Reaction Kinetics

Rate is how fast a concentration changes with time. For a reactant, write the rate as -Δ[A]/Δt, so the number comes out positive. For a product, write +Δ[P]/Δt. A tangent to a concentration-time curve gives the rate at one instant. A chord between two points gives the average rate over that interval. Always say which substance you measured. The coefficients in the equation link the rates for different substances.

A rate equation has the form rate = k[A]^m[B]ⁿ. Here m is the order with respect to A, and n is the order with respect to B. Their sum, m + n, is the overall order. The rate constant k stays the same for a reaction at a fixed temperature, with the same catalyst. When you change a concentration, the rate changes through the concentration terms. k itself does not change.

Sometimes two concentrations change at once. First use another pair of experiments to find one order. Then divide out the effect of that order. For example, suppose doubling both A and B makes the rate eight times faster, and B is first order. Doubling B alone would double the rate. The remaining factor of four must come from A, so A is second order. Finally, check your rate equation against every row of the table, not only the pair you used.

The half-life is the time taken for a reactant concentration to fall to half its starting value. In a first-order reaction, each equal time interval removes the same fraction of the reactant. So the half-life stays the same, whatever the starting concentration. To show first order on a graph, find several halvings in a row and check that each takes the same time. One halving alone is not enough evidence.

Sometimes one reactant is kept at an almost constant concentration, for example because it is in large excess. Then the decay you observe shows only how the rate depends on the other reactant. Say this condition before you give an overall order for the whole reaction. You do not need integrated rate equations for this syllabus.

A mechanism is a series of simple steps, called elementary steps. The rate-determining step is the slowest step, and it controls the overall rate. If the slow step is a collision between A and B, the rate equation is rate = k[A][B]. A later fast step can use up another reactant. That reactant then does not appear in the rate equation.

An elementary step cannot have an order that does not match the particles in that step. Also, zero order does not mean a substance takes no part in the reaction. It means its concentration does not control the slow step over the range measured. Always use the mechanism you are given. Do not invent a slow step from the overall coefficients.

To test the effect of one concentration, change only that starting concentration. Keep the total volume, the temperature and the other concentrations the same. A common way is to replace part of a solution with water. Measure the gradient at the start of each run, because concentrations fall as the reaction goes on. If a solid reacts, keep its particle size the same.

This is a sound way to plan an experiment on paper. The 9476 practical exam does not ask you to collect gas with a gas syringe or by displacing water. You may still be asked to interpret gas-volume data or to describe a suitable method.

The activation energy, E_a, is the minimum energy that colliding particles need in order to react. Particles must collide with enough energy and in the right orientation. At a higher concentration, there are more particles in the same volume, so collisions happen more often. The rate equation tells you how strongly the rate responds.

The Boltzmann distribution starts at zero and has a long tail. Particles have a spread of energies, and there is no sharp maximum energy. When the temperature rises, the curve shifts towards higher energies. A much larger fraction of particles now has enough energy to cross the barrier. Collisions also become a little more frequent. Both effects increase k, so the rate rises at the same concentrations. Heating does not lower the activation energy.

Adding more particles at the same temperature does not change the shape of the energy distribution. The fraction of particles with energy above E_a stays the same. What rises is the number of collisions that can succeed. So a temperature change and a concentration change work in different ways, even though both make a reaction faster.

A catalyst takes part in the reaction steps and is made again by the end of its cycle. It speeds up the forward and the reverse reactions, so equilibrium is reached sooner. It does not change ΔH, the equilibrium constant or the equilibrium composition at the same temperature. In practice a catalyst can still be poisoned or wear out, even though the overall equation does not use it up.

Use the same Boltzmann curve to explain the speed-up. The temperature has not changed, so the curve itself is the same. The catalyst gives an alternative pathway with a lower activation energy. The E_a line moves to the left on the energy axis. A greater proportion of particles now has energy equal to or greater than the new E_a, so the frequency of effective collisions increases. The rate constant k increases, and so does the rate. Each link in this chain usually earns its own mark.

NO and NO₂ change back and forth, so they cancel out of the overall equation. This gas-phase scheme shows homogeneous catalysis of SO₂ oxidation in the air. Real atmospheric sulfur chemistry also involves other oxidants and reactions in water droplets. Sulfur oxides that form and dissolve in water add to acid rain.

Without a catalyst, two negative ions, I^- and S₂O₈^2-, must collide. They repel each other, so this route is slow. Iron ions offer a different pathway. Each step now involves a positive iron ion and a negative ion, which attract. This route has a lower activation energy. Fe²⁺ is oxidised to Fe³⁺ and then made again, so only a small amount is needed.

The solid catalyst is in a different phase from the reacting gases. A larger surface area gives more active sites. A poison binds strongly to the sites and blocks them. Reactants must bind strongly enough to weaken their bonds, but products must not bind so strongly that they cannot leave. A catalyst gives a useful rate at a lower temperature. It does not, by itself, raise the equilibrium yield.

In a car catalytic converter, exhaust gases pass over a solid surface of metals such as platinum and rhodium. The gases adsorb, react on the surface and then leave. Carbon monoxide acts as a reducing agent and removes nitrogen oxides, for example 2NO(g) + 2CO(g) → N₂(g) + 2CO₂(g). Other reactions on the surface oxidise CO and unburnt hydrocarbons. The converter must be hot to work well, so it is less effective just after a cold start.

Enzymes are specific in two ways: which substrate binds, and which reaction is catalysed. A substrate that does not fit the active site is not converted efficiently. The lock-and-key model explains this selectivity. It is a simple model; real proteins are not perfectly rigid.

Find orders from rate data or from a given elementary mechanism, never from the overall coefficients. A zero-order concentration-time graph is a straight line. A first-order graph has a constant half-life. Work out the units of k from the rate units and the overall order.

A higher concentration makes collisions more frequent. A higher temperature also makes collisions more frequent, and, more importantly, raises the fraction with energy at or above E_a, so k increases. A catalyst gives a new pathway with a lower barrier. The energy distribution and the equilibrium constant stay the same at that temperature. Enzymes also show active-site specificity and are sensitive to temperature and pH.