 ## General Chemistry

Learn the toughest concepts covered in Chemistry with step-by-step video tutorials and practice problems by world-class tutors

15. Chemical Kinetics

# Integrated Rate Law

When we include the variable of time to our Rate Law then we obtain the Integrated Rate Laws.

Understanding the Integrated Rate Laws
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concept

## Integrated Rate Law Concept 1 2m
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now the integrated rate law describes the relationship between reactant and their concentrations as well as time. Now. This helps to determine how long it takes for X amount of moles per liter of reactant to become consumed or used up. And we're going to say here that the integrated rate law depends on the order of the reaction. Now, the first one we talked about is our zero order integrated rate law. Now here this is for reactions following zero order rate laws and with it we're going to use the following equation. Here it is A sub T equals negative Katie plus a sub zero here, a sub T equals the final reacting concentration. A sub zero equals the initial reacting concentration. Because zero time has passed, K equals your rate constant in. Now remember to figure out the units for K, we say K equals M2, the negative end plus one times time inverse. And we're going to say that n equals the order of the reaction. So if we're dealing with zero order, that means it is equal to zero. So zero negative zero plus one just means plus one. This would mean that for zero order reaction. The units for K are in polarities to the one times time inverse. Now, time here could be seconds, days, years or whatever. Here, T would be our time. So these are the components that make up our zero order integrated rate law equation. Now this equation is also connected to the equation of a straight line here, A. T would be connected to y, your rate constant, K would be connected to em which is your slope. And remember it's not just cade's negative K. T. Here will be connected to X. And then your initial concentration connected to be. If we look at this graphically, remember a graph is a plot of Y versus X. Your Y again would be your a sub T. So the concentration of your reactant and then T. Is your ex. And then remember here this is our initial concentration. When we first started our reaction notice that the slope is decreasing because remember slope is equal to M. Which is equal to negative K. Negative K. Meaning that it's decreasing over time. Also remember that slope is equal to change in Y over change in X. Which in this case we can say is really the change in concentration over the change in time. So keep these in mind when discussing any question dealing with a zero order reaction.
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example

## Integrated Rate Law Example 1 2m
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A plot of the concentration of nitrogen Tri Oxide vs time. With a slope of .260 gives a straight line. What was the initial concentration of nitrogen tri oxide? If after 35 seconds it's concentration dropped to 2.75 times 10 to the negative to moller. Alright, so here they're telling us it's a plot of concentration versus time. Remember if our plot is of your reacting concentration versus time? That would mean that it is a zero order reaction or zero order reaction. So we know it's zero order. So that means that our final reacting concentration equals negative Katie plus initial reacting concentration. Now here they're asking us to figure out the initial concentration. So we don't know what this portion is. We know that it drops to this final number here. So that's our final concentration. So that's 2.75 times 10 to the negative two equals. Now remember this equation. First order reaction is also equal to the equation for a straight line and here is our slope which is equal to our negative K. So when they tell me my slope is 0.260. They're really telling me what K. Is. So we're gonna plug that in for K. So negative 0.260 T. Here is time which we're told is 35 seconds. Alright, so then we're gonna have 2.75 times 10 to the negative two equals negative 9.1 plus the initial concentration of your reactant. Here, you're going to add 9.1 to both sides. And when we do that we're gonna get as our initial concentration 9.1275 moller. So this would represent our initial concentration here. In our question. This has three sig figs, three sig figs, two sig figs. So here, if we had it, in terms of two sig figs, it would come out to be uh 9.1 molar as our initial concentration, Right? So that would be our final answer.
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concept

## Integrated Rate Law Concept 2 2m
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example

## Integrated Rate Law Example 2 2m
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A certain reaction has a rate constant of .289 seconds inverse. How long in seconds would it take for the concentration of reaction reaction a to decrease from 1.43 moler. 2.850 moller. Alright, so they tell us our rate constant is .289 seconds inverse. Remember a dead giveaway for a first order reaction? Is that K is in units of time inverse. So that's unique to first order. The fact that it's seconds in verse tells us that its first order. So since its first order, we know that we're dealing with Ln. 80 equals negative Katie plus Ln A. Up Now here our initial is 1.43. so Ln of 1.43 And then our final is .850. So Elena .850 over here equals Alright, so here our rate constant is .289. And then here we're looking for time T What we're gonna do here is we're gonna subtract Ellen of 1.43 from both sides. When we do that we're going to get .520193 equals negative .289. T The viable size by -289. Okay, so here since K is in seconds. That means time would also be in seconds. Remember that their units must always match if Kay was in minutes inverse. Then time would be in minutes. Okay, they're gonna match in terms of the units. So here when we work all this out we get one point seconds here are answer has three sig figs because 30.289 has three sig figs, as well as 1.43 and 0.850. So again, 1.80 seconds will be the time for this first order process.
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concept

