Collision Theory and Arrhenius Equation Explained

[00:00:00]collision theory and the arrhenius equation going to be the topics in this lesson from my new general chemistry playlist my name is chad and welcome to chad's prep where my goal is to take the stress out of learning science now in addition to high school and college science prep we also do mcat dat and oat prep as well i'll be sure to leave links to those courses in the description below now i am posting several of these lessons a week so if you want to be notified every time i post one subscribe to the channel click the bell notification now we're going to start off with collision theory and this really gives us kind of a conceptual understanding of the requirements that need to be met at the molecular level for a chemical reaction to take place and then we'll finish this off with the uranus equation which is going to show the dependence of the rate constant that lowercase k on both the activation energy and the temperature so we'll kind of get a little bit of a conceptual understanding and put it in kind of a mathematical description but we're going to have some calculations to go with it as well so let's start with collision theory and again collision theory is the kind of at the molecular level the requirements that need to take place for a chemical reaction to occur in the first of those requirements is there's got to be a collision so if two molecules are going to be reacting with each other then they need to collide with each other so that's the first part and you can kind of see why higher concentrations of reactants generally lead to faster reactions because if you have more reactants closer together then they're going to collide more often which is going to lead to a faster reaction but that's not enough in addition to a collision the molecules actually have to collide when they have the proper orientation and so we like to think of that when this collision happens that kind of kick-start this reaction you're going to have bonds breaking and so it turns out you know in the parts of the molecule where these bonds are being broken those are the parts that have to be involved in the collision or else these molecules just bounce right off each other and no reaction happens and so again you've got to have a collision but also the molecules have to have the right orientation during that collision as well but even that's not enough turns out they've got to have sufficient energy during this collision so they've got to have been moving fast enough when they collide that they can break these bonds and what this translates into and related to something we talked about the last lesson is they've got to have enough energy to get over that activation energy barrier we often associate the activation energy in a reaction with the bonds that are being broken and so if two molecules don't have enough kinetic energy going into that collision then they're just going to bounce off each other again even if they have the right orientation so they got to have enough energy to get over that activation energy barrier so those are our three requirements we've got to have a collision we've got to have the molecules in the right orientation and they have to have enough energy to get over that activation energy barrier now you've got to understand these three parts to collision theory but you're also going to understand how we use this to explain from a conceptual level why reactions tend to go faster at higher temperatures and so at higher temperatures you've got to realize that on average molecules are moving faster their average kinetic energy goes up now are all molecules moving faster at higher temperatures no but the average shifts towards higher kinetic energy so what's really going on there if you have molecules moving faster well then they're going to collide more often and so that's the first place that we're going to see temperature play a role just you get a greater collision frequency is somehow sometimes how it's worded and that just means that collisions are happening more frequently and if you have collisions more frequently then you're going to have more of those collisions more frequently that lead to a chemical reaction but the other place temperature plays a role is in this last part with sufficient energy so it turns out you can't change the the proper orientation only a certain percentage of the collisions are going to have that right orientation and you can't really change that by temperature but what temperature is going to do is it's going to change uh it's going to increase the the percentage of molecules that have enough energy to get over that activation energy barrier enough change the percentage of molecule that have sufficient energy and so the idea is that at any given temperature you have really this distribution of molecules so and what you're going to do is shift that distribution to higher energies as you go to higher temperatures and so maybe at some low temperature only 10 percent of the collisions have enough energy to get over that activation energy barrier but maybe you raise the temperature substantially and now 30 percent of those molecules actually have enough energy to get over that activation energy when they collide and so you're going to get a faster reaction as a result and so that's kind of the two-fold effect of temperature you get a greater collision frequency the molecules are colliding more frequently and you get more a greater percentage of the molecules having enough energy so to get over that activation energy barrier so now we're going to take a look at the arrhenius equation here and the uranus equation relates the rate constant to the activation energy and the temperature at which a reaction is carried out so a couple things you should understand about this lovely rate constant and so the first part is that this is the same rate constant that showed up in a typical rate law and so typical rate laws looked something like this