All reaction rates increase with increased temperature
The vast majority of chemical reactions speed up as heat is added. There are two approaches to understanding this. One is to begin with first principles, that is, with the basic physics of collisions of particles in a gas or liquid and building up to more complicated collisions involving orientation, considering the effect of temperature on particle speeds, and incorporating a statistical approach in scaling up to an "ensemble" of many particles (like a mole), culminating in a model of the nature of a reaction which is good enough to make predictions. This is the statistical mechanics approach, and it is fascinating, but beyond the scope of this page.
The second approach is the phenomenological one, first proposed by Svante Arrhenius, a Swedish physical chemist, who won the Nobel Prize in chemistry in 1905. Arrhenius observed that
- Heat causes particles to move faster (have more kinetic energy) in a predictable way, and thus experience more collisions per unit time;
- The probability of a reaction occurring upon collision is likely to depend in some way upon the orientation of the colliding particles; and
- As a reaction mixture is heated, the number of particles that have enough kinetic energy to react (activation energy, Ea) rises.
This approach is often referred to as collision theory. From these observations, Arrhenius proposed the relation in the box below.
The cool thing about the statistical mechanics approach is that the model we develop is completely consistent with Arrhenius' phenomenological approach.
phenomenological
The phenomenological approach to a problem focuses on what happens that can be measured (seen, heard, smelled, temperature recorded, &c.), to find a model (often mathematical) that can be used to predict future behavior, or behavior under different conditions. This is contrasted with a first-principles or ab initio approach, which forms a model beginning with the simplest physical principles and building from there.
The Arrhenius Equation
Arrhenius observed that the rate of a reaction is exponentially proportional to the temperature and directly proportional to a second parameter that accounts for effects like alignment during a collision.
$$k = A \, e^{\frac{-E_a}{RT}}$$
where k is the rate constant with units of s-1 ("per second"), Ea is the activation energy, R = 8.314 J·mol-1K-1 is the molar gas constant, T is the temperature in Kelvins, and A is called the pre-exponential factor or the frequency factor.
The activation energy
Let's work through the elements of Arrhenius' equation. First, the activation energy, Ea. All (except for a special few) chemical reactions have an activation barrier that must be overcome before the reaction can proceed.
Think about burning methane or propane gas in a Bunsen or Tirrill burner in the chemistry lab. You can let the gas out of the burner, but it won't ignite unless you provide some small spark with an igniter. That small amount of energy is necessary for starting the spontaneous, self-sustaining reaction that then begins. For the rest of the time the flame is burning, the energy normally released by the (exothermic) reaction is sufficient to activate all further combustion reactions. The plot below shows a reaction profile for a typical exothermic (gives off heat) reaction.
Notice that the process of conversion of reactants to products proceeds through some sort of intermediate phase, called the transition state, which we usually designate with the "double dagger" symbol, ‡.
The transition state (which we'll explore more below) involves a structure that is something in between the structures of the reactants and products. It represents a point in a reversible reaction from which the reaction might proceed either way — toward products or back toward reactants.
The activation energy is the difference between the potential energy of the transition state and that of the reactants.
Notice that the reverse reaction of the one shown in the graph is endothermic, and that the activation energy is larger. Here's that same reaction reversed so you can see what I mean:
This is a general feature of reversible reactions that are endo- or exothermic.
One important requirement for a reaction to occur is that the particles simply need to slam into one another hard enough. "Hard enough" is the activation energy, Ea
How many particles have the kinetic energy it takes to react?
The graph below shows the distribution of molecular speeds at three temperatures. Particle speeds in a three-dimensional system (like a container of gas or a beaker of liquid) conform to a speed distribution called the Maxwell-Boltzmann distribution.
