrogenic atom in question.