# 17.3: Collision and Activation - Arrhenius' Law (2023)

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##### learning goals

Make sure you fully understand the following essential ideas presented above. It is especially important that you know the exact meaning of all terms highlighted in green that relate to this topic.

• Explain the meaning of a reaction mechanism and define the elementary step and the intermediate step.
• Describe the role of collisions in reaction mechanisms and explain why not all collisions result in the formation of products.
• Sketch activation energy diagrams for simple reactions that are endothermic or exothermic,
• Explain how an activated complex differs from an intermediate.
• Define the catalyst and sketch an activation energy diagram that illustrates how catalysts work.
• Explain the meaning of the different terms used in Arrhenius' law.
• Sketch a typical Arrhenius law diagram for a hypothetical reaction at higher and lower temperatures.
• Explain how the activation energy of a reaction can be determined experimentally.
• Explain the meaning of the different terms that appear in the pre-exponential factor of the Arrhenius equation.

Why are some reactions much faster than others, and why are reaction rates independent of the thermodynamic tendency for the reaction to occur? These are the central questions we will address in this unit. With that, we open the door to the important topicreaction mechanisms: What happens at the microscopic level when chemical reactions occur? We thank Prof.Svante Arrheniusfor unlocking this door! To keep things as simple as possible, we will limit ourselves to the reactions that occur in thegas phase.The same principles apply to reactions in liquids and solids, but with additional complications that we'll discuss in a later unit.

## reaction mechanisms

OMechanismA chemical reaction is the sequence of actual events that occur when reactant molecules are converted into products. Each of these events represents aelementary stepwhich can be represented as the coming together of individual particles ("collision") or as the breaking of a molecule ("dissociation") into simpler units. The molecular entity that results from each step may or may not be an end product of the reaction.intermediary— a species that is created in an elementary step and destroyed in a subsequent step and therefore does not appear in the net reaction equation.

A reaction mechanism must ultimately be understood as a “beat-by-beat” description of processes at the molecular level, whose sequence leads from reactants to products. These elementary steps (also calledelementary reactions) are almost always very simple with one, two or [rarely] three chemical species, each classified as

 unimolecular A → by far the most common bimolecular A+B → termomolecular A+B+C → Very rare

## Collision theory of chemical change: molecules must collide before they can react

This cardinal rule should guide any analysis of a common chemical reaction mechanism. This explains why thermomolecular processes are so unusual. The kinetic theory of gases tells us that for every 1000 binary collisions there is only one event where three molecules come together at the same time. Collisions in four directions are so unlikely that this process has never been demonstrated in an elementary reaction. Consider a simple bimolecular step

$\ce{A + B-> Products} \nonumber$

Of course, if two molecules A and B are going to react, they must come close enough to break some of their existing bonds and allow the necessary new bonds to form in the products. We call this meetingcollision.

The frequency of collisions between A and B in a gas is proportional to its concentration; If we double [A], the frequency of A-B collisions doubles, and doubling [B] has the same effect. Thus, if all collisions lead to products, then the rate of a bimolecular process is first-order in A and B, or second-order overall:

$\text{Tax} = k[\ce{A}][\ce{B}] \nonumber$

However, not all collisions are created equal. In a gas at room temperature and normal atmospheric pressure, it is about 1033Collisions in every cubic centimeter per second. If every collision between two reacting molecules produced products, all reactions would complete in a fraction of a second.

When two pool balls collide, they just bounce off each other. This is also the most likely outcome when the reaction between A and B requires significant breaking or rearrangement of the bonds between their atoms. To initiate an effective response,The collisions should be enoughenergetic(kinetic energy) to cause this bond break. More on that below. And there is often an additional requirement. In many reactions, especially in more complex molecules, the reacting species must be oriented appropriately for the process at hand. For example, in the gas-phase reaction of nitrous oxide with nitric oxide, the oxygen end of N2O must meet the nitrogen end of NO; reversing the orientation of one of the molecules prevents the reaction.

Due to the great randomness of molecular motion in a gas or liquid, there are always molecules properly oriented enough for some of the molecules to react. But the more critical this orientation requirement is, the fewer collisions will take effect.

