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Iicsad [QuilibriumFinding half life and rate constant from graph of concentration versus imcUse thls graph to answer the following questions: What the half Ilife th...

Question

Iicsad [QuilibriumFinding half life and rate constant from graph of concentration versus imcUse thls graph to answer the following questions: What the half Ilife the reacton?D-o DDRound your answersignificant diglts:Suppose the rate the reaction Knovin be first orderE HL Calculate the value the rate constant kk = 0Round your answer to 2 significant digits. Also Include the comect unit symbol;Predict the concentration of HI in the engineer' reacton Vaeca after 0.120 seconds havc DusscdAssume

Iicsad [Quilibrium Finding half life and rate constant from graph of concentration versus imc Use thls graph to answer the following questions: What the half Ilife the reacton? D-o DD Round your answer significant diglts: Suppose the rate the reaction Knovin be first orderE HL Calculate the value the rate constant k k = 0 Round your answer to 2 significant digits. Also Include the comect unit symbol; Predict the concentration of HI in the engineer' reacton Vaeca after 0.120 seconds havc Dusscd Assume no other reaction Important; and continue Haainnal the rate first order [=i] Qx Round Your 4naxei significant diglts_



Answers

Consider the hypothetical reaction $$ A+B+2 C \longrightarrow 2 D+3 E $$ where the rate law is $$ \text { Rate }=-\frac{d[\mathrm{A}]}{d t}=k[\mathrm{A}][\mathrm{B}]^{2} $$ An experiment is carricd out where $[\mathrm{A}]_{0}=1.0 \times 10^{-2} \mathrm{M}$ $|\mathrm{B}|_{0}=3.0 \mathrm{M},$ and $\left.\mathrm{IC}\right|_{0}=2.0 \mathrm{M} .$ The reaction is started, and after 8.0 seconds, the concentration of $\mathrm{A}$ is $3.8 \times$ $10^{-3} \mathrm{M}$ a. Calculate the value of $k$ for this reaction. b. Calculate the half-life for this experiment. c. Calculate the concentration of A after 13.0 seconds. d. Calculate the concentration of C after 13.0 seconds.

So Chemical kinetics is an area off our physical chemistry that is concerned with the understanding of rate chemical reactions as well as studying the chemical processes and transformations of reactant to products within a reaction vessel so fast we have l n concentration a equals negative k prime t at l n concentration A at time zero where we have a value for a value for T and a value for a zero. So then I have rearranged on plugged in my numbers undetermined what? K primers. And that is not 0.12 So moving on to the next part. Now we have the right que prime concentration A equals not 0.1 to 1 times 10 to the power minus two equals not point. Not not 12 miles per liters per second. So then the rate of K concentration a concentration be it's squared so K can be found using the following where K equals not point not 13 So next weaken Continue on to the second part. So we have our equation of concentration of a is the negative. Okay, prime t add L. A and concentration at Time zero, where we have k we have concentration of a concentration off a at time zero. We re arrange for tea, which is time, and it is time to be 5.8 seconds. Next, we've got concentration of a cause. Negative K prime t add Ln concentration A at time zero where we have our cake. Prime value R T value on a North valley. We re arrange for concentration of a on that is determined to be no point, not not to one. So this is just the last part. Here we have the concentration off. See, that is concentration of T C. Subtract two Delta concentration of a where the concentration of a is one point not times 10 to the power minus two subtract nor point nor nor to one which generates a value off no point, no, No. Eight. So that's concentration of C is concentration of c. Subtract two Delta A, which is to

Because we know that a plot of one over the concentration gives us a straight line. This is a second order reaction because the slope of the line is 6.9 times 10 to the negative to This is also the rate constant K for this reaction. So the differential rate laws simply going to be equal to rate is equal to K multiplied by the concentration raised to the second power. Because it's second order, we can then write the first or the second order integrated rate law because we know it's second order and then the K value will be the slope in units of one over morality. Second, the first half life will be won over the concentration at time zero multiplied by the rate constant concentration was given to us a 00.1. This gives us 145 seconds. The half life doubles each time because the concentration is cut in half. So the second half life would begin with the concentration of 0.5 and that will give us double the half life of the first for the second half. Life 290 seconds

So this is the sort of question that if you don't really know the tricks, it's going to feel impossible. But it's really not too challenging, if you do know. And so the first big thing you have to realize is that if the plot of something of one over a versus time where a is the concentration of reactive is linear, that tells us that this is second order with respect to a and since the reaction just a becoming be that also immediately tells us what the rate lot should be. The rate it's always first eagle toe, a rate constant. But since it's the only thing we have is a and we know it's to the second order who take a we square it. This is our rate lock. And so now we're already really along that well, since we also know this is a second order reaction, we can just copy the integrated rate law one over the concentration of a she looks a k T plus one over the concentration of a not where que is the rate constant. A. Not is the amount we started with, and this constellation of a is the amount at Time T. And so that's our integrated rate long and finally to figure out the rate constant K. We don't actually to do any work because we're told with the slope of the line is and another fact that comes associate with that's in this thing. You should really study the plots and when they're linear, meaning what? But anyways, the slope of this, if it's linear, is equal. Okay, that's the case for a second reaction. So we're told with the slope is and so we can just read off what Kate should be. It's 3.6 times 10 to the negative. Second leader squared her moles. It should not be leaders where it's just the leaders leaders promote for a second. It's now we have K. We want to figure out the House life for this reaction when he'd use the half life formula for a second order reaction, which is one over K times. The initial concentration we have all these values. The initial concentration were given as 2.8 times attention any of third. I have k right here and I can just plug this in and pump out a number, and it's equal to 99 to 0 0.6 seconds part of it. And so from here, if we want to figure out how long it would take to get to a certain concentration, we use the integrated rate lock. So remember, let's go right here. What's bugging? We're looking to get to the concentration of seven times 10 to the negative fourth. And we, as in all these cases, were starting at 2.8 times tense. Negative third. Okay, we know that. And so be careful with how you plug this in, but you can find the time pretty easily. It is equal to 29761 seconds. That's our final answer.

Is this the sore question? Where you really need to know about what? Linear plots, plots result with what order? If they're linear And so on this sizzling hot into with you integrate rate law, you know? Know what? I'm time. Well, you should definitely read your book on this. Anyways, if we take the concentration of a and plotted versus time and it ends up linear, we know this is first order with respect to a And with that, we can just pump out a bunch of information that counts from the integrated rate law. For instance. We know that what the integrate ray law is gonna be It's natural log of a equals negative. Katie, were case the rate constant plus the natural log of the initial concentration? We also can know what the half life is. Jiggles the natural log of two over K that's derived from that integrated ray lock. And we can also say pretty quickly here if the rate law is always first rate equals k, the rate constant and times the concentration of a the first power. So that Cindy is concentration of a because it's first order with respect to a and there's no other reactions. So this is our rate lock. And so that answers the first question. Mostly, um, now to figure out the value of the rate constant. We also can use the fact that for US plot like this in its first order, that slope is equal to negative K. And so we're told what the slope is, and so we can easily pump out cages by copying it from the problem. It's 2.97 times, 10 to the negative, second in verse minutes. And so we're in business. Certainly we want to figure out the half life well. It's legal to the outline of two over K, and that equals 23.3 minutes plugging k from right up here. And, uh, finally we want to figure out how much time it's gonna take to get to a certain concentration. Well, we're gonna use the integrated rate law. It's so that concentration working, looking for is 2.5 times tense, and I give third our initial concentration. There's two times 10 today in a second we have K. It's still 2.97 times 10 Judy negative second and all of that. We consult for the time, which will come out of 69.9 minutes. Fantastic


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