WEBVTT chemistry/ap-chemistry/hovasapian
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Hello, and welcome back to Educator.com; welcome back to AP Chemistry.
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Today, we are going to continue our discussion of integrated rate laws.
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Last lesson, we introduced the integrated rate law for a first-order reaction, and also the half-life formula for a first-order reaction.
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Today, we are going to talk about second-order reactions and zero-order rate laws.
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So, let's just jump right on in.
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OK, so again, we are talking about (in order to simplify matters for ourselves, just so we get a good sense of the kinetics) a single reactant that decomposes to products.
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That is what we have been doing this entire time.
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Our fundamental reaction that we have been working with is the following products.
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OK, now, the rate law for a second-order is the following--this is the differential rate law; and again, when we say "rate law," they mean differential rate law.
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When they mean integrated rate law, they will specifically say "integrated rate law."
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So, the rate is equal to -Δ;A/Δt; this rate symbol, which is the differential part in differential rate law, is equal to some constant, K, times the concentration of A to the second power.
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That is the differential rate law for a second-order reaction.
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Now, when we integrate this particular function, we end up with the following.
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Let me write; this is the differential rate law; now, the integrated, which relates concentration as a function of time.
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It is going to be 1 over the concentration of A, is equal to Kt, plus 1 over the initial concentration.
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For a second-order reaction, your integrated rate law says 1 over the concentration of A is equal to the rate constant, times the time, plus 1 over the initial concentration of A.
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So again, in this particular case, we can set it up as y=mx+b, but this time, the y is 1 over the concentration--the data that we take.
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That is what is important; and the slope of the line that we get is the rate constant.
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Notice, this time it is a positive slope--positive K instead of negative K.
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What we are going to do is: we are presented with some standard raw kinetic data; we are going to find...we have the time; we have concentration; we are going to also change those concentrations to 1/concentration, and we are also going to do logarithm of concentration, because now, we have to check; now, we have 2 different types of equation, 2 different types of reaction orders.
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Now, we have to check to see whether it is first-order or second-order.
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We have to have 2 columns: one with the logarithm of A, one with the reciprocal of the concentration of A.
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And again, you will see that in just a minute, when we do the example.
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So, now, this is the integrated rate law; let's go ahead and work out the half-life for a second-order reaction.
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And again, remember: half-life means that the concentration of A is equal to the initial concentration, over 2; it is when half of it is gone.
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We will go ahead and write; that implies using this equation with those values: 1 over A₀ over 2 equals Kt, plus 1 over A₀; this ends up being 2 over A₀, minus 1 over A₀.
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I flip this and move this over; it equals Kt; 2 over A₀ minus 1 over A₀ is 1 over A₀, equals Kt.
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Now, I will divide through by K, and I get that t of 1/2 is equal to 1/K, times the concentration (the initial concentration).
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This is the formula for the half-life of a second-order reaction.
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Notice, in this particular (you know what, this is not too clear; let me write this out; I need to erase this, and let me write it bigger over to the side; I think it will be a little more clear; let me get rid of this thing; A₀...so, we get t is equal to 1/K, times the initial concentration)...
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Now notice: in this case, the half-life of a second-order reaction not only depends on K, but it also depends on the initial concentration.
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That was not the case with the first-order reaction; a first-order reaction does not depend on the concentration--it just depends on K; it's constant.
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This is not constant; so the half-life--if you start with a certain amount, half of it is going to take a certain amount of time to get rid of.
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Well, now that you have gotten through half of it, now that is a new initial amount.
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In order to release half of that, that half-life is going to change; it is actually going to end up being longer.
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So now, it depends on two things: the rate constant and the concentration.
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There you go: we have that equation, which is the integrated rate law; we have this equation, which is the half-life; and, of course, we have this equation, which is the differential rate law.
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These are the things that matter in kinetics: the differential rate law, the integrated rate law, and the half-life formula.
