Arrhenius Equation: Study with Video Lessons, Practice Problems & Examples

The Arrhenius equation is pivotal in understanding how temperature affects chemical reaction rates. It posits that the rate constant (K) is equal to the frequency factor (A), multiplied by an exponential factor that includes the activation energy (Ea), the gas constant (R), and the temperature (T). The equation is expressed as:

K=A⋅e^-EaRT where R is 8.314 J/(mol·K), used when discussing speed, energy, or velocity. When comparing two sets of conditions, the two-point form of the Arrhenius equation,

Ln(K2K1)=-EaR⋅(1T2-1T1) allows for the calculation of changes in the rate constant with varying temperatures. Additionally, the linear form of the Arrhenius equation, used when plotting ln(K) against 1/T, helps determine activation energy by fitting the data to a straight-line equation,

Y=mx+b where Y is ln(K), m is -Ea/R, x is 1/T, and b is ln(A). This linear relationship is crucial for interpreting graphs and extracting meaningful kinetic data about a reaction.

Arrhenius Equation Concept 1

The Arena's equation investigates the temperature dependence of chemical reaction rates. Here we're going to say that the Arrhenius equation has an Arrhenius equation formula and that is

K = A × e − E A R T

Here K = R rate constant A. Here we're going to say equals our frequency factor or also called our pre-exponential factor or sometimes called the Arrhenius factor. So these are just a bunch of different names from the same variable. EA equals our activation energy or energy of activation or energy barrier. Again, this also has names as well.

R = R gas constant, which here equals 8.314 Joules over moles times K. Now here this R constant we used 8.314 anytime we're referring to speed, energy or velocity. So keep that in mind. R becomes 8.314 with the units of Joules over moles times K anytime one of these three ideas is being discussed. OK, so keep this in mind.

The Arrhenius equation is

K = A × e − E A R T

Arrhenius Equation Example 1

The gas phase reaction of nitrogen monoxide with chlorine to form NO CL and CL has an activation energy of 7.2 kilojoules per mole and a frequency factor of 8.9 * 10⁹ molarity's inverse times seconds inverse. Here we need to calculate the rate constant at 110°C. All right, so here we have to figure out what our rate constant is. So we're looking for K. And remember with the Arrhenius equation:

K = A ⁢ e - E A R T

we're told what our frequency factor is. So this is A and that is 8.9 * 10⁹ molarity is inverse times seconds, inverse times e to the negative.

Now here R uses joules, so activation energy also needs to be in joules. So we're going to say we have 7.2 kilojoules per mole. And remember that one KJ is equal to 10³ joules. So that's 7200 joules per mole. So that's what we're going to put here, 7200 joules per mole. We're going to say R is our gas constant 8.314 Joules over moles times K and then our temperature needs to be in Kelvin. So add 273.15 to our degree Celsius and that gives us our Kelvin, which comes out as 383.15 K.

So we plug that in 383.15 Kelvin. If we look here, we're going to say equals. So Joules cancel out with joules, Moles cancel out with moles, Kelvin cancel out with Kelvins. So there are only no units on this portion, so the units for our rate constant here will be in molarities inverse times seconds, inverse when we plug all this in. So we're going to do here our frequency factor times e raised to this power. When we do that, we're going to get 9.285 * 10⁸ molarities inverse times seconds inverse.

Now within the question, this has two sig figs. This has two sig figs. This has two sig figs. So we need two sig figs as our final number. Significant figures. So we have 9.3 * 10⁸ molarities inverse times seconds inverse. So this will be our final answer.

Arrhenius Equation Concept 2

Now the two point form of the Arrhenius equation shows how changing the temperature can impact the rate constant, which uses the variable K. Now here we're going to say that the higher the reaction temperature, this causes an increase in the rate constant K. Now here when do we use a two point form of the Arena's equation? Well, it's used whenever we're dealing with two rate constants and two temperatures for a given reaction.

Now the Arrhenius equation 2 point form formula is as follows. It is ln⁡(K2K1)=-EAR⁡(1T2-1T1). Here K1 is our initial rate constant, K2 is our final, T1 is our initial and T2 is our final. Now the thing about this is this is the version I want you to remember. This is the one you need to memorize. Professors sometimes can be a little bit tricky. They'll give you 2 forms of the same exact equation. It doesn't matter which one you use because mathematically you'll get the same answer.

Now you might see it written as ln⁡(K2K1)=EAR⁡(1T1-1T2). The equations are pretty similar. Where's the difference? The difference is here we're using negative EA and here it is a positive EA. Here we're using T2 first, here we're using T1 first. Again. It doesn't matter which one of the two that you use, both will give you the same exact answer whether you're solving for one of the KS, one of the TS, or activation energy itself.

