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all right welcome back so this is another Alex topic called deducing a rate law from initial rate data this is a great topic it's definitely a little confusing though so we're gonna go ahead and do two different examples of this topic we'll see how timing goes so either it will be after this video or I will link to another video with another example of this problem so in this problem the the sort of key thing the thing that we need to sort of keep in mind is what the question is asking and this type of rate equation or rate law this is really the the key aspect and we need to understand what's going on with this before we really do the problem because if we don't then it doesn't really make sense what we're doing so this rate law is going to tell us what the reaction rate is the initial rate of reaction that's this rate here and that's going to equal some constant K that's specific to that chemical reaction times the concentration of the reactants raised to some power X or Y so these are my reactants n2 and h2 so this would be some reaction where I'm reacting nitrogen with hydrogen maybe to form ammonia and the rate of that reaction is going to be determined by the order of these x and y so we call these the order in X the order in n2 and the order in h2 and if you add those together that's the overall order of the reaction so I'm not specifically important for this problem but that's sort of what we're looking for in a problem like this we're trying to solve for this value X and trying to solve for this value Y using this information over here so now let's orient ourselves to to this information over here what what this is sort of saying is we're going to essentially do three different experiments experiment one we're going to start with an initial concentration of n2o4 8 molar an initial concentration of H 2 of 1.11 molar and when you mix those two things together this is the initial rate of the reaction this is how fast the reaction is going to go if we look at experiment 1 to experiment to in experiment 2 I'm going to increase the concentration of n2 I'm gonna put more n2 into my reaction mixture so going from point four eight molar to one point four molar but I'm going to keep the concentration veyts you the same so when I do this when I make this change whatever change I observed in my initial rate of the reaction that's going to be due to this change in n2 since this h2 stays the same as I increase the concentration of n2 then any increase I observe over here is due to that increase in n2 that's going to tell us what this x value is so the way that we'd actually do that the way that we'd sort of you know solve for X I'm going to compare experiment two over experiment one and I'm choosing to put experiment two on the top because one point four is a bigger number than zero point four it's gonna make my numbers make a little bit more sense so as you're doing this comparison one you need to figure out which two experiments have only one thing changing so the h2 stays the same here between experiments one in experiments two but my n2 concentration is changing so any again any change I see over over here I can correlate that to that change in n2 so I'm gonna figure out for experiment 1 or excuse me for experiment 2 I started 1 point 4 molar zero point 4 8 molar for experiment 1 and that equals two point nine two so what this is telling me what this information here is telling me is that I've increased the concentration by two point nine two times typically you know if we're doing these experiments on our by ourselves you'll be do like a nice doubling or a nice tripling but in this Alex topic it wants us to sort of look at some numbers that aren't exactly the easiest numbers to deal with but suffice to say going from experiment one to experiment to I've increased my concentration of n2 by two point nine two times so then what we're gonna do is we're gonna say well what happens to the rate of the reaction the rate of the rate of the reaction goes from zero point five six molar per second and zero point one nine one molar per second so that ratio equals two point nine three so these two numbers are essentially the same right if you look at this two point nine two two point nine three what that's telling us is that for every time we increase the concentration of n2 by a factor of two point nine two we see the same increase in our rate so this you know these are our rates so two point nine three that would be sort of this increase in the rate the increase in the rate is the same as the increase in my concentration that tells us that the x value is one and the sort of way that that you might write that out to sort of calculate it I'm gonna take my rate to point nine three equals so the K and the h2 concentration that's gonna be constant remember our h2 concentration between these is constant between experiments one experiments - and so this all right here would be constant our K values obviously gonna be a constant so we can actually relate our change in our n2 which is two point nine two two our change in our rate so basically we're solving this equation for x and since we can see that these are essentially the same number that tells us that x equals one so we figured out our order in n2 let's do the same thing for solving for y so solve for y so for y that's our order in h2 I'm going to be looking for two experiments where the only thing that changes is the conversation of h2 so experiment one and experiment three I've got a change in my h2 concentration from one point one one molar to four point two seven molar and my n2 concentration stays the same so because my n2 concentration stays the same between experiments one experience three that's a good sort of comparison to make so I'm going to take experiment three on the top and experiment one on the bottom because again I want to put the the larger numbers on the top it just gives you know it makes these numbers make a little bit more sense it's easier to deal with I guess so if I do that then I'm gonna say four point two seven molar