## Anatomy of a collision

Energetic collisions between molecules cause the interatomic bonds to stretch and bend even more, temporarily weakening them and making them more susceptible to fission. Distortion of bonds can expose their associated electron clouds to interactions with other reagents that can result in the formation of new bonds. Chemical bonds have some of the properties of mechanical springs, whose potential energy depends on how much they are stretched or compressed. Any atom-to-atom bond can be described by a potential energy diagram, showing how its energy varies with length. When the bond absorbs energy (from heating or collision), it is raised to a more quantized vibrational state (indicated by the horizontal lines) that weakens the bond as its length fluctuates between the expanded boundaries corresponding to the curve. A given collision will normally excite a series of bonds in this way. within about 10–13Second, this excitation is distributed to the other bonds of the molecule in a very complex and unpredictable way, which allows the added energy to be concentrated in a particularly vulnerable spot. The affected bond may continue to stretch and bend, making it more likely to crack. Even if the bond is not broken by pure stretching, it can be distorted or twisted to expose nearby electron clouds to interactions with other reagents that can promote a reaction.

For example, consider the isomerization of cyclopropane to propylene, which occurs at relatively high temperatures in the gas phase. We can think of the collision sequence with the product as [roughly simplified]:

(Video) Collision Theory - Arrhenius Equation & Activation Energy - Chemical Kinetics Observe that

• For simplicity, we don't show the hydrogen atoms here. This is reasonable, as C-C bonds are weaker than C-H bonds and therefore less likely to be affected.
• the collision in it will normally be with another cyclopropane molecule, but as no part of the colliding molecule is incorporated into the product, it can, in principle, be an inert gas or other unreactive species;
• Although the C-C bonds in cyclopropane are all identical, the instantaneous location of the collision energy can distort the molecule in several ways ( ), resulting in a configuration sufficiently unstable to initiate product rearrangement.

The cyclopropane isomerization described above is typical of many decomposition reactions that follow first-order kinetics, implying that the process is unimolecular. Until about 1921, chemists did not understand the role of collisions in unimolecular processes. It turns out that the mechanisms of such reactions are quite complicated and follow second-order kinetics at very low pressures. Such reactions are described in more detail aspseudounimolecular.

## activation energy

### Higher temperatures, faster reactions

The chemical reactions associated with most food spoilage are catalyzed by enzymes produced by bacteria that mediate these processes. It is well known that chemical reactions occur faster at higher temperatures. Everyone knows that milk goes sour much faster at room temperature than in the fridge, butter goes rancid faster in summer than in winter, and eggs boil faster at sea level than in the mountains. For the same reason, cold-blooded animals such as reptiles and insects are noticeably sluggish on cold days.

It's not hard to see why this should be so. Thermal energy relates direction to motion at the molecular level. As temperature increases, molecules move faster and collide more violently, greatly increasing the likelihood of bond scissions and rearrangements, as described above.

### Activation energy diagrams

Most reactions involving neutral molecules cannot occur until they acquire the energy necessary to stretch, bend, or distort one or more bonds. This critical energy is calledactivation energythe reaction.Activation energy diagramsof the type shown below represent the total energy input into a reaction system as it progresses from reactants to products.

In particular, when examining such charts, note the following:

• O "reaction coordinate" Plotted on the abscissa represents the changes in atomic coordinates as the system progresses from reactants to products. In the simplest elementary reactions, this may correspond to the stretching or twisting of a specific bond and be represented to scale. In general, however, The reaction coordinate is a rather abstract concept that cannot be tied to a single measurable and scalable quantity.
• Oactivated complex(also known astransition state) represents the structure of the system as it appears at the peak of the activation energy curve. he doesnocorrespond to an identifiable intermediate structure (which would be best seen as the product of a separate elementary process), but rather to an atomic configuration that exists during the collision, which lasts only about 0.1 picosecond.
• Activation energy diagrams always include the energetic (Δyouor DH) of the net reaction, but it is important to understand that the latter quantities depend only on the thermodynamics of the process, which is always independent of the reaction path. This means that the same reaction can have different activation energies if allowed to follow alternative pathways.
• With a few exceptions for very simple processes, activation energy diagrams are largely conceptual constructs based on our standardcollision modelfor chemical reactions. It would be unwise to read too much into them. ### Activation Energy Diagrams Gallery

