Strand 2  Energy and Rates
Unit 4 Chemcial Kinetics

Reaction Rates

    You may be familiar with acid-base titrations that use phenolphthalein as the endpoint indicator. You might not have noticed, however, what happens when a solution that contains phenolphthalein in the presence of excess base is allowed to stand for a few minutes. Although the solution initially has a pink color, it gradually turns colorless as the phenolphthalein reacts with the OH- ion in a strongly basic solution.

    The graph below shows what happens to the concentration of phenolphthalein in a solution that was initially 0.005 M in phenolphthalein and 0.61 M in OH- ion. As you can see  the phenolphthalein concentration decreases by a factor of 10 over a period of about four minutes.

Phenolphthalein Rate

    Experiments such as the one that gave us the data in the above table are classified as measurements of chemical kinetics (from a Greek stem meaning "to move"). One of the goals of these experiments is to describe the rate of reaction which is the rate at which the reactants are transformed into the products of the reaction.

    The term rate is often used to describe the change in a quantity that occurs per unit of time. The rate of inflation, for example, is the change in the average cost of a collection of standard items per year. The rate at which an object travels through space is the distance traveled per unit of time, such as miles per hour or kilometers per second. In chemical kinetics, the distance traveled is the change in the concentration of one of the components of the reaction. The rate of a reaction is therefore the change in the concentration of one of the reactants(X) that occurs during a given period of time (t):

Rate Equation.

    The rate of the reaction between phenolphthalein and the OH- ion isn't constant; it changes with time. Like most reactions, the rate of this reaction gradually decreases as the reactants are consumed. This means that the rate of reaction changes while it is being measured. 

    To minimize the error this introduces into our measurements, it seems advisable to measure the rate of reaction over periods of time that are short compared with the time it takes for the reaction to occur. We might try, for example, to measure the infinitesimally small change in concentration, d(X), that occurs over an infinitesimally short period of time, dt. The ratio of these quantities is known as the instantaneous rate of reaction.

Rate Equation 2

   To determine the instantaneous rate, we need to measure changes over very small time intervals. In fact, the time interval can be made infinitesimally small and then the instantaneous rate is the slope of a tangent line to the curve of a concentration versus time plot. The slope provides the instantaneous rate at that moment in time.

Instantaneous Rate

    As seen in the diagram above, the drawn tangent line (blue line) intersects the curve at a specific time t.  Hence the instantaneous rate at this point is the slope of this line.


    For example, consider the reaction in which C4H9Cl reacts as follows:


The following graph is obtained:

Reaction Graph

    The instantaneous rate at 600 second can be found by finding the slope of the tangent at 600 seconds. 


             = (0.04 - 0.02)/(450 - 750)
             = 0.02/-300
             = -0.000067 mol/Ls

    The negative is left out because the rate is 0.000067 mol/Ls and all the negative indicates is that the reaction is slowing down which is true of all reactions.

    The average rate of reaction between two points can be found by drawing a tangent or secant line between those two points on the curve.  For example in the phenolphthalein reaction above, the average rate of reaction between 40 and 80 seconds can be found by finding the slope of the appropriate line bisecting these two points.

Average Rate


                = (0.0035 - 0.0025)/(40-80)
                = (0.001)/(-40)
                = -0.000025 mol/Ls

    Again, you get rid of the negative sign to indicate that we are just looking at the rate and are not worried about the fact that it is slowing down.

Factors Affecting Reaction Rates: 

Surface Area

    In the lab, powdered calcium carbonate reacts much faster with dilute hydrochloric acid than if the same mass was present as lumps of marble or limestone.

Calcium Carbonate Reaction

    Or consider the catalytic decomposition of hydrogen peroxide:

Hydrogen Peroxide Decomposition

    This is another familiar lab reaction. Solid manganese(IV) oxide is often used as the catalyst. Oxygen is given off much faster if the catalyst is present as a powder than as the same mass of granules.

    Catalytic converters use metals like platinum, palladium and rhodium to convert poisonous compounds in vehicle exhausts into less harmful things.  For example, a reaction which removes both carbon monoxide and an oxide of nitrogen is:

Catalytic Converter Reaction

    Because the exhaust gases are only in contact with the catalyst for a very short time, the reactions have to be very fast. The extremely expensive metals used as the catalyst are coated as a very thin layer onto a ceramic honeycomb structure to maximize the surface area.