The Maxwell Boltzmann distribution is a plot of the number or fraction of particles in a system that have a certain speed vs. those speeds, where speed is directly proportional to kinetic energy (E = ½mv2). The plot below shows Maxwell-Boltzmann distributions for three hypothetical temperatures, T1 < T2 < T3
Notice that each curve is peaked at the most-probable speed — the speed that most particles are likely to have, but it is also skewed toward higher speeds; that turns out to be a consequence of working in three dimensions.
The area under each curve represents the total number of particles under consideration, so as we elongate each curve (going from T1 to T2 to T3) it must shorten to keep that area the same.
The activation energy of a hypothetical reaction is shown as a dashed vertical line. Notice that at the lowest temperature (T1) shown, no particles have enough energy to react — there is not enough heat in the system. At T2 a small fraction of all particles have speeds exceeding the activation energy, therefore some reactions can proceed. And at T3, over half of the particles have enough kinetic energy to react upon collision.
It's worth remembering this graph and being able to sketch it yourself when you need to think about such things. It clearly shows that as a reaction mixture is heated, the number of particles with sufficient kinetic energy to react increases.
This is reflected in the Arrhenius equation:
$$k = A e^{\frac{-E_a}{RT}}$$
To get a feel for how it works, take just the exponential term and put the negative in the exponent to work:
$$e^{\frac{-E_a}{RT}} = \frac{1}{e^{\frac{E_a}{RT}}}$$
Now if we set Ea = R = 1 just for the purpose of figuring out the behavior of the exponential term with respect to temperature, we get this proportionality:
$$e^{\frac{-E_a}{RT}} \propto \frac{1}{e^{\frac{1}{T}}}$$
Now it's fairly clear to see that as temperature increases, the exponential term gets close to 1 (e0 = 1), meaning that every particle has sufficient energy to react. Conversely, as T → 0, the denominator (e1/T) gets large, therefore the exponential term approaches zero, and no reactions occur.
So the Arrhenius equation correctly captures the basic idea behind activation energy.
The pre-exponential factor (frequency factor), A
The Arrhenius equation can really be viewed as the product of two probabilities:
- The probability that colliding particles have sufficient energy to react (given by the exponential part of the equation), and
- The pre-exponential factor, which gives the probability that particles with sufficient energy to react do react. The factor, A, mostly represents orientation effects: Colliding particles must be properly aligned to react, although the sensitivity to alignment can vary a lot from reaction to reaction.
In a more detailed treatment of the Arrhenius equation (working from basic physics principles), all of the orientation information is included, and we simply group it into the factor A. Here it will have to suffice that for every reaction, there is some orientation-dependent probability. The good news is that A isn't too difficult to find experimentally, as we shall see.
An alignment example
As an example of the effect of alignment, take the conversion of carbon monoxide, CO to carbon dioxide (CO2), a reaction that can happen inside the catalytic converter in a gas automobile. In order for the reaction to occur, a collision between CO and O2 must occur with an alignment that allows a collision between one of the oxygen atoms of O2 and the carbon of CO. A collision 180˚ apart produced no reaction:
For most reactions, alignment has some flexibility. In this figure, CO remains stationary and we approach it from various directions around a circle with an O2 molecule. The depth of color of the O2 in the diagram gives the probability that a reaction will occur.
Some reactions are quite sensitive to alignment, and others not so much. These factors are encapsulated in the pre-exponential factor in a way that doesn't allow us to tease them out specifically, but they are there and can be determined by careful experimentation, such as the work done in Professor Y.T. Lee's lab at UC Berkeley, for which he won the Nobel Prize in Chemistry.
Determining Ea and A from experimental data
If we begin with the Arrhenius equation,
$$k = A e^{\frac{-E_a}{RT}}$$
and take the natural log of both sides (noting that the log of a product is a sum of logs), we have
$$ln(k) = ln(A) - \frac{E_a}{RT}$$
A subtle rearrangement of this formula gives an equation that looks linear:
Notice the correspondence between terms of the Arrhenius expression and y = mx + b. This means if we measure a set of rates (k) and temperatures, T, we can create a plot of ln(k) vs. 1/T. The slope of that plot will be -Ea/R and its intercept (if we're comfortable extrapolating that far) is ln(A).