and the rate was directly proportional to the value of the rate constant so a higher rate constant led to a faster rate and a smaller rate constant led to a slower rate that's the first part you can understand so but you should also understand the effects of activation and energy and temperature on the value of this rate constant now in this case if you understand oh yeah you look at this lovely equation and you see a graph that's exponential decay going down well then props to you you are in the vast minority but you should know the impact of both activation energy on the k value and temperature on that k value and as we just talked about a little bit ago reactions tend to go faster at higher temperatures and so as they go faster at higher temperatures it's because you get a higher value of the rate constant so it turns out a larger temperature here means you're going to have a smaller negative number for the exponential which actually leads to a higher value well again i don't care if you really understand that mathematically but you do need to know the result here that a higher temperature leads to a higher k value and a higher k value is going to lead to a higher rate so there's our mathematical influence of temperature on that rate now if we look at the activation energy it's the exact opposite a larger activation energy means you have a bigger energetic hill to get over so and less molecules have enough energy to get over that hill and so you're going to have a smaller rate a lower rate and so in this case it's actuated through the rate constant yet again and so a larger activation energy is going to lead to a smaller rate constant and so the idea now is that with a larger activation energy you have a more negative exponential which is going to make your k value go down and again if you can't see that mathematically i don't care but you definitely need to understand this relationship that a larger activation energy leads to a smaller rate constant and that smaller rate constant leads to a slower rate all right we take a little closer look at this lovely arrhenius equation there's really two parts to this you've got this first part so in this first part is called the arrhenius constant or some people call it the pre-exponential factor since it's written before the exponential factor so and it turns out that this guy is a constant for a given reaction so but it is related to the proper orientation of the molecules when they collide there's some different you know geometrical uh spatial factors that kind of factor into it but the big thing i once realized it's constant for a given reaction so and it is related to the proper orientation of the molecules and so you can kind of think of it as the fraction of molecules that actually have the proper orientation when they collide and so you know if for a given reaction only 10 percent of the collisions have the right orientation well then this would be 0.1 that's kind of how it works and you can't really change it for that reaction under that particular mechanism now the other part here is this exponential term and this exponential term is really the fraction of molecules that have enough energy to get over that activation energy barrier so we saw earlier we looked at our reaction coordinate diagrams so that we had an energy hill to kind of get over and the idea here is that this is the fraction of molecules that actually have enough energy to get over it and so if you know at a given temperature only 20 percent of the molecules have enough energy to get over that hill great we factor that in right here and over here if only 10 of the molecules are in the proper orientation that gets factored in right here and both of those combined give us effectively the rate constant and so the rate is dependent upon the concentrations of your reactants but also on this rate constant which factors in uh again your proper orientation and sufficient energy and so all of this kind of relates back to that collision theory with higher concentrations you get more collisions so and then this describes the proper orientation and this guy the fraction of molecules with sufficient energy so that's where that's kind of coming from now you might see this written in a couple of different ways so this is 1 but we might take the natural log of both sides here and if we take the natural log of this you're going to get ln of k is equal to and if we take the ln here it's really the the product of two terms here and when you've got logs of a product you can actually split it out into separate log terms and so we've got the sum of two separate log terms like this and i'm going to change the order here but it turns out that natural log and exponential are inverse functions they cancel each other out and all you're left with is simply just negative e a over rt for this term and then this one's just going to be ln of a now i'm going to change the order here i'm going to do that for a reason i'm going to rewrite this equation as ln of k equals negative ea and instead of writing rt all together i'm actually going to separate this out into 1 over t but this essentially is just negative ea over rt and then plus ln of a cool and that's the way it'll commonly be presented so most of the time students are going to be provided with both versions here of the uranus equation it's not something we often have students memorize i apologize if you're one of those that has to memorize it so that's the form it's going to be often given in and we do that for a reason so oftentimes uh we can now get a linear graph out of this that helps us actually solve these things you know like we do an experiment in the lab both to get the uranius constant as well as to solve for the activation energy so this top graph here this is not a linear graph it's an exponential and you get exponential decay but for this bottom one here if you graph the right things you can get a straight line and the idea is that we're going to kind of match this up to