    You are only going to get a reaction if the particles in the gas or liquid collide with the particles in the solid. Increasing the surface area of the solid increases the chances of collision taking place.  Imagine a reaction between magnesium metal and a dilute acid like hydrochloric acid. The reaction involves collision between magnesium atoms and hydrogen ions.

Surface Area Effect

    Increasing the number of collisions per second increases the rate of reaction.

Nature of the Reactants

    Different chemical reactions take place at different rates. The study of reaction rates is the study of the forming and breaking of chemical bonds, a very complex subject. The nature of the reactants involved in a chemical reaction will determine the kind of bonding that occurs. Reactions with bond rearrangement or electron transfer generally take longer than ionic reactions. Ionic reactions are almost instantaneous because of the strong attraction between the charged particles. 

    This is another way of saying that different chemicals may react at different rates even though the overall chemistry may look the same. A simple example is that lithium, sodium, potassium and cesium all react with water to give hydrogen. Lithium reacts somewhat slowly and is unexciting. Sodium reacts quickly, skimming along the surface of the water on a layer of hydrogen. If the sodium piece gets stuck on the wall of a container the temperature may rise enough for the hydrogen to be ignited. Potassium reacts so vigorously with water that the hydrogen bursts into flame with the oxygen in the air, while cesium effectively explodes when it hits water.

    Since reactions proceed due to bond making and breaking, ionic bond changes occur faster than covalent bond changes. Ionic bond changes occur in precipitation and neutralization reactions.  In general, if more bonds have to be broken and reformed in order for the reaction to take place, the reaction will be slower.  Thus the more more complex the substances involved or the more substances involved, the slower the reaction.

Concentration of the Reactants

    For many reactions involving liquids or gases, increasing the concentration of the reactants increases the rate of reaction. In a few cases, increasing the concentration of one of the reactants may have little noticeable effect of the rate.  Don't assume that if you double the concentration of one of the reactants that you will double the rate of the reaction. It may happen like that, but the relationship may well be more complicated.

    In the lab, zinc granules react fairly slowly with dilute hydrochloric acid, but much faster if the acid is concentrated.

Zinc Experiment

    Solid manganese(IV) oxide is often used as a catalyst in this reaction. Oxygen is given off much faster if the hydrogen peroxide is concentrated than if it is dilute.

Hydrogen Peroxide Decomposition

    When a dilute acid is added to sodium thiosulphate solution, a pale yellow precipitate of sulphur is formed.

Sodium Thiosulfate Experiment

    As the sodium thiosulphate solution is diluted more and more, the precipitate takes longer and longer to form.

    The same argument applies whether the reaction involves collision between two different particles or two of the same particle.  In order for any reaction to happen, those particles must first collide. This is true whether both particles are in solution, or whether one is in solution and the other a solid. If the concentration is higher, the chances of collision are greater.

Concentration Effect

    If a reaction only involves a single particle splitting up in some way, then the number of collisions is irrelevant. What matters now is how many of the particles have enough energy to react at any one time.  Suppose that at any one time 1 in a million particles have enough energy to equal or exceed the activation energy.  If you had 100 million particles, 100 of them would react.  If you had 200 million particles in the same volume, 200 of them would now react.  The rate of reaction has doubled by doubling the concentration.

    Suppose you are using a small amount of a solid catalyst in a reaction, and a high enough concentration of reactant in solution so that the catalyst surface was totally cluttered up with reacting particles.  Increasing the concentration of the solution even more can't have any effect because the catalyst is already working at its maximum capacity. 

    Suppose you have a reaction which happens in a series of small steps. These steps are likely to have widely different rates - some fast, some slow.  For example, suppose two reactants A and B react together in these two stages:

Reaction Mechanism

    The overall rate of the reaction is going to be governed by how fast A splits up to make X and Y.  This is described as the rate determining step of the reaction.  If you increase the concentration of A, you will increase the chances of this step happening for reasons we've looked at above.  If you increase the concentration of B, that will undoubtedly speed up the second step, but that makes hardly any difference to the overall rate. You can picture the second step as happening so fast already that as soon as any X is formed, it is immediately pounced on by B. That second reaction is already "waiting around" for the first one to happen.

    A reaction mechanism describes the one or more steps involved in the reaction in a way which makes it clear exactly how the various bonds are broken and made. The following example comes from organic chemistry. This is a reaction between 2-bromo-2-methylpropane and the hydroxide ions from sodium hydroxide solution:

Reaction Mechanism Example

    The overall reaction replaces the bromine atom in the organic compound by an OH group.  The first thing that happens is that the carbon-bromine bond in a small proportion of the organic compound breaks to give ions:

Slow Step

    Carbon-bromine bonds are reasonably strong, so this is a slow change. If the ions hit each other again, the covalent bond will reform. The curly arrow in the equation shows the movement of a pair of electrons. If there is a high concentration of hydroxide ions present, the positive ion stands a high chance of hitting one of those. This step of the overall reaction will be very fast. A new covalent bond is made between the carbon and the oxygen, using one of the lone pairs on the oxygen atom.