Our plot might look something like this hypothetical one:
vs1OverTGraphSample.png)
In order to find the activation energy from the slope, we multiply by -R. If we're working in moles, R = 8.314 J/mol·K, and if we're working in particle numbers, R becomes the Boltzmann constant*, k = 1.381 × 10-21 J/K. The pre-exponential factor, A is just the value of the graph intercept raised as a power of e.
*
Example 3 below gives a nice example of finding Ea and A from such a plot using real data.
Important note:
Some Arrhenius plots are not linear. Such exceptions are often concave upward in such a way that two sections are nearly linear (think hockey stick). In some cases they are concave downward. There are competing theories, beyond the scope of this section, for these exceptions.
If the region you are most interested in is at higher temperatures (smaller 1/T), then simply tossing out the non linear portion of the data and fitting a line to the linear data might be a solution. How you deal with any nonlinearity is case dependent.
Example 1
The activation energy of a reaction is Ea = 80 KJ/mol. By what factor will the rate of the reaction increase by heating this reaction from 300K to 320K ?
Begin a problem like this by taking logs of both sides of the Arrhenius equation. It's the most convenient form for this purpose:
$$ln(k) = ln(A) - \frac{E_a}{RT}$$
Now the pre-exponential factor and the activation energy are properties of the reaction that don't depend on temperature, so we can write two equations keeping those constant, like this:
$$ \begin{align} ln(\color{#E90F89}{k_1}) &= ln(A) - \frac{E_a}{R\color{#E90F89}{T_1}} \\[5pt] ln(\color{#E90F89}{k_2}) &= ln(A) - \frac{E_a}{R\color{#E90F89}{T_2}} \end{align}$$
That gives us two equations and two unknowns. One way to solve them simultaneously is to subtract them: ln(k2) - ln(k1) is
$$ln(k_2) - ln(k_1) = ln(A) - \frac{E_a}{RT_2} - ln(A) + \frac{E_a}{RT_1}$$
Now using the laws of logs on the left (a difference of logs is the log of a quotient) and noticing that the ln(A) terms vanish on the right, we have
$$ln\left( \frac{k_2}{k_1} \right) = \frac{E_a}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right)$$
Now we can plug in the numbers and cancel the units. Notice that all units cancel. This is necessary; the argument of a log function should have no units, and the number we're seeking is a ratio of quantities (rates) with the same units.
$$ln\left( \frac{k_2}{k_1} \right) = \frac{80,000 \, J\cdot mol^{-1}}{8.314 J \cdot mol^{-1} K^{-1}} \left( \frac{1}{300} - \frac{1}{320} \right) \, K$$
We have
$$ln\left( \frac{k_2}{k_1} \right) = 2.0$$
Raising both sides as a power of e gives the ratio of k2 to k1:
$$\frac{k_2}{k_1} = e^{2.0} = 7.4$$
So k2 is a factor of 7.4 greater than k1,
$$k_2 = 7.4 \, k_1$$
which means that a 20˚C rise in temperature leads to a seven-fold increase in the rate of the reaction. If the activation energy were higher, the temperature effect would be greater. This plot shows that dependence for several activation energies:
Example 2
A catalyst increases the rate of a reaction by reducing the activation energy. If the activation energy of a particular reaction is 50 KJ/mol and we want to run it at 300 K, by how much must a catalyst reduce that activation energy in order to cause a ten-fold increase in the rate?