Activation energy diagrams can describe exothermic and endothermic reactions: ... and the activation energies of the forward reaction can be large, small, or zero (regardless of the value of Δ, of courseH): Processes without activation energy usually involve the combination of oppositely charged ions or the pairing of electrons in free radicals, as in the dimerization of nitric oxide (which is an electron-odd molecule). In this diagram for the dissociation of bromine, theEAit's just the enthalpy of atomization

Br2(G)→ 2Br(G)

and the reaction coordinate corresponds approximately to the elongation of the vibrated bond. The "activated complex", if considered present, is only the last and longest "distance". The reverse reaction, the recombination of two radicals, occurs immediately after contact. ### Where does the activation energy come from?

In most cases, the activation energy is provided by thermal energy, either through intermolecular collisions or (in the case of thermal dissociation) through thermal excitation of a bond stretching vibration at a sufficiently high quantum level. (Video) Arrhenius Equation Activation Energy and Rate Constant K Explained

When the products are formed, the activation energy is returned in the form of vibrational energy, which is quickly dissipated into heat. However, it is worth noting that sometimes other sources of activation energy apply:

• Absorption of light by a molecule(Photo suggestion) It can be very clean and efficient, but it doesn't always work. It is not enough that the wavelength of the light corresponds to the activation energy; it must also be within the absorption spectrum of the molecule, and (in a complex molecule) enough it must end up in the right part of the molecule, say at a specific bond.
• electrochemical activation. Molecules capable of donating or accepting electrons on the surface of an electrode can be activated by an additional potential (known asOverload) between electrode and solution. The electrode surface usually plays an active role, so the process is also calledElectrocatalysis.

### Catalysts can reduce activation energy

ACatalystis generally defined as a substance that accelerates a reaction without being consumed by it. More specifically, a catalyst provides an alternative pathway with lower activation energy between reactants and products. As such, they are vital to chemical technology; approximately 95% of industrial chemical processes involve catalysts of various types. Furthermore, most of the biochemical processes that occur in living organisms are mediated byDifficult, these are catalysts made of proteins. It is important to understand that a catalyst only affects thisKinetica reaction; he doesnochange the thermodynamic trend for the reaction to proceed. Thus, there is a unique value of ΔHfor the two paths shown in the diagram on the right

### temperature and kinetic energy

An overview of the principles of molecular velocities of gases and the Boltzmann distribution can be found inPage "KMT classic".

In the vast majority of cases, we rely on thermal activation, so we must primarily consider what fraction of molecules have sufficient kinetic energy to react at a given temperature. According to kinetic molecular theory, a population of molecules at a given temperature is distributed over a variety of kinetic energies, which is described by the Maxwell-Boltzman law of distribution. The two distribution diagrams shown here are for a lower temperatureT1and a higher temperatureT2. The area under each curve represents the total number of molecules whose energies fall within a given range. The shaded areas indicate the number of molecules with sufficient energy to meet the requirements dictated by the two values ​​ofEAthat are shown.

It is clear from these diagrams that the proportion of molecules whose kinetic energy exceeds the activation energy increases rapidly as the temperature increases. This is why virtually all chemical reactions (and all elementary reactions) proceed more rapidly at higher temperatures.

### 2 A Lei by Arrhenius

By 1890, it was well known that higher temperatures accelerated reactions, often doubling the rate for every 10 degree increase, but the reasons for this were unclear. Finally, in 1899, the Swedish chemist Svante Arrhenius (1859-1927) combined the concepts of activation energy and Boltzmann's law of distribution into one of the most important relationships in physical chemistry:  Take a moment to focus on the meaning of this equation, neglecting theAfactor for now.