Fast Step

    Because carbon-oxygen bonds are strong, once the OH group has attached to the carbon atom, it tends to stay attached.  The mechanism shows that the reaction takes place in two steps and describes exactly how those steps happen in terms of bonds being broken or made. It also shows that the steps have different rates of reaction - one slow and one fast.

    The overall rate of a reaction (the one which you would measure if you did some experiments) is controlled by the rate of the slowest step. In the example above, the hydroxide ion can't combine with the positive ion until that positive ion has been formed. The second step is in a sense waiting around for the first slow step to happen. The slow step of a reaction is known as the rate determining step. As long as there is a lot of difference between the rates of the various steps, when you measure the rate of a reaction, you are actually measuring the rate of the rate determining step.

    Here is the mechanism we have already looked at. How do we know that it works like this?

Slow Step

Fast Step

    By doing rate of reaction experiments, you find this rate equation:

Rate Law

    The reaction is first order with respect to the organic compound, and zero order with respect to the hydroxide ions. The concentration of the hydroxide ions isn't affecting the overall rate of the reaction.  If the hydroxide ions were taking part in the slow step of the reaction, increasing their concentration would speed the reaction up. Since their concentration doesn't seem to matter, they must be taking part in a later fast step.  Increasing the concentration of the hydroxide ions will speed up the fast step, but that won't have a noticeable effect on the overall rate of the reaction. That is governed by the speed of the slow step.  In a simple case like this, where the slow step of the reaction is the first step, the rate equation tells you what is taking part in that slow step. In this case, the reaction is first order with respect to the organic molecule - and that's all.  This gives you a starting point for working out a possible mechanism. Having come up with a mechanism, you would need to find more evidence to confirm it. For example, in this case you might try to detect the presence of the positive ion that is formed in the first step.

    Changing the concentration of substances taking part in a reaction usually changes the rate of the reaction. A rate equation shows this effect mathematically. Orders of reaction are a part of the rate equation. There are several simple ways of measuring a reaction rate. For example, if a gas was being given off during a reaction, you could take some measurements and work out the volume being given off per second at any particular time during the reaction.  A rate of 2 cm3 s-1 is obviously twice as fast as one of 1 cm3 s-1.  However, for this more formal and mathematical look at rates of reaction, the rate is usually measured by looking at how fast the concentration of one of the reactants is falling at any one time. For example, suppose you had a reaction between two substances A and B. Assume that at least one of them is in a form where it is sensible to measure its concentration - for example, in solution or as a gas.  For this reaction you could measure the rate of the reaction by finding out how fast the concentration of, say, A was falling per second.  You might, for example, find that at the beginning of the reaction, its concentration was falling at a rate of 0.0040 mol dm3 s-1.  This means that every second the concentration of A was falling by 0.0040 moles per cubic decimeter. This rate will decrease during the reaction as A gets used up.

   Orders of reaction are always found by doing experiments. You can't deduce anything about the order of a reaction just by looking at the equation for the reaction.  So let's suppose that you have done some experiments to find out what happens to the rate of a reaction as the concentration of one of the reactants, A, changes. Some of the simple things that you might find are:

1. One possibility: The rate of reaction is proportional to the concentration of A which means that if you double the concentration of A, the rate doubles as well. If you increase the concentration of A by a factor of 4, the rate goes up 4 times as well. You can express this using symbols as:

First Order

    Writing a formula in square brackets is a standard way of showing a concentration measured in moles per cubic decimeter (liter). You can also write this by getting rid of the proportionality sign and introducing a constant, k.

First Order Rate Law

2. Another possibility: The rate of reaction is proportional to the square of the concentration of A which means that if you doubled the concentration of A, the rate would go up 4 times (22). If you tripled the concentration of A, the rate would increase 9 times (32). In symbol terms:

Second Order Rate Law

    Thus by doing experiments involving a reaction between A and B, you would find that the rate of the reaction was related to the concentrations of A and B in this way:

Rate Law

    This is called the rate equation for the reaction. The concentrations of A and B have to be raised to some power to show how they affect the rate of the reaction. These powers are called the orders of reaction with respect to A and B.