## Integrated Rate Law Concept 3 1m
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for reactions that with second order processes, we use the following equation and it is 1/80 equals positive K. T plus one over a. Oh so 80 here is the final reacting concentration. A. O. Is the initial reaction concentration Now Ks our rate constant in units of remember K equals M two. The negative end plus one times time inverse. And equals the order of our process here. Since it's second order, it's negative two for n Plus one times time inverse. This would mean that when it comes to second order processes, the units for K would be in polarities inverse times time, inverse. So always be on the lookout for that when looking to see if a reaction is second order or not. T is of course time. Now the second order integrated rate law equation is related to the equation of a straight line here, 1/80 is connected to Y que here is equal to our slope. M. T. Is X and one over A O. Is B. Now, if we took a plot of one over A versus time, that's a dead giveaway. That's the second order process. Remember, plots are of Y versus X. So one over reacting concentration matches with this for why that's why it's on the y axis and then the X axis has time because T. Is equal to X. Here. Now here our initial starting is down here and notice that the line is increasing over time. That's because for second order processes, K is positive here it's a positive Katie. We're customizing nada que when it deals with zero order and first order processes. Second order is unique. It's the only one with the increasing slope over time, Right? So just keep these little things in mind that are giving you clues on second order processes.
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example

## Integrated Rate Law Example 3 2m
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here we're told that the reacting concentration for a second order reaction was 0.7670 moller after 300 seconds and 7.3 times 10 to the negative to moller after 750 seconds. What is the rate constant K. For this reaction? Alright so they tell us it's second order. So that means 1/18 equals K. T. Plus one over A. L. Here we're told that we initially have .670 moller after 300 seconds. That's gonna be our initial. Okay. No one says that initial has to start at zero seconds. It's just in this case we're starting at 300 seconds And then one over the final concentration is 7.3 times 10 to the -2. Here we're looking for. Kay but we need to figure out what our time is, how much time has elapsed. Well we're starting at 300 seconds and this is the initial concentration And we go to 750 seconds. If we subtract those two numbers that tells us how much time has elapsed And that's 450 seconds have elapsed. Alright so we're going to subtract one over .670 from both sides. Alright so then when we do that we're gonna get 12. equals K. 4 50. So then we're gonna divide both sides now by 4 50. And when we divide both sides by 4 50 that's gonna give us our K value K. Here will equal 2.71 times 10 to the -2. Now here's the thing where the units for K. Remember for K it's M to the negative end plus one times time inverse since its second order and is too so negative two plus one times time inverse. Here. The units for a time where we see are in seconds. So this would be seconds inverse. So negative two plus one is negative one, so M to the negative one times seconds to the negative one. So this would represent the value for our rate constant K as well as its units.
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Problem

For the reaction A →  B, the rate constant is 0.0837 M–1•sec–1. How long would it take for [A] to decrease by 85%?

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Problem

The following reaction has a rate constant of 3.7 × 10–3 M•s–1 at 25°C:

A → B + C

Calculate the concentration of C after 2.7 × 103 sec where [A]0 was 0.750 M at 25°C; assume [C]0 = 0 M.

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Problem

For the decomposition of urea, NH2CONH2 (aq) + H+(aq) + 2 H2O (l) → 2 NH4+ (aq) + HCO3 (aq), the rate constant is 3.24 × 10–4 s–1 at 35°C. The initial concentration of urea is 2.89 mol/L. What fraction of urea has decomposed after 3.5 minutes?

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Problem

Iodine-123 is used to study thyroid gland function. As this radioactive isotope breaks down, after 5.7 hrs the concentration of iodine-123 is 56.3% complete. Find the rate constant of this reaction. 