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Pretty much everything else can be worked out from this information, and all of this, of course, comes from raw kinetic data: time, concentration.
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OK, now, let's go ahead and do an example, because that is the best way that these things work.
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Example: we have butadiene; butadiene dimerizes (and dimerization means that two molecules of something stick together); butadiene dimerization was studied, and the following kinetic data were obtained.
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OK, so let's go ahead and write out the formula.
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2 C₄H₆ turns into C₈H < font size="-6" > 12 < /font > .
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OK, so now, we have our time value; I'm going to go ahead and write it as one big plot.
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Our time value; we have our concentration of C₄H₆, which is our reactant.
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Now, what we want to do here is: we want to find the rate constant; we want to find the order of the reaction; and we want to find the half-life (I'm sorry about that; I should have actually told you what it is that we are actually going to be doing here).
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So, the first thing we want to do is: we want to find the order of this particular reaction, based on the data that we are given.
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Then, we want to find out what the rate constant is; and then, we want to find out what the half-life of the reaction is, based on this data.
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So, t (let me go ahead and write this down) is 0, 1000 seconds, 1800, 2800, 3600, 4400, 5200, and 6200.
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And what we have is 0.0100, 0.0625, 0.0476...no, that is not right; that is 0.0100; so this is going to be 0.00625--the butadiene is diminishing--0.00476, 0.00370, 0.00313, 0.00270, and we are almost done--no worries--0.00241; and 0.00208.
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OK, given this raw data, find the order of the reaction; find the rate constant; and find the half-life of this reaction, based on the fact that we started with .0100 moles per liter of this butadiene.
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All right, so we need to check: is it first-order or second-order?
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We need to plot both the logarithm of C₄H₆ versus time, and the reciprocal of C₄H₆ versus time, to see which one of these gives us a straight line.
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If this gives us a straight line--the logarithm--it is first-order; if this one gives us a straight line, it's second-order.
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That is how we do it.
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OK, so here is what the data looks like.
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Let's see; let's do logarithm first.
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We get -4.605, -5.075, -5.348, -5.599, -5.767, -5.915, -6.028, and our last one is -6.175; reciprocals are a lot easier.
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What we have is 100; 160; 210; 270; 320; 370; 415; and 481.
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OK, so now we are going to plot (I'll do it in blue): we are going to plot this versus time, and we are going to plot this versus time, to see which one gives us a straight line.
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Let's go ahead and take a look at what these plots look like.
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I'm going to go ahead and put them next to each other; so, I'm not going to give you too much detail on these graphs--you can actually go ahead and plot them yourself on a piece of graph paper, or using a software like Excel or something like that.
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What you are going to end up with, as it turns out--let's see here, this is going to be 200...yes; so let's do the logarithm of the butadiene versus time here; and let's do the 1 over the reciprocal of butadiene (which I'll just call A) over time here.
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When I do this, the logarithm graph is something like this--definitely not a straight line.
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When I do the reciprocal, I get a straight line; so I will have you confirm this, but this is exactly what happens.
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So, because it is the reciprocal that gives us a straight line, it is a second-order reaction.
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Second-order: so, we can go ahead and write the differential rate law: C₄H₆/Δt equals the constant times C₄H₆ to the second power.
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We know it now; it is a second power; we derived it from the graph now, so we have an order of 2.
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Now, the next thing we want to know is: what is the rate constant?
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OK, well, the rate constant--the best thing to do is to go ahead and (let's see, how shall we do the rate constant)...let's just go ahead and use the integrated rate law.
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We know that the second order is: 1, over the concentration of A, equals -K (actually, you know what?--what am I saying?--let me just...write it out, so we have it)...equals *positive* K times t, plus 1 over A₀.
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This is y=mx+b.
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We just graphed the reciprocal over that; we got a straight line, so now, what we want to do is: we want to pick two points on that straight line, take the Δy over the Δx of those two points...
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And again, the points need to be on the line; if the data points that we have in the original data, when we calculated it, are on the line, that is fine--you can use those data points.
