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RK3. Activation Barriers

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  • Why do reactions take place at different rates? Why do some happen quickly, and others proceed very slowly? Why might the same reaction proceed at different rates under different conditions? There are a number of factors that influence reaction rates, but this article focuses on the activation barrier.

    An activation barrier is a sort of energetic hurdle that a reaction must bypass. Some reactions have higher hurdles and some have lower hurdles. It is easier to overcome lower hurdles, so reactions with low activation barriers can proceed more quickly than ones with higher activation barriers:

    • A low activation barrier allows a reaction to happen quickly.
    • A high activation barrier makes a reaction proceed more slowly.

    A reaction can be exergonic overall, but still have an activation barrier at the beginning. Even if the system decreases in energy by the end of the reaction, it generally experiences an initial increase in energy.

    • Even if a reaction gives off energy overall, energy must be added initially to get the reaction started.

    This situation is similar to investing in a business. A business generally requires a financial investment to get started. If the business is successful, it will eventually make products and pay money back to the investors. If the business is unable to make back its initial investment, it may fail.

    Reactions require an initial investment of energy. This energy may come from surrounding molecules or the environment in general. If the reaction is successful, it will proceed to make products and it will emit energy back to its surroundings.

    • It always "costs" a molecule energy to enter into a reaction; it "borrows" that energy from its environment.
    • That initial investment of energy may be "paid back" as the reaction proceeds.

    All reactions must overcome activation barriers in order to occur. The activation barrier is the sum of the energy that must be expended to get the reaction going. An activation barrier is often pictured as a hill the reactants must climb over during the reaction. Once, there, it can slide down the other side of the hill to become products. At the top of the hill, the molecule exists in what is called the "transition state." At the transition state, the structure is somewhere between its original form and the structure of the products.

    The type of diagram shown above is sometimes called a "reaction progress diagram." It shows energy changes in the system as a reaction proceeds. One or more activation barriers may exist along the reaction pathways, due to various elementary steps in the reaction.

    In order to understand more concretely the terms "reaction progress" and "transition state," consider a real reaction. Suppose a nucleophile, such as an acetylide ion, donates its electrons to an electrophilic carbonyl. The π bond breaks and an alkoxide ion is formed.

    "Reaction progress" refers to how far the reaction has proceeded. The transition state refers specifically to the highest energy point on the pathway from reactants to products. It refers to the structure at that point, and the energy associated with that structure.

    In the following diagram, the term "reaction progress" has been replaced by an illustration that matches the status of the reaction with the corresponding point in the energy curve. The structure in the square brackets is the transition state, corresponding to the maximum of the curve. The "double dagger" symbol indicates a transition state structure.

    The transition state is not a true chemical structure. It does not necessarily obey the rules of Lewis structures, because some new bonds have started to form and some old bonds have started to break; partial bonds have no place in a Lewis structure.

    Physically, the transition state structure cannot be isolated. Because it sits at the top of an energy curve, the transition state tends to convert into something else. A change in either direction will lower its energy. The tendency is to proceed to lowest energy if possible. As soon as the transition state forms, it will either slide back into the original starting materials or slip forward into the final products.

    • The transition state is inherently a high-energy, unstable structure with a very short lifetime. As soon as it is formed, it disappears.

    Problem RK3.1.

    Draw the transition state for the following elementary reactions.

    More commonly, reaction progress diagrams are not drawn like the one above. Instead, structures of reactants, transition states, and products are shown along the potential energy curve, as shown below.

    Not all reactions take place in one step. Sometimes there are one or more intermediate species. An intermediate differs from a transition state in that it has a finite lifetime. Although it is not as stable as the reactants or the products, an intermediate is stable enough that it does not immediately decay. Going either forward to products or back to reactants is energetically uphill.

    Problem RK3.2.

    Draw reaction progress diagrams for the following reactions. Note that the reactions may be composed of more than one elementary step.

    Rate Constant

    The rate constant is a measurable parameter that can be used to form an idea about the activation barrier of a reaction. The rate constant for a reaction is related to how quickly the reaction proceeds. A large rate constant corresponds to a very fast reaction. A very small rate constant corresponds to a slow one.

    • The rate constant is an index of the speed of the reaction.

    Rate constants have different units depending on how the reaction proceeds; a reaction with a "first order" rate constant of 0.001 s-1 would run to completion in about an hour. A reaction with a first order rate constant of 10-6 s-1 might take a couple of weeks.

    The rate constant gives direct insight into what is happening at the transition state, because it is based on the energy difference between the reactants and the transition state.

    The rate constant can be broken down into pieces. Mathematically, it is often expressed as

    \[ k = \left ( {RT \over Nh } \right ) e^{-\Delta {G \over RT}} \]

    In which R = the ideal gas constant, T = temperature, N = Avogadro's number, h = Planck's constant, and ΔG = the free energy of activation.