the slope intercept equation of a line y equals mx plus b and so as long as i plot the natural log of k not just plane k but the natural log of k on the y axis and then 1 over temperature in kelvin on the x-axis i will get a straight line and the slope of that line is going to equal the negative activation energy over r and from that slope i can determine the activation energy and then the y-intercept here is going to equal the natural log of the irradiance constant and i can determine the uranius constant from that now much more commonly in gen chem we're going to have students actually determine the activation energy but in principle again you could totally determine the uranius constant and sometimes in some of the lab work we have you do we'll have you get both and we'll have you you know run a reaction at a bunch of different temperatures and get different values for the uh for the uh arrhenius i'm sorry for the rate constant at different temperatures then we'll have you plot ln of k versus one over t and use the slope to get the activation energy and that y-intercept to get the uranius constant let's take a look at this graph real quick again ln of k on the y-axis 1 over temperature and you've got to make sure that's in kelvin super important so and you're going to get a graph with a negative slope and in this case this point right here is going to equal ln of a and you can get your radius constant from it and then the slope on this graph m is going to equal negative e a over r and so routinely we'll give students a question on an exam in this section and we'll give you a graph like this so and then we'll either give you the equation of the line so and if we give you the equation of the line this would be the part to really focus on what's the numerical value for that slope or we'll just say the slope is and give you the slope directly so but whether we give you the equation of the line or just give you the slope specifically it's that slope that you really want because you're going to set that slope equal to negative ea over r and then rearrange it to solve for that activation energy and so whatever numerical value this is you're just going to plug it in but if you rearrange that you're actually going to get that the activation energy so we're going to multiply the other side by r and bring the negative sign over as well is going to equal negative slope times r and that's how you get your activation energy now one thing you should know here super important is the value of r that's your universal gas constant and we often give students this in a couple of different sets of units and when dealing with energy you're most commonly going to see it given as 8.314 joules per mole kelvin whereas when you're dealing with like gases like in the ideal gas chapter you're more commonly going to see it presented as 0.08206 liter atmospheres per mole kelvin instead so but at dealing with equations involving energy it's much more common to use the si unit for energy the joule per mole kelvin instead but that being the case then if you use that value of r here which is often provided for students then your activation energy is going to come out in units of joules per mole but it's most commonly either provided for you in kilojoules per mole or asked to view on an exam as an answer choice in kilojoules per mole so you may have to convert it and again to go from joules to kilojoules you'd end up dividing by a thousand so just keep that in mind super important on units here that'll become important in one other place as we'll see so here's the iranians equation here's another version of the uranus equation it's this one that allows us to kind of see the relationship being linear between l and k and one over t and how we could use that to solve for the activation energy for sure uh but maybe even that arrhenius constant now there's one other version of this equation so that you might be given usually not something you have to memorize but it's one other version of this equation and essentially it's kind of looking if you if you kind of look at where it comes from you could kind of look at as connecting two points on that line and and calculating the slope between them and you might recall that slope is rise over run or change in y over change in x and well change in y would be ln of k and change in x in this case would be the change in 1 over t and ultimately that's where this next version of the equation comes from and so this has a few different forms it might take but i'm going to give you this one ln of k1 over k2 equals ea over r times 1 over and in this case i gave it to you as t2 minus 1 over t1 cool now this is tricky so but you might look at this as kind of being related to delta y and rise over run and this being related to delta x so this is like the change in ln k and you might be like well it doesn't look like much of a change now this might look like the change in 1 over t the change being like final minus initial so 1 over t2 minus 1 over t1 that makes sense but this doesn't look like much of a change until you realize that this is the same thing written as ln of k1 minus ln of k2 and a property of logs is when you subtract separate log terms you can combine them into one log term with division and so this is the same thing as ln of k1 over k2 you might also realize too that ln of k1 minus ln of k2 is exactly the negative of ln of k2 minus ln of k1 and so ln of k1 over k2 is this the exact negative of ln of k2 over k1 and you might be like well chad i remember the slope was not actually equal to ea over r but negative ea over r and well i could factor in that negative sign by flipping these over i could factor in that negative sign by flipping these around as well and so there are a few different versions of this equation depending on what you do with that negative sign and either leaving it into the equation right here or implementing it in one of these two places so my apologies i'm giving you just one version of this