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But, if they are not, make sure that you are using points that are on the line.
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When we calculate Δy/Δx from this graph, based on the kinetic data that we derived, we end up with the following.
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Let's see: let's use a couple of points, actually; so let's use Δy over Δx; it equals...as it turns out, a couple of the data points do fall on the line.
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I pick the first and the last; so it's going to be 481 minus 100, over 6200 minus 0, which gives me 0.06145, and the unit is going to be liters per mole-second.
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I know the units for the rate constants don't really make sense; they tend to change, depending on what order it is; it is actually the number that matters.
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This is our rate constant; and now, we want to find t < font size="-6" > 1/2 < /font > .
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OK, well, t < font size="-6" > 1/2 < /font > of a second-order reaction is 1/K, times the initial concentration.
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It is equal to 1 over 0.06145, and the initial concentration was 0.0100; so, when we do that, we get 1627 seconds.
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There you go: raw data: time and concentration; we calculated the logarithm of concentration; we calculated the reciprocal of the concentration; we plotted ln of concentration versus time for one graph; we plotted reciprocal of concentration versus time for a second graph.
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We saw which one of those graphs gave us a straight line; in this particular case, it was the reciprocal versus time that gave us a straight line: second-order.
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Because it is second order, I go ahead, and I can write the differential rate law, if I need it.
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We need the rate constant; so I just found a couple of points on that straight line; I took the slope, y₂-y₁ over x₂-x₁.
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I ended up with the rate constant; and then, because I have the half-life formula, and I have the rate constant which I found, and I have the initial concentration (which is the initial concentration from raw data), I was able to calculate the half-life of this reaction--very, very straightforward.
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Nothing altogether too complicated: again, differential rate law, integrated rate law, half-life--those three things are what is important, as far as the kinetics of a reaction are concerned.
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OK, that was second-order reaction.
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Let's talk about a zero-order reaction.
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All right, so a zero-order reaction is exactly what you think it is; it says that the rate (well, let me actually write it in the top here)...
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Zero-order reaction: well, the rate (which is equal to -Δ of some reactant/Δt) equals KA to the 0 power, which is K times 1, which is K.
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In other words, the rate itself is constant; the rate doesn't change.
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The integrated rate law: when we integrate this particular function, you end up with the following.
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You end up with the concentration of A (you know what, I am not even going to write these anymore; they are driving me crazy), equals -Kt, plus A₀.
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The concentration at any given time is equal to minus the rate constant, times time, plus the initial concentration.
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Or, it equals the initial concentration, minus rate constant, times time; it's diminishment.
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Again, y=mx+b.
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So now, if you had some raw kinetic data, and you just plotted the concentration versus the time, and you got a straight line, that is a zero-order reaction; there you go.
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Now, half-life: A=A₀/2; so let's put it in here.
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A₀/2=-K times t, plus A₀.
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A₀/2 minus A₀=-Kt.
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-A₀/2=-Kt.
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Therefore, t < font size="-6" > 1/2 < /font > equals A₀ (these lines--they always show up in the most inopportune moments), divided by 2K.
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Let's do it in red.
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We have this; we have this; and we have this.
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For a zero-order reaction, the differential rate law says that the rate is equal to K, the rate constant; it is a constant.
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The integrated rate law says that the concentration at any time A is equal to -Kt, plus A₀.
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The half-life formula says that the initial concentration, divided by twice the rate constant, will give you the amount of time it takes for half of whatever you started with to disappear.
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Notice, in this case: again, it depends on the initial concentration, and it depends on K.
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Now, you might be wondering what a zero-order reaction is; where would you run across a zero-order reaction?
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When would you have a reaction that doesn't actually depend on the concentration?
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Zero-order reactions show up in places where...usually in things that involve catalysis; and, extended to the biological realm, that means enzymes.