First, notice that this is another form of the exponential decay law we discussed in the previous section of this series. What "breaks down" here is not the concentration of a reactant as a function of time, but the magnitude of the rate constant as a function of the exponent–EA/RT. And what is the meaning of this size? if you rememberRTand theaverage kinetic energy, it can be seen that the exponent is just the ratio of the activation energyEAto the average kinetic energy. The higher this ratio, the lower the speed (hence the negative sign). This means that high temperature and low activation energy favor higher rate constants and therefore speed up the reaction. And since these terms appear in an exponent, their impact on the rate is quite significant.

The following two diagrams show the effects of activation energy (indicated here byE) in the rate constant. Even a modest activation energy of 50 kJ/mol reduces the rate by a factor of 108.  The logarithmic scale in the diagram on the right results in nice straight lines, as described in the next heading below. Looking at the role of temperature, we see a similar effect. (If thexaxis were in "kilograms", the slopes would be more comparable in magnitude to those on the kilojoule graph in the upper-right corner.)

## Determination of activation energy

The Arrhenius Equation

(Video) Collision Theory and Activation Energy (A-Level IB Chemistry)

$k=A \mathrm{e}^{-E_{a}/RT}$

can be written non-exponentially, which is generally more convenient to use and interpret graphically. Taking the logarithm of both sides and separating the exponential and pre-exponential terms gives

$\ln k=\ln \left(A \mathrm{e}^{-E_{\mathrm{a}} / R T}\direita)=\ln A+\ln \left(\mathrm{e}^{ -E_{\mathrm{a}} / R T}\direita)$

$\ln k=\ln A-\dfrac{E_{a}}{RT}$

this is the equation of a line whose slope is $$–E_a /R$$. This provides an easy way to determine activation energy from values ​​of $$k$$ observed at different temperatures; we simply plot $$\ln k$$ as a function of $$1/T$$.

##### Example $$\PageIndex{1}$$: Isomerization of cyclopropane

So, for the isomerization of cyclopropane to propylene The following data were obtained (calculated values ​​shaded in pink): T, ° C 1/T, k–1× 103 k, S–1 lnk 477 523 577 623 1.33 1.25 1.18 1.11 0,00018 0,0027 0,030 0,26 –8,62 –5,92 –3,51 –1,35

From the calculated slope we have

– (EA/R) = –3,27 × 104k

EA=– (8.314 J mol–1k–1) (–3,27 × 104K) = 273 kJmol–1

Comment:This activation energy is quite high, which is not surprising since a carbon-carbon bond must be broken to open the cyclopropane ring. (C-C binding energies are typically around 350 kJ/mol.) For this reason, the reaction must be carried out at high temperature.

### It doesn't always have to be a plot

(...if you're willing to live a little dangerously!) From lnk-vs.-1/Tdiagram produces a straight line, it is often convenient to estimate the activation energy of experiments at just two temperatures. To see how this is done, consider this

$\ln k_{2}-\ln k_{1}=\left(\ln A-\frac{E_{a}}{R T_{2}}\right)-\left(\ln A-\ frac{E_{a}}{R T_{1}}\right)=\frac{E_{a}}{R}\left(\frac{1}{T_{1}}-\frac{1}{ T_{2}}\right) \nonumber$

(...in which weAterm disappear using the expressions for the two In-kterms.) Solving the expression on the right for the activation energy yields

$E_{a}=\dfrac{R\ln \dfrac{k_{2}}{k_{1}}}{\dfrac{1}{T_{1}}}-\dfrac{1}{T_{ two}). }}} \no number$

##### Example $$\PageIndex{2}$$

A common rule of thumb for the temperature dependence of a reaction rate is that a 10°C increase in temperature approximately doubles the rate. (Of course, this is not true in general, especially when a strong covalent bond needs to be broken.) But what would be the activation energy for a reaction that showed this behavior?

###### Solution

We'll center our ten-degree range at 300K. Substituting in the above expression gives = (8,314)(0,693) / (0,00339 - 0,00328)

= (5,76Jmol–1k–1) / (0,00011 K–1) = 52400Jmol–1=52,4 kJmol–1

##### Example $$\PageIndex{3}$$

In Los Angeles, it takes about 3.0 minutes to boil a hard-boiled egg, but at the higher altitudes of Denver, where water boils at 92°C, the boil time is 4.5 minutes. Use this information to estimate the activation energy for the coagulation of ovalbumin protein.