We start again with this form of Arrhenius' equation.
$$ln(k) = ln(A) - \frac{E_a}{RT}$$
Now rewrite the equation twice, one with the original rate (k1) and the original activation energy (Ea1), and one with the faster rate (k2 = 10 k1) and the lower activation energy, Ea2.
$$ \begin{align} ln(\color{#E90F89}{k_1}) &= ln(A) - \frac{\color{#E90F89}{E_{a1}}}{RT_1} \\[5pt] ln(\color{#E90F89}{k_2}) &= ln(A) - \frac{\color{#E90F89}{E_{a2}}}{RT_2} \end{align}$$
Now subtract one equation from the other:
$$ln(k_2) - ln(k_1) = ln(A) - \frac{E_{a2}}{RT} - ln(A) + \frac{E_{a1}}{RT}$$
Using the laws of logs (a difference of logs is the log of a quotient), and factoring out 1/RT on the right gives
Now plugging in those numbers gives
$$E_{a2} = 50,000 - RT \, ln(10)$$
We finish by plugging in R and T to get the final activation energy:
$$ \begin{align} &= 50000 \,\frac{J}{mol} - \left( 8.314 \, \frac{J}{mol\cdot K} \right) 300 \, K \cdot ln(10) \\[5pt] &= 44,257 \; J \end{align}$$
This reduction of about 5.7 KJ/mol represents an 11.4% drop in the activation energy. That's not much of a decrease for a ten-fold increase in rate.
If only we could design custom catalysts to reduce activation energies by any arbitrary amount. While there has been much progress in research on catalysts, we're still a way away from that dream. Catalysts are mostly still found through trial and error.
Example 3
Use the data below to estimate the activation energy, Ea and the pre-exponential factor for the decomposition (by hydrolysis) of chloroacetic acid.
The reaction is this one, in which a hydroxyl (OH) group from water replaces the chlorine on chloroacetic acid:
In 1884 van't Hoff and others measured the rate of this reaction at several temperatures. Here is a portion of the data:
| T (˚C) | Rate (s-1) |
|---|---|
| 80 | 2.22 |
| 90 | 6.03 |
| 100 | 17.3 |
| 110 | 43.6 |
| 120 | 105 |
| 130 | 237 |
We can use data like this to determine activation energy and the pre-exponential factor by forcing the non-linear equation to behave like a line, like this:
In this form, if we plot ln(k) on the y-axis vs. 1/T on the x, the slope will be -Ea/R and the y-intercept will be ln(A).
Our plot of ln(k) vs. 1/T looks like this:
Whether with a spreadsheet program or your graphing calculator, it's a fairly simple matter to perform a best fit of a line to this data. The result gives a slope of -973.4, so a good estimate of the activation energy is
$$E_a = 973.4 \cdot 8.314 = 8.1 \; KJ/mol$$
The (estimated) pre-exponential factor is the y-intercept raised as a power of e, or
$$A = 344,552$$
Linear fits can be done on your graphing calculator
Linear fits or regressions can be done right on your graphing calculator (like a TI-84, or similar). Here's briefly how to do it, but you really learn by doing it yourself and making some mistakes.
How to
To plot the points, press [STAT], then select EDIT. You'll see columns of "lists," labeled L1, L1, ... L6. You can access these in various list operations using the [2ND]→[1] through [6] buttons. Edit one of the columns, filling it with your x-axis (1/T) data. You can do math as you enter the data into the column. For example, if your first temperature is 300K, just enter "1/300" as your first list item. Finish the second list with the ln(rate) values.
Press [2ND]→[STATPLOT] and activate Plot #1. You can choose the type of plot to use. I've used a scatter plot above, the best choice for most applications like this. Make sure to go to [Y=] and clear any functions there.
Hit [GRAPH]. You may have to punch [ZOOM] and go to "Zoom stat" to center your plot on the screen.
Now to perform a fit of a line to the data, again push [STAT], arrow over to CALC and hit [ENTER]. You'll want 2-VARIABLE STATS for this application. Hit that and select your x and y lists — make sure to keep them straight. Select LIST 1 by entering [2ND]→[1], and so on.
You don't need to worry about FREQLIST for these simple fits. Just select CALCULATE and hit enter. Using the data from Example 3, you should get the result above. The output is a linear equation along with its parameters a (the slope) and b (the y-intercept).
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