equation they're all a little bit related use whichever one you're given but you do need to be careful uh and the big thing is that k1 is the rate constant at temperature 1 and k2 is the rate constant at temperature 2. there are five variables in this equation there is k1 k2 t1 t2 and the activation energy r is a constant with five variables in this equation it is really common for us to give you four of them and then say solve for the fifth and if you get this question in your exam my apologies it is a pain in the butt there are several places for you to go wrong here we're going to do an example calculation but up front what you need to know is that t1 and t2 have to have to have to be in kelvin noaa fans or butts and then activation energy and r have to have their units of energy match activation energies are often provided to you in kilojoules per mole but r is often given to you in units of joules per mole kelvin either make them both joules or both kilojoules because those joules or kilojoules need to cancel when you're doing the calculation here so those units have to match cool so there's the the two most common places to make errors besides just plain old algebra and calculator errors so let's look at an example here all right the question we're going to work out here if the rate constant of a reaction is 0.0014 s to the minus 1 seconds to the minus 1 at 25 degrees celsius for reaction with an activation energy of 40 kilojoules per mole then what is the value of the rate constant at 50 degrees celsius so in this case again there are five variables in this equation t1 and t2 we are given both of those we're gonna have to convert them to kelvin but we're provided with both of those there's two rate constants and we're given one of those two and then there's an activation energy and we're provided with that as well and so in this case it turns out k1 and k2 it doesn't matter which is which it's not like k2 has to be higher than k1 it's actually arbitrary and so you can choose whichever and if you're solving for k this is the worst possible scenario because you're going to solve for something under a natural log and that might involve a little algebra that's a little fuzzy in your mind but in this case if you've got the choice though it is going to be a little easier to solve for k1 in the numerator than it would be for k2 in the denominator just purely from an algebra standpoint and so i'm going to make k1 the rate constant at 50 degrees celsius and i'm going to make k2 the rate constant at 25 degrees celsius and so i just have to be careful then if k2 is the rate constant 25 degrees celsius then i have to make sure that i convert 25 degrees celsius to 298 kelvin and plug that in for t2 right here let's see how this works out so ln of k1 all over 0.0014 s the minus 1 equals ea over r and in this case i've got 40 kilojoules i'll write this a little different per mole over 8.314 joules per mole kelvin we'll stop there for a second this is again one of those common places students make a mistake because the activation energy as it was supplied was in kilojoules per mole the r value is typically given to you is going to be in joules per mole kelvin these need to match you need to convert one or the other and whether you want to convert 40 kilojoules per mole to 40 000 joules per mole or if you want to convert 8.314 joules per mole kelvin to 0.008314 kilojoules per mole kelvin it is your choice my my preference though is to convert 40 kilojoules to 40 000 joules instead and that's what i'm going to do here so that this works out properly so we'll make this 40 000 joules per mole and then we've got 1 over t2 minus 1 over t1 and again we said k1 was going to match up with t1 and then k2 is going to match up with t2 and the k2 is the one i plugged in that corresponds to 25 degrees celsius which corresponds to 298 kelvin oh that's yeah that's the right one i almost got that backwards myself see i told you it's tricky and then finally 1 over t1 well in this case 50 degrees celsius is 323 kelvin and here we go so let's pull out the calculators and i have one here somewhere all right so first thing i'm going to do here is do 1 over 298 minus 1 over 323 so 1 divided by 298 minus 1 divided by 323 and then i'm going to get an answer but i'm not going to stop there i'm going to then multiply this by this ratio so times 40 000 and divided by 8.314 and now i'm going to stop there and that's 1.25 and that's what the ln of k1 over 0.0014 is going to equal okay so now we've got to solve for something that is under the natural log and so we got to get rid of that natural log and the way we get rid of that is by remembering that the exponential was the inverse function here and so we're going to take e to this power here and e to this power here on both sides but here it's going to cancel and so we're going to be left with k1 over 0.0014 equals e to the 1.2496 so i'm going to work that out so in this case e to that last answer in my calculator is 3.49 cool and then i'll just multiply that up by the 0.0014 and we're going to get k1 equal to point zero zero four eight eight i probably actually should only give this in two sig figs since the original one was only given in two sig figs so we'll round this to point zero.0049 cool so by going from 25 degrees celsius to 50 degrees celsius an increase in temperature yes the k value got bigger and in this case roughly three and a half times bigger so once again lots of places you can go a skew on this one make sure your temperatures are in kelvin make sure activation energy and r are in the same units and then if you're solving for something under the log you got to remember that it's the exponential that's going to allow you to do that now if you found this lesson helpful then smile at a random stranger and have a great day and if you're looking for any practice on problems in kinetics check out my general chemistry master course i'll leave a link in the description free trials available happy studying