Arrhenius Equation - Video Tutorials & Practice Problems
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The Arrhenius Equation illustrates the temperature dependence of the rate constant k.
Arrhenius Equation
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Arrhenius Equation Concept 1
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Arrhenius 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 equals a times here e is the inverse of the natural log taken to the negative e a over r t. Here, k equals our 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 for the same variable. Ea equals our activation energy or act energy of activation or energy barrier. Again, this also has names as well. R equals our gas constant, which here equals 8 0.314 joules over moles times k. Now here, this r constant, we use 8.314 anytime we're referring to speed, energy, or velocity. Okay. 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. Okay? So keep this in mind, the Arrhenius equation is k equals a times e to the negative e a over r t.
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Arrhenius Equation Example 1
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The gas phase reaction of nitrogen monoxide with chlorine to form NOCl and Cl has an activation energy of 7.2 kilojoules per mole and a frequency factor of 8.9 times 10 to the 9, molarities inverse times seconds inverse. Here we need to calculate the rate constant at a 110 degrees Celsius. Alright. 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 equals a times e to the negative ea over rt. We're told what our frequency factor is, so this is a, and that is 8.9 times 10 to the 9, molarities 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 gonna say we have 7.2 kilojoules per mole. And remember that 1 kilojoule is equal to 10 to the 3 joules. So that's 7,200 joules per mole. So that's what we're gonna put here, 7,200 joules per mole. Next, we're gonna 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 degrees 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 gonna say equals. So joules cancel out with joules, moles cancel out with moles, kelvin cancel out with kelvins. So there will be 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 gonna do here our frequency factor times e raised to this power. When we do that, we're gonna get 9.285 times 10 to the 8, molarities inverse times seconds inverse. Now within the question, this has 2 sig figs, this has 2 sig figs, this has 2 sig figs, so we need 2 sig figs as our final number of significant figures. So So we have 9.3 times 10 to the 8, molarities inverse times seconds inverse. So this will be our final answer.
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Arrhenius Equation Concept 2
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Now the 2 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 the 2 point form of the Regis equation? Well, it's used whenever we're dealing with 2 rate constants and 2 temperatures for a given reaction. Now the Arrhenius equation 2 point form formula is as as follows. It is lnk2overkoneequalsnegativeeaoverr times 1 over t 2 minus 1 over t 1. Here, k 1 is our initial rate constant, k 2 is our final, t 1 is our initial, and t 2 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 two 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 lnk2overkoneequalseaoverr times 1 over t 1 minus 1 over t 2. Equations are pretty similar. Where's the difference? The difference is here we're using negative e a, and here it is a positive e a. Here we're using t 2 first. Here we're using t 1 first. Again, it doesn't matter which one of the 2 that you use, both will give you the same exact answer, whether you're solving for 1 of the k's, 1 of the t's, or activation energy itself. I just prefer to use this top one here because this is the one that is shown most often. K? So just keep that in mind. You might see it in 2 different ways, in class, on a formula sheet, what whatever. But use the top one. Alright. So keep in mind, this is the 2 point form of the Arrhenius equation. We use it anytime we're dealing with 2 rate constant k's or 2 temperatures t.
When dealing with TWO rate constants or TWO temperatures then we use the Two-Point Form of the Arrhenius Equation.
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example
Arrhenius Equation Example 2
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A chemical reaction has rate constants of 4.6 times 10 to the negative 2 seconds inverse and 8.1 times 10 to the negative 2 seconds inverse at 0 degrees Celsius and 20 degrees Celsius respectively. What is the value of the activation energy? Alright. So they're giving us 2 rate constants, since this one is stated first, this is k 1, and this would have to be k 2. Respectively would mean that 0 degrees is connected to k 1, and therefore it would be t 1, and the second temperature will be connected to k 2, and therefore it is t 2. Now we'd say here that l n of k 2 over k 1 equals negative e a over r, times 1 over t 2 minus 1 over t 1. Here we're going to place the numbers that we have, so it's ln of so k2 is 8.1 times 10 to the negative 2, divided by 4.6 times 10 to the negative 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 over moles times k times, we're gonna say 1 over t 2, so remember we have to add 273 0.15 to each one of these Kelvin temperatures. So when we do that this becomes 293.15 k minus 1 over 273.15k. Here we're going to do ln of 8 of k 2 over k 1. When we do that, we're gonna get as our answer 0.0 or 0.5 65808 equals Alright. So e a is gonna be multiplying by what's in here. So when we do that we're gonna get a negative value in here, so this is gonna be what's in here is equal to negative 2 point 49769 times 10 to the negative 4. A negative times a negative gives me a positive, so that'd be positive e a times 2.49-769 times 10 to the negative 4, divided by still 8.314. Here we're going to multiply both sides by 8.314. So when we do that, we're gonna get 4.70413 equals e a still times 2.4969 times 10 to the negative 4. Divide both sides now by 2.4979 times 10 to the negative 4. So when we do that we're gonna get our e a, which I'm gonna write over here. So e a here equals 18,833.91 joules per mole. Now if we go back up here, this has 2 sig figs, 2 sig oh, 2 sig figs for 4.68.1. Let's not worry about the So let's just do this in terms of 2 sig figs. So that will become 1.9 times 10 to the 4 joules per mole. So this would be our value for our activation energy abbreviated as ea.
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Arrhenius Equation Concept 3
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So we use the linear form of the Arrhenius equation when a plot of l n k versus inverse temperature is given. And we're gonna say it's used to determine the activation energy or e a of the reaction. Now recall that the equation for a straight line is y equals m x plus b, and what we're gonna do is we're going to rearrange the Urgent's 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 ln k. Ln a represents our starting point, and remember our slope is change in y over change in 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 equals m x plus b. Here, our y is l n k, our slope m is related to negative e a over r, one over t, our inverse temperature is equal to x, and l n a 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 pre exponential factor, e a is our activation energy, and r is our gas constant which is equal to 8.314joules over moles times k. So remember, we use this linear form of the Arrhenius equation anytime we're given a plot of l n k versus inverse temperature.
In order to relate the plot of a graph to the Arrhenius equation then we manipulate it into the Linear Form of the equation.
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example
Arrhenius Equation Example 3
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A plot of l and k versus inverse temperature has a slope of negative 8,313. 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 gonna deal with the linear form of the Arrhenius equation. So here we're going to say that lnk equals negative e a over r times one over t, plus ln of a. And remember, this is equal to the formula for a straight line, which is y equals 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 gonna set them equal to each other. Here, this is negative 8313 kelvin equals negative e a over 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. Kelvins cancel out and we'll have our answer in joules per mole. Now here, initially, we're gonna have is negative 69,114.282 joules per mole equals negative e a. Just divide both sides by negative one and we'll get e a equals Here, this has 4 sig figs in it, so our answer should have 4 sig figs as well. So we're gonna say 69,100 and 10 joules per mole as our final answer.
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Problem
Problem
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
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Problem
Problem
A reaction with an activation energy Ea = 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
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Problem
Problem
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.