I just prefer to use this top one here because this is the one that is shown most often. OK, so just keep that in mind. You might see it in two different ways in class, on a formula sheet, what, whatever. But use the top one, all right? So keep in mind this is the two point form of the Arena's equation. Use it anytime we're dealing with two rate constant KS or two temperatures T.

Arrhenius Equation Example 2

A chemical reaction has rate constants of 4.6 * 10^-2 seconds^-1 and 8.1 * 10^-2 seconds^-1 at 0°C and 20°C respectively. What is the value of the activation energy? All right, so they're giving us two rate constants since this one is stated first. This is K₁, and this would have to be K₂ respectively. Would mean that 0°C is connected to K₁ and therefore would be T₁ and the second temperature will be connected to K₂ and therefore it is T₂.

Now we'd say here that ln(K₂/K₁) = -E_A/R * (1/T₂ - 1/T₁). Here we're going to place the numbers that we have so it's ln(K₂/K₁). So K₂ is 8.1 * 10^-2 / 4.6 * 10^-2. We're looking for activation energy, so E_A is what we need to find. R here is our gas constant, which is 8.314 Joules/(mole·K). We're going to say 1/T₂. So remember we have to add 273.15 to each one of these Kelvin temperatures. So when we do that, this becomes 293.15 K - 1/273.15 K.

Here we're going to do ln(K₂/K₁). When we do that, we're going to get as our answer, 0.565808. All right, so E_A is going to be multiplying by what's in here. So when we do that, we're going to get a negative value in here. So this is going to be what's in here is equal to -2.49769 * 10^-4. A negative times a negative gives me a positive that be positive E_A times 2.49769 * 10^-4 divided by still 8.314.

Here we're going to multiply both sides by 8.314, so when we do that, we're going to get 4.70413 equals E_A, still times 2.49769 * 10^-4. Divide both sides now by 2.49769 * 10^-4. So when we do that, we're going to get our E_A, which I'm going to write over here. So E_A here equals 18,833.91 joules per mole.

Now if we go back up here, this has two sig figs, 2 sig figs for 4.6 and 8.1. Let's not worry about the temperatures, so let's just do this in terms of two sig figs. So that will become 1.9 * 10⁴ joules per mole. So this would be our value for our activation energy, abbreviated as E_A.

Arrhenius Equation Concept 3

So we use the linear form of the Arrhenius equation when a plot of lnK versus inverse temperature is given, and we're going to say it's used to determine the activation energy or EA of the reaction. Now recall that the equation for a straight line is Y=MX+B. And what we're going to do is we're going to rearrange the Arrhenius equation to fit this equation for a straight line.

Now if we take a look at a plot, remember our X here is represented by inverse temperature, our Y by lnK. lnA represents our starting point. And remember our slope is changing Y over changing X, which is rise over run. By basically rearranging our Arrhenius equation we're able to get the linear form of it. Here this linear form is related to a straight line, it's equal to Y=MX+B.

Here our Y is lnK, our slope M is related to -EA/R, inverse temperature is equal to X and lnA is B, which is again our starting point. Now here we'd say remember these variables that K is our rate constant, A equals our frequency factor or preexponential factor, EA is our activation energy and R is our gas constant, which is equal to 8.314 Joules over moles times K.

So remember we use this linear form of the Arrhenius equation anytime we're given a plot of lnK versus inverse temperature.

Arrhenius Equation Example 3

A plot of LNK versus inverse temperature has a slope of -8313. What is the activation energy for this reaction? So remember when they give us a plot, it's always of Y versus X, and the fact that we're dealing with LMK versus inverse temperature means that we're going to deal with the linear form of the Arrhenius equation.

So here we're going to say that lnK=-EA/R∙1T+lnA. And remember this is equal to the formula for straight line, which is Y=MX+B. M represents our slope and it's also equal to negative activation energy divided by R, our gas constant. So them telling us our slope here is them really giving us M? We're going to set them equal to each other here.

This is -8313 Kelvin equals -EA/R. R is our gas constant 8.314 Joules over moles times K. We're going to multiply both sides by 8.314 joules over moles times K. Kelvin's can't slap and we'll have our answer in joules per mole. Now here, initially we're going to have is -69114.282 joules per mole equals -EA.

Just divide both sides by -1 and we'll get EA equals here. This has four sig figs in it. Or answer should have four sig figs as well, so we're going to say 69110 joules per mole as our final answer.

The rate constant of a reaction at 32°C is 0.060/s. If the frequency factor is 3.1 × 10¹⁵ s^–1, what is the activation barrier?

• A. 9.1 x 10⁴J/mol
• B. 1.0 x 10⁴J/mol
• C. 9.8 x 10⁴J/mol
• D. 8.3 x 10⁴J/mol

A reaction with an activation energy E_a = 55.00 kJ/mol is run at temperature of 30ºC. Determine the temperature required to increase the rate constant 3 times.