on the top over one point one one molar on the bottom so experiment three experiment one and then this equals three point eight five so going from experiment one to experiment three we're going to increase our concentration of H two by three point eight five times and then we want to see what happens to our rate so when I do that two point eight three molarity per second over zero point one nine one molarity per second so this ratio you know when I go from experiment one right here to experiment three right here this change in rate it goes from point one nine one to two point eight three well what you know how much is it changing in rate so that if you do that calculation you get a value of fourteen point eight so this is now we've got a difference in numbers so now we're gonna do a similar thing that we did here to figure out what that order in in H two is so I would say my rate increases by a factor of fourteen point eight when my h2 concentration increased by a factor of three point eight five I'm gonna solve this for y and if you solve this for y there's a number of ways you know like mathematically we could solve it somehow but what I would do it's gonna either be one two or three realistically so I would just sort of plug in well what's three point eight five squared it's exactly fourteen point eight so you know figuring out okay looking at this how do I solve for y I will just do guess and check personally it's it's most likely gonna be one or two might be three but we can sort of you know judge but I would just test it out see what would happen three point eight 5 to the second power gives you exactly fourteen point eight so we know that the order in h2 is two so what does this mean basically what this means is that when you change the concentration of H to the rate is going to change by a bigger factor right it's going to instead of when I double or excuse me when I when I multiply this by three point eight five when I want to increase my h2 concentration by three point eight five percent not percent by three point eight five times there we go my rate is gonna increase by fourteen point eight times that's what we've you know figured out by doing our comparisons up here so this should make sense hopefully if it doesn't definitely no and I can I can try to fix that that explanation up a little bit but the the take-home message is we want to compare experiments where only one thing changes and then we're going to relate that one change so between experiments one and experiments three the only thing that's changing in terms of our concentrations is my h2 concentration so this changed my h2 concentration from one point one one to four point two seven that's what's causing this change in my initial rate of the reaction so you know we want to choose those experiments carefully so now we've got we've got the answer for for this part here the rate equals I'm just gonna I'm just going to write the whole thing K times the concentration of n2 and that's to the first power so we can just leave it like that times the concentration of H two to the second power and then for our K value we can really choose any of these experiments to sort of plug in to solve for our actual K value so what I mean there is I'm just gonna choose experiment one and you don't have to choose experiment one you can choose any experiment you want I'm gonna plug in the values so zero point one nine eleven that's my initial rate of reaction my concentration of n2 is zero point four eight my concentration raised to the first power conservation of h2 is one point 1 1 raised to the second power and I'm gonna solve for K right so just solve this for K you should end up for this problem with the value of zero point three two now the units so the unit's are very important so these units on K they're gonna be variable they're gonna change based on what my order of reaction is basically and they need to be whatever units that are that are you know gonna be on K need to work out essentially so that our units for our rate are molarity over second so we know that this you know so if we write out rate equals we're just up here actually I'll save us some time so we know that the unit's on on the rate are molarity per second we need to figure out what our units on K are we know what our units on n2 and h2 are that's molarity to the first power times molarity to the second power right so this could this stands for molarity so basically in here I need to figure out well what do my units need to be so that you know it cancels out and I get the proper units over here so it's always going to be one over molarity such that it cancels out two of them leaves one of them times seconds so one over molarity squared times seconds that set of units will allow us to end up with these units the proper units for our rate so the way that we're gonna we're gonna calculate that or show that I guess we can get rid of that would be molarity to the minus two times seconds to the minus one so that's the easiest way to enter that on Alex so when I actually you know enter my my answer here 0.32 two sig figs because I've got two sig figs in my value here molarity to the minus two and then you need to use this little symbol to put a dot seconds to the minus one so that would be the proper proper units for for this problem so again these units are gonna vary based off of whatever our orders are so if our order was two and two that would change it to molarity to the minus third times seconds to the minus one it's always gonna be something like this but we need to make sure we know what's going on with this molarity here but we you know if we do a treatment like this we should be able to figure out well what do I need to have there to make it so that my units on the rates for work out and get me where I want to go okay so I will do another example in another video maybe not right away but I'll do another example in a little bit to go over just one more example for how to solve this problem but hopefully that makes sense hopefully that works for you if not just let me know and we'll we'll keep going all right