    If the order of reaction with respect to A is 0 (zero), this means that the concentration of A doesn't affect the rate of reaction. Mathematically, any number raised to the power of zero (x0) is equal to 1. That means that that particular term disappears from the rate equation. 

    If the order of reaction with respect to A is >1, this means the concentration of A is causing the reaction to speed up by some factor.  However, if the order of reaction with respect to A is <0 or negative, this means that the concentration of A is causing the reaction to slow down by some factor.  Note that a negative exponent could have just been written as a decimal or fraction in which case the order of reaction would be between 0 and 1 but not including 0 or 1.  The exponents are not usually written in this manner though.

    The overall order of the reaction is found by adding up the individual orders. For example, if the reaction is first order with respect to both A and B (a = 1 and b = 1), the overall order is 2. We call this an overall second order reaction.

    It doesn't matter how many reactants there are. The concentration of each reactant will occur in the rate equation, raised to some power. Those powers are the individual orders of reaction. The overall order of the reaction is found by adding them all up.

    A fundamental challenge in chemical kinetics is the determination of the reaction order (or, in general, the Rate Law) from experimental information. We know that the rate law closely related to the reaction mechanism, and the knowledge of the mechanism of a given reaction allows us to control that reaction. But how does one actually determine a rate law? The first step is to control the conditions under which a reaction occurs, and then determine the rate of that reaction. If we measure the reaction rate just after we define the reaction conditions, then this is a measurement of the initial rate of the reaction. 

    Consider a reaction A + B  Products. An experimenter prepares a reaction mixture of 1.00 M of each of the reactants and measures the rate of reaction initially (at very early times, i.e. before the reactant concentrations have had a chance to change much due to the progress of the reaction) to be 1.25 x 10-2 M/s. What is the rate constant for the reaction?  You cannot determine this from the above information, because first you have to know the reaction order!  The reaction stoichiometry does NOT determine the reaction order except in the special case of an ELEMENTARY reaction. (An elementary reaction is one in which the reaction takes place with a mechanism implied by the reaction equation. Most reactions are NOT elementary)

    More information about the reaction rate law is needed than a single measurement of an initial rate! In order to determine the rate law experimentally, we must use more than one measurement of rate versus concentration! 

    So in our reaction of A + B  Products the experimenter prepares several reaction mixtures and determines the initial reaction rates under these different conditions. The data obtained for several experimental runs in tabulated below 

Run #
Initial [A]
Initial [B]
Initial Rate
1.00 M
1.00 M 
1.25 x 10-2 M/s
1.00 M
2.00 M
2.5 x 10-2 M/s 
2.00 M
2.00 M
2.5 x 10-2 M/s 

    Before we do anything, we must determine the order of the reaction with respect to every reactant, i.e. we must determine the rate law for the reaction. In the above example, the order can be easily determined by inspection. Find a pair of experimental runs that the concentration of only one reactant changes. This is called the method of isolation and a good experimental design always has one such pair of experimental runs. We can see that in runs 1 and 2, only the initial concentration of B has been varied. In fact, [B]0 has doubled from run 1 to run 2 and the reaction rate has also doubled. Therefore the reaction must be First Order in B. Examination of runs 2 and 3 show that in these experimental runs the concentration dependence of A has been isolated. In this case the doubling of the initial concentration of A has no effect on the reaction rate so the reaction must be Zeroth Order in A. 

    Thus the rate law for the reaction is 

rate = k[B]1[A]0 = k[B] 

    Now, the rate constant can be determined from any of the experimental runs. Let's substitute the values of trial # 3 into our rate law.

rate = 2.5 x 10-2 M/s= k (2.0 M)   and thus k = 1.25 x 10-2 s-1

    But some rate laws are more complex.  Consider a reaction A + B + C  Products  Several initial conditions of this reaction are investigated and the following data are obtained: 

Run #
0.151 M
0.213 M
0.398 M
0.480 M/s
0.251 M
0.105 M
0.325 M
0.356 M/s
0.151 M
0.213 M
0.525 M
1.102 M/s
0.151 M
0.250 M
0.480 M
0.988 M/s

    Isolate the effect of the concentration of one of the reactants. In runs number 1 and 3, the only change of initial concentrations is that of reactant C, a change from 0.398 M to 0.525 M with a change in rate from 0.480 M/s to 1.102 M/s. These numbers are not so easy as to be able to guess the order by inspection, so lets think mathematically. If we define the order of the reaction in reactants A, B, and C as a, b, and c, we can write down the rate law (with unknown orders): rate = k [A]a [B]b [C]c

