Kinetics

Avr. Rate = Change in conc/change in time = deltaC/deltaT

Rate = d(conc.)/dt = derivative of concentration wrt time i.e. derivative of the line of a conc-time graph

+ve ROR means the conc. of the species is increasing with time -ve ROR means the conc. of the species is reducing with time

ROR has units of moldm^-3s^-1 usually [A] is the shorthand for conc. of whatever is inside the square brackets

2H2 + O2 —> 2H2O ROR = r = (1/va) (d[A] / dt) where va is the stoichiometric coeff va has -ve sign for reagents, +ve sign for products e.g. using example ROR = r = -1/2 d[H2]/dt d[H2O]/dt = -2 d[O2]/dt

Rate = k[A]^a [B]^b k = Rate constant or coefficient a, b are constants, these are powers which give the ‘orders’ wrt different species

r = k[A] is a first order rate equation, in this case k has units of s^-1 r = k[A]^2 or r = k[A][B] are examples of second order rate equations, the units of k are dm^3mol^-1s^-1

1) If we know the rate law and the constants in it then we can use it to predict the rate for any set of conditions 2) the form of the rate law can tell us something about the mechanism of the reaction 3) knowing the rate law enables us to sperate the conc. dependence from underlying fundamental effect which is size of rate const.

Rate constants (and so rates overall) are heavily temperature dependent so ALWAYS state temperature generally size of rate coefficient increases with temp

• Many rate constants will follow the Arrhenius equation k = A exp(-Ea / RT) k = rate constant (variable units) A = pre-expo factor (same units as k) Ea = activation energy (Jmol^-1) T = temp (T) R = ideal gas const. (J K^-1 mol^-1)

ln(k) against 1/T will give a straight line graph of gradient = -Ea/R Y-int = ln(A)

Assume we have some equil. of A+B —REV.—> C+D assume the forwards direction follows rate law rate(f) = kf [A] [B] and the reverse follows rate(r) = kr[A][B] At equil. these have the same value as the forwards and backwards reactions have the same rate so kf [A(eq)] [B(eq)] = kr [B(eq)] [C(eq)] K = [C(eq)] [D(eq)] / [A(eq)] [B(eq)] = kf / kr so K = kf / kr

MOs need to interact: - this is usually HOMO-LUMO interactions - Bonds are broken, Bonds are formed

• a potential energy surface uses the fact that in theory, using MOs, we can calculate the energy of any arbitrary arrangement of molecules • in principle we can calculate the energy of any given arrangement of A and B as they interact • we end up with a potential energy surface (PES) which gives the PE of the system as a function of the positions of all the atoms in the system • in reality this is monumentally difficult to form and usually requires multi-dimensional plots

we can use the PE surface to imagine the atoms ‘moving’ over it, they start out in a position which corresponds to reactants then move along some path over the surface as they rearrange themselves and end up in a position corresponding to products the stable molecules, products/reactants exist in potential energy minima

• when going from reactants to products over the PES, some energy will always have to be put in because the reactants and products exist in potential energy minima • the route which involves the least expenditure of energy is the most favoured route • it’s the most favoured route because the energy E which has to be put in in order to reach the energy of the transition state is only present in some cases, the lower this energy is, the more reactions can occur • we can safely assume as very few molecules have sufficient energy to react, that they all go via the favoured route

• The transition state is the state of the molecule(s) at the highest energy point on the PES on the route between the reactants and products • it is not a molecule in the conventional sense because it is very unstable as molecules exist in potential energy minima and intermediates exist on potential energy maxima (unstable equil.) so any slight change and there’s a driving force to remove the intermediate • they exist for a very small amount of time, in the order of a single molecular vibration

an intermediate is a standard molecule but one which exists in a metastable equilibrium between the reactants and products a transition state is not stable

• the minimum energy pathway from reactants to products is called the reaction coordinate • it can be plotted (on X) against potential energy • this means we have Reactants (low energy) rising up to maxima (transition state) lowering in energy again to products • the activation energy is the key feature of this plot, Ea is the energy difference between the maxima and the reactants

• Collision theory is a theory about the motions and energies of the particles of gases, its entirely classical • the theory treats particles as objects where size «distance between them • the particles have kinetic energy which causes them to move in random directions • they collide elastically with each other and the walls of the container

• the energies and speeds of the particles can be given by the Maxwell Distribution • the mean relative speed (cbar(rel)) cbar(rel) = (8kbT / pi*mu)^1/2 kb = Boltzmann constant T = temp mu = reduced mass of system

mu = (mA*mB) / (mA + mB)

cbar(rel) = (8kbT / pi*μ)^1/2

Collision rate = Z(AB), collisions m^-3 s^-1 Z(AB) = cAcBpi(rA + rB)^2 * cbar(rel) = cAcBpi(rA+rB)^2 * (8kbT / pi*mu)^1/2 cA = conc. of A in molecules per m^3 cB = conc. of B in molecules per m^3 rA = radius of A rB = radius of B cbar(rel) = mean relative speed this expression can be simplified by defining the collision cross-section, σ, as pi*(rA+rB)^2, giving Z(AB) = cAcBσ(8kbT / pimu)^1/2