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So, for example, if you have--let's say just 10 enzymes, catalyzing a reaction; well, if all of those 10 enzymes are busy (meaning if they are full, doing what they are supposed to do), it doesn't matter how much more reactant you actually put in there.
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It is called a substrate--the thing that the enzyme grabs onto and fills with; so it doesn't matter--once those 10 enzymes are full, you can't make the reaction go any faster, because it doesn't matter how much concentration--how much more of the reactant--you actually put in; the reaction rate is controlled by how many enzymes are actually doing the work and how fast they are doing it.
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So, under catalysis conditions, the reaction rate actually depends on the catalyst, not so much the concentration of the reactant.
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Now, this isn't always the case; we're just saying that, when you run across a zero-order reaction, that more often than not, it is going to be in a catalysis situation.
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That is just a qualitative bit of information that you should know, if it happens to come up.
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OK, so now, let us take a global view of what it is that we have done with zero-, first-, and second-order reactions.
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Let's summarize what we have, and it will give us a good, nice one-page summary of reaction kinetics and how to deal with them.
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OK, so this is going to be a summary for the kinetics associated with the reaction aA going to products.
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So again, this entire time we have been talking about a single reactant decomposing into products.
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It diminishes, and it forms something.
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How can we figure out the rate?
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All right, so here, let's go ahead and make ourselves a little table; and we will go ahead and put a line here, and we will say first-order, second-order, zero-order reaction.
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OK, the differential rate law for a first-order reaction is: the rate is equal to K, times the concentration of A to the first power.
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The rate is equal to K, times the concentration of A to the second power.
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And, the rate equals just plain old K.
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That is the differential rate law.
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OK, our integrated rate law: we have, for a first-order reaction--we have: ln of A is equal to -Kt, plus (oops, I always do an h for ln; I don't know why) ln of A₀; again, concentrations, kinetics--we are talking about concentrations.
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Now, the rate law for a second-order is reciprocal: 1 over A is equal to Kt, plus 1 over A₀.
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And, this one is: A is equal to -Kt, plus A₀.
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These are the integrated rate laws.
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OK, a plot giving a straight line is going to be ln of A versus t, y versus x.
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So, the plot of ln(A) for a first-order reaction--the plot of ln(A) versus t gives a straight line.
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Here, it is 1/A versus t that gives a straight line.
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Here, it is A versus t that gives a straight line.
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And again, it is just the left-hand side of the integrated rate law: ln(A), 1/A...you check the data.
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Slope: well, the slope is -K, K, -K of the particular line, if you happen to get one.
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In other words, the slope is that; so you can find K; that is how you find K.
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And, half-life: the t < font size="-6" > 1/2 < /font > is equal to ln(2) over K; notice, for a first-order reaction, it is kind of interesting; it does not depend on concentration at all; it is constant.
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The half-life for a second-order reaction is equal to 1/K times the concentration of A₀.
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It depends on the initial concentration.
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And, t < font size="-6" > 1/2 < /font > of this one equals the initial concentration over 2K.
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So, this is everything that we have done: first-order, second-order, zero-order reaction rates.
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When we are given kinetic data, which involves time and concentration, we have to plot the logarithm of the concentration versus time, the concentration versus time, and 1 over the concentration versus time.
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Whichever one of these graphs gives us a straight line, that is the order of the reaction.
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When we have the order of the reaction (let's say, for example--boom!--we come up with second-order), we know the differential rate law; we can write the integrated rate law; we can find K from the slope; and we have the half-life formula.
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We can tell you what the concentration is at any time; we can tell you what the concentration is going to be at any percentage that we want along the way, instead of the time--all of this information right here.
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This is a summary of standard kinetic operating procedure.
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We'll go ahead and leave it at that; next time, we will talk about temperature dependence of reaction rates, and talk about something called the Arrhenius equation, and activation energy, and things like that.
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But until then, thank you for joining us for AP Chemistry and kinetics, and thank you for joining us at Educator.com.
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Take care; goodbye.