    The ratio of the initial rates of runs 1 and 3 is then: 
Complex General Rate Law
    Note that we have chosen this ratio so that many terms on the right hand side cancel, i.e. k and the concentrations of species A and B. The ratio reduces to: 
Complex Rate Law
    Note that none of the numbers in this equation have any units anymore because we have divided them out in a dimensionless ratio. Now, how to solve for c? Take the natural log or ln of both sides of the equation 

ln B. S.
which allows us to determine that 

Value of c
    So the reaction is Third Order in reagent C!  To continue, we pick another pair of runs that have a change in the initial concentration of just one reactant and repeat the math we just did; however, there are no such cases in this experiment.  So instead we pick another pair of runs that changes in just one variable besides C. We already know the exponent for C so it is no longer an unknown.  This simplifies our work.  Let us choose the runs 1 and 4. The ratio of the initial rates of these runs is: 
Complex Rate Law Second Variable
    Note that everything cancels, is known, or is the order b. So: 
Complex Rate Law Second Variable
    We can determine b as before by taking the ln of both sides of this equation: 
Complex Rate Law Second Variable
and b = 1.00. The reaction is First Order in B! 

   Now that we know the order of Reactants A and B, we can use another pair of experimental runs to determine the remaining unknown order. We must use a pair of runs where the initial concentration of A changes, so we pick runs 1 and 2. As before, the ratio of initial rates is 
Complex Rate Law Third Variable
The rate constant cancels as before and we can solve for a: 
Complex Rate Law Third Variable
and, as usual, take the ln of both sides: 
Complex Rate Law Third Variable
which results in a = 2.00 and the reaction is Second Order in A. 

The rate constant for the reaction may now be evaluated from any of the experimental runs: 
Complex Rate Law
or k = 1.57 x 103    L5 mol-5s-1

    These units are appropriate for a reaction that is sixth order overall. Use another experimental run to check your answer. If the rate law is correct, every experimental run will give the same value of the rate constant. 

    A first-order reaction is one where the rate depends on the concentration of the species to the first power. For a general unimolecular reaction, A  products the reaction rate expression for a first order reaction is rate = k[A]0 = k[A].  We also can write that rate = change in [A]/change in time or R = -d[A]/dt (negative because we are slowing down).  Thus we get

R = -d[A]/dt = k[A]

    Rearranging this we get:

d[A]/[A] = -kdt

        This equation defines any population change in nature and so we recognize it as a factor of the natural exponent e.  That is, things changing according to this equation are governed by the rate of change corresponding to the natural log or e which is a log to the base of 2.718.  You may be used to log which is to the base of 10.  Thus the log of 10 = 1 and the log of 100 = 2.  Note that this log really is written as log10 but the 10 is assumed.  We could also look at log2 of 8 = 3 or the log7 of 49 = 2.  In the case of first order reactions we are looking at the loge which s really ln.  Thus the reverse ln = e just as the antilog7 of 2 = 49.

    This gives us an equation for the first order decay:

[A]/[A]0 = e-kt

where [A]0 is the initial concentration, [A] is the concentration after some time = t and k is the rate law constant.  If t = the time for 50% of [A]0 to decompose, the equation becomes 0.5 = e-kt.  This time span is referred to as the half life for that substance.  Taking the ln of both sides reduces this equation to 0.693 = kt.  Thus if we know the time for one half life, we can always determine the value of k which can then be used to predict any concentration change over time.

     Consider this example.  If the half life of a sample is 12.5 hours, what percentage of a 0.25 g sample remains after 13.7 hours. 

0.693 = kt

0.693 = k(12.5)
k = 0.693/12.5 = 0.05544 hours-1

substituting into [A]/[A]0 = e-kt

[A]/0.25 = e-(0.05544)(13.7)
[A]/0.25 = e-0.759528
[A]/0.25 = 0.4679

[A] = (0.4679)(0.25) = 0.117 g.

Temperature of the Reaction

    As you increase the temperature the rate of reaction increases.  As a rough approximation, for many reactions happening at around room temperature, the rate of reaction doubles for every 10°C rise in temperature. You have to be careful not to take this too literally.  It doesn't apply to all reactions. Even where it is approximately true, it may be that the rate doubles every 9°C or 11°C or whatever.  The number of degrees needed to double the rate will also change gradually as the temperature increases.  Some reactions are virtually instantaneous - for example, a precipitation reaction involving the coming together of ions in solution to make an insoluble solid, or the reaction between hydrogen ions from an acid and hydroxide ions from an alkali in solution.  So heating one of these won't make any noticeable difference to the rate of the reaction.  Almost any other reaction you care to name will happen faster if you heat it - either in the lab, or in industry.

