Rates and reaction dynamics

What the exam asks

The item you will meet most is a table of initial rates: three or four experiments, concentrations in the columns, rate in the last. It is worth practising until it is mechanical. Find two rows where one concentration changed and the others did not, take the ratio of the concentrations and the ratio of the rates, and ask what power connects them. Then repeat for the other reactant. The trap the exam relies on is reading the orders off the balanced equation, and it sets it by giving the reaction coefficients that disagree with the orders. If your rate law matches the stoichiometry without your having looked at the data, you have not answered the question. The second trap is the catalyst, which candidates let change ΔH or the equilibrium yield. It changes neither. It only changes when you get there.

Kinetics asks how fast, and answers it with two things: how often molecules collide hard enough and in the right orientation, and how big the barrier is that they have to clear. Concentration and temperature are the two handles, and they work differently. More concentration means more collisions. More temperature means a rapidly growing fraction of collisions that are violent enough to matter, which is why heat does far more to a rate than crowding does.

The rate law is where the marks are, and the one thing to know about it is that it is experimental. You cannot read the orders off the balanced equation, because the equation describes the overall bookkeeping while the rate law describes the slowest step. A reactant with a coefficient of 2 can be first order, zero order, or anything else. The only way to find out is to change its concentration and watch what the rate does, which is exactly what an initial rates table is.

Energy profiles are the picture behind all of it. Reactants on the left, products on the right, a hill in between whose height is the activation energy and whose top is the transition state. Once you can read three heights off that diagram, most of the qualitative questions in this topic answer themselves.

What to hold

• The rate law is rate = k[A]^m[B]^n, and the orders m and n are found by experiment, never by reading the coefficients, because the coefficients describe the overall stoichiometry while the rate law describes the slowest step.
• The method of initial rates works by holding every concentration fixed but one, so any change in rate belongs to the one you moved, and the factor the rate changes by equals the factor the concentration changed by raised to the order: double a concentration for a fourfold rate and it is second order.
• Zero order in a reactant means its concentration does not appear in the rate law, so the reactant is real and involved but takes no part in the slowest step or anything before it.
• The units of k depend on the overall order, running M/s for zero order, per second for first order and 1/(M s) for second order, so the units alone give away the overall order.
• Rates are measured initially because at that moment the concentrations are still the values you set, before any reactant has been appreciably consumed.
• Raising the temperature raises the rate mostly because the fraction of molecules with enough energy to clear the barrier grows exponentially, not because collisions get more frequent, which they do only slightly.
• A catalyst gives an alternative pathway with a lower activation energy, and it is regenerated, so it does not appear in the overall equation.
• A catalyst lowers the barrier for the forward and reverse reactions by the same amount, because both are measured to the same peak, so it speeds both equally, moves no equilibrium, and leaves the reactant and product energies untouched so it cannot change ΔH.
• On an energy profile the forward activation energy is the peak height above the reactants, the reverse activation energy is the peak height above the products, and ΔH is the products minus the reactants.
• A transition state is a peak and cannot be isolated; an intermediate is a dip between two peaks and is a real species with a finite lifetime.
• In a multi-step reaction the rate-determining step is the one with the highest barrier to climb, and speeding up any other step changes nothing.
• A collision leads to reaction only if it carries at least the activation energy and the molecules are oriented correctly, which is why not every energetic collision reacts.