• A. 319 K
• B. 46 K
• C. 288.6 K
• D. 303 K

The following data shows the rate constant of a reaction measured at numerous temperatures. Use the Arrhenius plot to determine the frequency factor for the reaction.

• A. 1.08×10⁴
• B. 3.00×10^-3
• C. 30.4
• D. 1.57×10¹³
• E. 6.45×10^-6

Here’s what students ask on this topic:

Considering the Arrhenius equation, what is the slope of a plot of $\ln k$ versus $\frac{1}{t}$ equal to?

The Arrhenius equation is a formula that describes how the rate constant ($k$) of a chemical reaction depends on temperature. It can be written as:

$k=A\cdot e^{-\frac{Ea}{R\cdot T}}$

where:

• $k$ is the rate constant,
• $A$ is the pre-exponential factor,
• $Ea$ is the activation energy,
• $R$ is the gas constant, and
• $T$ is the temperature in Kelvin.

Taking the natural logarithm of both sides gives:

What is "a" in the Arrhenius equation?

In the Arrhenius equation, "A" represents the pre-exponential factor or frequency factor. It's a constant that provides the frequency of collisions with the correct orientation for a chemical reaction to occur. The full Arrhenius equation is:

$k=A{e}^{-\frac{Ea}{RT}}$

where:

• k is the rate constant of the reaction,
• A (the pre-exponential factor) is indicative of the number of collisions that result in a reaction,
• e is the base of the natural logarithm,
• E_a is the activation energy required for the reaction (expressed in joules per mole or kilojoules per mole),
• R is the universal gas constant, and
• T is the temperature in Kelvin.

The pre-exponential factor "A" essentially relates to the likelihood that reactant particles will collide in the correct orientation with sufficient energy to overcome the activation energy barrier and react. It's influenced by factors such as the frequency of collisions and the probability that collisions are successful in leading to a reaction.

Which of the following is true about the Arrhenius equation? Can you select all that apply?

The Arrhenius equation is a formula that describes how the rate constant (k) of a chemical reaction depends on the temperature and activation energy of the reaction. Here are some true statements about the Arrhenius equation:

1. It shows that the rate constant (k) increases exponentially with an increase in temperature.
2. The equation includes a factor (A), known as the pre-exponential factor or frequency factor, which is related to the frequency of collisions and the orientation of reacting particles.
3. It contains the term $e^{-\frac{E{a}_{}}{RT}}$, where E_a is the activation energy, R is the universal gas constant, and T is the temperature in Kelvin. This term represents the fraction of molecules that have enough energy to overcome the activation energy barrier at a given temperature.
4. The equation can be written in two forms: the exponential form $k=A{e}^{-\frac{E{a}_{}}{RT}}$

Which form of the Arrhenius equation can be conveniently used to calculate E_a for a reaction?

The Arrhenius equation in its logarithmic form is most convenient for calculating the activation energy (E_a) of a reaction. This form is given by:

$\ln \left(k\right)=\ln \left(A\right)-\frac{E{a}_{}}{R}\left(\frac{1}{T}\right)$

where:

• k is the rate constant of the reaction,
• A is the frequency factor, which represents the number of times that reactants approach the activation barrier per unit time,
• E_a is the activation energy,
• R is the universal gas constant (8.314 J/mol*K), and
• T is the temperature in Kelvin.

To calculate E_a, you can conduct experiments to determine the rate constant k at different temperatures T. Then, plot ln(k) versus 1/T on a graph. The slope of the line obtained from this plot will be -E_a/R. Multiply the slope by -R to find the activation energy E_a for the reaction.

How can the frequency factor be found in the Arrhenius equation?

The frequency factor (also known as the pre-exponential factor) in the Arrhenius equation is a term that represents the frequency of collisions with the proper orientation for a reaction to occur. It's denoted by the symbol A and has units of s^-1.

To determine the frequency factor, you typically need experimental data. You can find it by plotting the natural logarithm of the reaction rate constant (k) against the inverse of the temperature (1/T) in Kelvin. This plot should yield a straight line if the reaction follows the Arrhenius behavior, and the equation for this line is derived from the Arrhenius equation:

$\ln \left(k\right)=\ln \left(A\right)-\frac{Ea}{R\frac{1}{T}}$

where:

• k is the rate constant,
• A is the frequency factor,
• E_a is the activation energy,
• R is the gas constant,
• T is the temperature in Kelvin.

The y-intercept of this plot will be ln(A), so by taking the antilogarithm of the y-intercept, you can determine the value of the frequency factor A. This method requires that you have a series of rate constants measured at different temperatures.