(Video) 14.4 Collision Theory and the Arrhenius Equation | General Chemistry

Solution:

The ratio of the rate constants in the LA and Denver Ranges is 4.5/3.0 = 1.5, and their respective temperatures are373KE365.000. With indexes 2 and 1 referring to LA and Denver, respectively, we have

EA= (8,314)(ln 1,5) / (1/365 – 1/273) = (8,314)(0,405) / (0,00274 – 0,00366)

= (3,37Jmol–1k–1) / (0,000923 K–1) = 3650Jmol–1=3,65 kJmol–1

comment: This rather low value seems reasonable, since protein denaturation involves the destruction of relatively weak hydrogen bonds; no covalent bond is broken

barbecue and popcorn

• Many biological processes exhibit a temperature dependence that obeys the Arrhenius law and can therefore be characterized by an activation energy. Check out this interesting page from Dartmouth U. which looks at the kinetics ofthe zirpen grill.
• In an article about thePopcorn popping kinetics(Cereal Chemisty 82(1): 53-59) J. Byrd and M. Perona found that the burst obeys a first-order rate law with an activation energy of 53.8 kJ/mol.

## The pre-exponential factor

Now it's time to focus on the pre-exponential termAin the Arrhenius equation. We neglect it because it is not directly involved in the relationship between temperature and activation energy, which is the main practical use of the equation. but sinceAmultiplies the exponential term, its value clearly contributes to the value of the rate constant, and therefore to the speed. and by temperature.

If this fraction were one, Arrhenius' law would reduce to

$k = A$

In other words,Ais the fraction of molecules that would react if the activation energy were zero or if the kinetic energy of all molecules were exceededEA– Admittedly, an unusual scenario.

So what would limit the rate constant if no activation energy was needed? The most obvious factor would be the rate at which reactant molecules come into contact. This can be calculated from kinetic molecular theory and is known asFrequency-orZ collision factor. For some reactions, the relative orientation of the molecules at the point of collision is important, so we can also define a geometry orsteric factor(commonly denoted by ρ (small Greek letterro). In general we can expressAas the product of these two factors:

$A = Zρ$

Values ​​of $$ρ$$ are generally very difficult to estimate; They are sometimes estimated by comparing the observed rate constant with the one at whichAit's supposed to be the sameZ.

### Direction makes the difference

The more complicated the structures of the reactants, the more likely it is that the value of the rate constant depends on the paths along which the reactants approach each other. We've shown an example of this at the top of the page, but for another let's consider adding a hydrogen halide such as HCl across the double bond of an alkene, converting it to a chloroalkane. This kind ofelectrophilic addition reactionis a term familiar to all students of organic chemistry. Experiments have shown that the reaction only occurs when the HCl molecule approaches the alkene with its hydrogen end in a direction approximately perpendicular to the double bond, as shown in Fig. sob. The reason for this becomes clear if we remember that HCl is highly polar due to chlorine's high electronegativity, so the hydrogen end of the molecule is slightly positive. The double bond of ethene consists of two clouds of negative charge corresponding to σ (Sigma) e p (pi) molecular orbitals. The latter, extending above and below the plane of the C2H4Molecule, interacts with and attracts the HCl molecule.

If, instead, the HCl is approached with its chlorine end first, as in electrostatic repulsion between like charges causes the two molecules to collide before a reaction can take place. The same happens in ; The electronegativity difference between carbon and hydrogen is too small to make the CH bond polar enough to attract the incoming chlorine atom.

(Video) Activation Energy The lesson to be learned from this example is that once you start combining a variety of chemical principles, you gradually develop what might be called a "chemical intuition" that can be applied to a variety of problems. This is much more important than memorizing concrete examples. Now that you know what it takes to get an answer, you're ready for the next lesson, which describes your actual answer.mechanisms.

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