    The ideal gas constant, Planck's constant, and Avogadro's number are all typical constants used in modeling the behavior of molecules or large groups of molecules. The free energy of activation is essentially the energy requirement to get a quantity of molecules to undergo the reaction.

    Problem RK3.3.

    For each of the following pairs, use < or > to indicate which quantity is larger.

    a) e2 or e10

    b) e1/4 or e1/2

    c) e-3 or e-4

    d) e-1/2 or e-1/3

    Note that k depends on two variables:

    • ΔG, or the energy required for the reaction
    • T, or the temperature of the surroundings, which is an index of how much energy is available

    The ratio of activation free energy to temperature compares the energy needed to the energy available. The more energy available compared to the energy needed, the lower this ratio becomes. As a result, the exponential part of the function becomes larger. That makes the rate constant bigger, and the reaction proceeds faster.

    Large groups of molecules behave like populations of anything else. They have averages, as well as outliers on the high and low end. However, the higher the temperature, the more energy a group of molecules will have on average.

    In the following figure, the blue curve represents the energy content in a population of molecules at low temperature. The peak of the curve is near the average energy for this collection of molecules. Some of the molecules have more energy than average (they are further to the right on the blue curve) and some have less (further to the left).

    The black slice through this curve indicates how much energy is needed to get over the activation barrier for a particular reaction. Notice that, at low temperature, not that many molecules have enough energy to get over the barrier at any one time. The reaction will proceed very slowly. Nevertheless, more energy is available from the surroundings, and so after some time most of the molecules will have obtained enough energy so that they can eventually surpass the barrier.

    The yellow curve represents molecules at a higher temperature, and the red curve is a population at a higher temperature still. As the temperature is increased, larger and larger fractions of the molecules have enough energy to surpass the activation barrier, and so the reaction proceeds more quickly.

    • the rate constant compares energy needed to energy available

    • based on that comparison, a specific fraction of the population will be able to react at a time

    Free Energy of Activation

    The activation free energy is constant for a given reaction at a given temperature. However, at different temperatures, ΔG changes according to the following relation:

    \[ \Delta G = \Delta H - T \Delta S \]

    where ΔH = activation enthalpy and ΔS = activation entropy.

    The activation enthalpy is the element that corresponds most closely with the energy required for the reaction as described. The activation entropy concerns how the energy within the molecule must be redistributed for the reaction to occur. One of the major factors influencing energy distribution over the course of the reaction is molecular geometry.

    For example, suppose two molecules must collide for a reaction to take place.However, a reaction might not happen each time the molecules collide. The molecules may be pointing the wrong way, preventing the reaction from occurring. Atoms must often be oriented in the proper place in order to form a new bond.

    When molecules are restricted to only certain orientations or geometries, they have fewer degrees of freedom. With fewer degrees of freedom, energy can be stored in fewer ways. As a result, there is often an entropy cost in initiating a reaction.

    On the other hand, a reaction might start in a different fashion, with one molecule breaking a bond and dividing into two species. Because each individual piece can move independently from the other, the degrees of freedom increase. Energy can be stored in more ways than they could be before the reaction started. As a result, although this reaction would still have an activation barrier, it may actually be lowered due to the entropy component.

    Problem RK3.4.

    In the following drawings, one orientation of reactants is more likely to lead to the product shown. Select which one will be most successful in each set and explain what is wrong with each of the others.

    Problem RK3.5.

    Because the activation barrier depends partly on the energy needed to break bonds as the molecule heads into the transition state, comparative bond strength can be a useful factor in getting a qualitative feel for relative activation barriers.

    The metal-carbonyl (M-CO) bond strengths of the coordination complexes M(CO)6 have been estimated via photoacoustic calorimetry and are listed below, by metal.

    Cr: 27 kcal/mol Mo: 32 kcal/mol W: 33 kcal/mol

    1. Based on that information, sketch qualitative activation barriers for the loss of a CO ligand from Cr(CO)6, Mo(CO)6 and W(CO)6.
    2. Predict the relative rates for these three reactions (fastest? slowest?).

    Problem RK3.6.

    Comparing the strengths of bonds that will be broken in a reaction is often a good way to get a first estimate of relative activation barriers.

    1. Use the following bond strengths to estimate the barriers to addition of a nucleophile (such as NaBH4) to the following double bonds: C=O (180 kcal/mol); C=N (147 kcal/mol); C=C (145 kcal/mol). Make a sketch of the three reaction progress diagrams.
    2. In general, C=O bonds are the most reactive of these three groups toward electrophiles, followed by C=N bonds. Are these relative barriers consistent with this observation?
    3. What other factor(s) might be important in determining the barrier of the reaction?
    4. Modify your reaction progress diagram to illustrate these other factors.