    Particles can only react when they collide. If you heat a substance, the particles move faster and so collide more frequently. That will speed up the rate of reaction. That seems a fairly straightforward explanation until you look at the numbers!  It turns out that the frequency of two-particle collisions in gases is proportional to the square root of the kelvin temperature. If you increase the temperature from 293 K to 303 K (20°C to 30°C), you will increase the collision frequency by a factor of 1.7% for a 10° rise. The rate of reaction will probably have doubled for that increase in temperature - in other words, an increase of about 100%. The effect of increasing collision frequency on the rate of the reaction is very minor. 

    The important effect is quite different, that being activation energy.  Collisions only result in a reaction if the particles collide with enough energy to get the reaction started. This minimum energy required is called the activation energy for the reaction.  You can mark the position of activation energy on a Maxwell-Boltzmann distribution to get a diagram like this:

Reaction Energy Curve

    Only those particles represented by the area to the right of the activation energy will react when they collide. The great majority don't have enough energy, and will simply bounce apart.  To speed up the reaction, you need to increase the number of the very energetic particles - those with energies equal to or greater than the activation energy. Increasing the temperature has exactly that effect - it changes the shape of the graph.  In the next diagram, the graph labeled T is at the original temperature. The graph labeled T+t is at a higher temperature.

Shifted Energy Curve

    If you now mark the position of the activation energy, you can see that although the curve hasn't moved very much overall, there has been such a large increase in the number of the very energetic particles that many more now collide with enough energy to react.

Shifted Energy Curve Effect

    Remember that the area under a curve gives a count of the number of particles. On the last diagram, the area under the higher temperature curve to the right of the activation energy looks to have at least doubled - therefore at least doubling the rate of the reaction.  Increasing the temperature increases reaction rates because of the disproportionately large increase in the number of high energy collisions. It is only these collisions (possessing at least the activation energy for the reaction) which result in a reaction.

Addition of a Catalyst

    A catalyst is a substance which speeds up a reaction, but is chemically unchanged at the end of the reaction. When the reaction has finished, you would have exactly the same mass of catalyst as you had at the beginning.  Consider some examples:

  • Decomposition of hydrogen peroxide by manganese(IV) oxide, MnO2 
  • Nitration of benzene by concentrated sulphuric acid 
  • Manufacture of ammonia via the Haber Process by iron 
  • Conversion of SO2 into SO3 during the Contact Process to make sulphuric acid by vanadium(V) oxide, V2O5 
  • Hydrogenation of a C=C double bond by nickel 
     Again the critical factor is activation energy.  Collisions only result in a reaction if the particles collide with enough energy to get the reaction started. This minimum energy required is called the activation energy for the reaction. You can mark the position of activation energy on a Maxwell-Boltzmann distribution to get a diagram like this:

Reaction Energy Curve

    Only those particles represented by the area to the right of the activation energy will react when they collide. The great majority don't have enough energy, and will simply bounce apart.  To increase the rate of a reaction you need to increase the number of successful collisions. One possible way of doing this is to provide an alternative way for the reaction to happen which has a lower activation energy.  In other words, to move the activation energy on the graph like this:

Catalytic Effects

    Adding a catalyst has exactly this effect on activation energy. A catalyst provides an alternative route for the reaction. That alternative route has a lower activation energy. Showing this on an energy profile:

Catalyst Effect and Energy Profile Graph

    Thus a catalyst provides an alternative route for the reaction with a lower activation energy.  It does not actually lower the activation energy of the reaction.  There is a subtle difference between the two statements that is easily illustrated with a simple analogy.  Suppose you have a mountain between two valleys so that the only way for people to get from one valley to the other is over the mountain. Only the most active people will manage to get from one valley to the other.  Now suppose a tunnel is cut through the mountain. Many more people will now manage to get from one valley to the other by this easier route. You could say that the tunnel route has a lower activation energy than going over the mountain. But you haven't lowered the mountain! The tunnel has provided an alternative route but hasn't lowered the original one. The original mountain is still there, and some people will still choose to climb it. In the chemistry case, if particles collide with enough energy they can still react in exactly the same way as if the catalyst wasn't there. It is simply that the majority of particles will react via the easier catalyzed route.