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# Zero-order reaction (with calculus)

## Video transcript

let's say we have a zero order reaction where a turns into our products and when time is equal to zero we're starting with our initial concentration of a and after some time period t we would have the concentration of a at that time so we can express the rate of our reaction one way to do it would be to say that the rate of the reaction is equal to the negative change in the concentration of a over the change in time and another way to do this would be to write the rate law so the rate of our reaction is equal to the rate constant K times the concentration of a and since I said this is a zero order reaction this would be a to the zero power and any number to the zero power is equal to one therefore the rate of the reaction would be equal to K times one or the rate is just equal to the rate constant K so the rate of a zero order reaction is a constant it's independent of the concentration of a next we can set these equal right we can say that K is equal to the negative change in the concentration of a over the change in time so we have the negative change in the concentration of a over the change in time is equal to the rate constant K and next we can think about our calculus right instead of writing change in a over change in time we're going to write the rate of change of the concentration of a with respect to time we have our negative sign in here and then we have our K on this side so we're ready to think about our differential equation and we're going to multiply both sides by DT let's go ahead multiply both sides by negative DT so then we would have da on the left side all right and then we would have negative K DT on the right side and ready to integrate alright so we're going to integrate on the left k is a constant so we can pull it out of our integral on the right and go back up here to refresh our memories about what we'd be integrating from so we'd be going from time is equal to 0 to time is equal to T and from our initial concentration to our concentration at time T so we plug those in so we're going from time is equal to zero to time is equal to t and then for our concentration we've going from our initial concentration to our concentration at time T so we have some easy integrals here all right what's the integral of DA right that'll be of course a or the concentration of a so we have the concentration of a alright we are evaluating this from our initial concentration to our concentration of a at time T on the right side we have another easy integral integral of DT that's just T so we have negative K T all right from zero to T next fundamental theorem of calculus right so we would get the concentration of a alright minus the initial concentration of a is equal to on the right side that would just be negative KT so we get negative KT here and so this is one form for the integrated rate law for a zero order reaction all right we could rearrange this we could we could move the initial concentration to the right side so we would get the final concentration is equal to negative KT plus the initial concentration of a and so this is just another way to write our integrated rate law so here's the integrated rate law for a zero order reaction and if we look at the form of that right it's y is equal to MX plus B so y is equal to MX plus B so we put time on the x-axis right if we graph time on the x-axis on the y-axis we have the concentration of a we're going to get a straight line and the slope of that line alright the slope of that line M the slope is equal to negative K the slope is equal to the negative of the rate constant and the y-intercept would be the initial concentration of a so if we just sketch out a quick little graph here all right let me put our axes down so we put time on x-axis right so down here we'd have time and on the y-axis we have our concentration of a we're going to get a straight line all right we can get a straight line and you see if I can let me just go ahead and switch to a straight line I want to draw one in here so we got out a straight line like this okay and the slope of this line alright the slope of our line here the slope would be equal to negative K all right so once again we can see that from up here right the slope is equal to negative K so M is equal to negative K and our y-intercept of course would be the initial concentration right this would be the initial concentration of a so this is the integrated rate law for a zero order reaction next let's think about the half-life so remember the half-life is the time required for the concentration of a reactant to decrease to half of its initial concentration so when the time is equal to the half-life T 1/2 all right we'd plug this in for T the concentration of a one time is equal to the half-life would be half of the initial concentration all right so the initial concentration divided by two so that would go into here so let's rewrite what we have we now have the initial concentration divided by two is equal to negative K times the half-life plus the initial concentration of a so let's solve for half-life on the left side we have the initial concentration of a divided by two or one half of the initial concentration of a on the right side we have the initial concentration of a so this would be one half minus one which is negative one half so negative one half of the initial concentration of a is equal to negative K times T one half so solve for T one half our half-life T one half is equal to this would be the initial concentration of a and divided by two at times k2 times the rate constant so here is here is the half-life for a zero order reaction notice if you increase the initial concentration of a right if you increase the initial concentration of a what happens to the half-life they're directly proportional so if you increase the initial concentration of a the half-life should increase so the half-life should increase let's look at an example of a zero order reaction and this will help us understand this idea of half-life a little bit better so our example is the decomposition of ammonia so the decomposition of ammonia into nitrogen and hydrogen and this reaction occurs on the surface of a metal catalyst so let's say we're using platinum here so on the Left let's call this situation one on the Left we have a piece of platinum right so imagine this is our is our platinum surface and since the reaction occurs on the surface of the platinum the ammonia molecules have to be in contact with our platinum surface so let me draw in some ammonia molecules in blue here in contact with our platinum surface remember this is a zero order reaction right so the rate is equal to the rate constant times the concentration of ammonia to the zero power so increasing the concentration of ammonia doesn't affect the rate and this picture helps us to understand that a little bit better if you increase the concentration of ammonia and right I'm adding some more ammonia molecules in here you're not affecting the overall rate of the reaction because the ammonia molecules I just added are not in contact with our platinum surface alright so only these down here are in contact and are able to react so we increase the concentration of ammonia but we didn't have any effect on the rate because the rate of the reaction is limited by how much surface area of our metal that we have all right so that's situation number one let's think about situation number two over here we have our platinum surface right let's let's draw on a bunch of ammonia molecules so here we have some in contact with our platinum surface and then here we have even more all right we have even more ammonia molecules and obviously most of them are not touching the surface of our metal catalysts all right so we've increased the concentration we've increased the initial concentration of a right of course a here is our ammonia and so we must increase the half-life that's what we learned up here if you increase the initial concentration of a right you have a longer half-life and that makes sense because our reaction has a constant rate the more molecules we have present the longer it takes to consume half of them so we have all these molecules up here of ammonia that aren't touching our metal surface and so we're limited by our constant rate and so it's going to take a lot longer right to consume half of the molecules in blue or in situation number two so that's comparing situation number one with situation number two we've increased the initial concentration of a and therefore we have a longer half-life
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