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Consicer tha following stucies of kinerics Botn Pans anc C will #ouie analysis Witn Excel Numoerz Gocgl Shee s, LoggarPro come comparaole apreacaneet application.Th...

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Consicer tha following stucies of kinerics Botn Pans anc C will #ouie analysis Witn Excel Numoerz Gocgl Shee s, LoggarPro come comparaole apreacaneet application.The tabl ceow contain? Intal Rares Cata fortne TeactonofNtrogen Ioxide an Caroon mcnoxicle snownbelow. Determina tne orcer 0f 233n reac-ant,the magntuce tner ?constant and precic: the Inial Rare fortne last lina oithe .a0NOzz COtNOa coraTrial [NO: W [CO] @) Rate (Wa}0zJ00.sC20.0137O.svlloci0.05460.2500.7500.01330.7500.780The dara pelow

Consicer tha following stucies of kinerics Botn Pans anc C will #ouie analysis Witn Excel Numoerz Gocgl Shee s, LoggarPro come comparaole apreacaneet application. The tabl ceow contain? Intal Rares Cata fortne TeactonofNtrogen Ioxide an Caroon mcnoxicle snownbelow. Determina tne orcer 0f 233n reac-ant,the magntuce tner ?constant and precic: the Inial Rare fortne last lina oithe .a0 NOzz COt NOa cora Trial [NO: W [CO] @) Rate (Wa} 0zJ0 0.sC2 0.0137 O.svl loci 0.0546 0.250 0.750 0.0133 0.750 0.780 The dara pelow dlescnbestne decompositicn of Nizrogen cicx de accorcling reacticn belov Use tne data ; ceterminetne order tne Teact0n, tha magntude [72 rate Cconsant andineinria Concentiaticr NOz Precictine concentration NOz ater the reaction has been running for 76 seconc? NOzyi NOd Time [NOz] (M) 0,703 0.417 0.353 0.202 0.280 0.235 0.212 0.197 200 0.180



Answers

The tabulated data show the concentration of AB versus time for this reaction: $\mathrm{AB}(g) \longrightarrow \mathrm{A}(g)+\mathrm{B}(g)$ $$ \begin{array}{cl} \text { Time (s) } & {[\mathrm{AB}](\mathrm{M})} \\ \hline 0 & 0.950 \\ \hline 50 & 0.459 \\ \hline 100 & 0.302 \\ \hline 150 & 0.225 \\ \hline 200 & 0.180 \\ \hline 250 & 0.149 \\ \hline 300 & 0.128 \\ \hline 350 & 0.112 \\ \hline 400 & 0.0994 \\ \hline 450 & 0.0894 \\ \hline 500 & 0.0812 \\ \hline \end{array} $$ Determine the order of the reaction and the value of the rate constant. Predict the concentration of AB at $25 \mathrm{~s}$.

So, in order to figure out the order with respect to each of these reactions were looking for, uh, basically using two experiments where two of the reactors are held constant one that has changed. And we look at the painting, the rate. So put some values on that idea. And so I have the age to SC Um 03 If we want to figure out the order of this, we're looking for two experiments where the contradiction of H plus the concentration of I minus our constant between them but the concentration of H two s, c 03 changes and that one of the experiments that size by that are the 1st and 2nd 1 And so the concentration of H two at C 03 goes from one time center Negative fourth to being too time sensitive before and so obviously that's a two times increase. Now we want to look at how the rate is gonna change. And so it goes from 1.6 600 cents the negative seven to 3.33 time sensitive. Your seventh in this to you can see pretty obviously or you plug into a calculator is a two times increase. Instantly compare these numbers since they're the same. In other words, we're taking you to to the first power. This is going to be a first order reaction with respect to the H two s, c 03 We're gonna do this process for all of them. Now we look at the age plus looking for two experiments where the concentration of the other two reactions are constant and so one that would satisfy that is Experiment one experiment four The concentration of H plus goes from being too some sense in a second being four times tense negus second. And so this is obviously a two times increase. And the rate simultaneously I went from being 1.66 I'm sentinel your seventh being 6.66 times since the negus seventh. And so you can listen Jo Cox later, or just know and you'll realize just a four times increase in the rate. So he compared these two numbers. If we take to square, we get four. And so that tells us that this number this is the order of the reaction with respect to H plus. And so this is a second order reaction with respect to age plus. So we're on our way to, um over already locked. It's tedious, certainly. And finally we have I minus the experiments we could use. We could use Experian number one and then experiment number six. So the concentration of I minus between those two goes from two times 10 today, a second and then 24 I'm sensing a second, and the rate simultaneously goes from 1.66 times Centenario. Seven, 2, 13 0.2. Time center Negative. Seven. You'll see that this is a two times increase. This is an eight times increasing soap, just like before. We have to to the third power equals eight now. And so third power tells us the order with respect to I minus. And so it's first started with respect to age two at 03 second with H plus and third with I'm ice and we can use. That's right. The rate locks the rail always equals a rate constant first, and then we take H two F C 03 to the first power. So just like that, an H plus, but it's squared and then I minus. But it's cute. And so this is our a lot right here. We want to figure out OK, we can plug it any of the experience, find a rate plug in all of these, and we confined. Okay, if you that for multiple of them and find an average, you'll find that the cake now you should get is right around 5.2 times 10 to the fifth. And the units are gonna be definitely unusual. Leaders to the fifth Moles to the fifth per second. This is necessary in order to cancel out all these concentration use that your rate ends up with the commandments. So this is our final answer.

To answer this question and determine the approximate concentration of a after 110 seconds for the zero first and second order reactions, we need to know what the rate constant is. So if you've done problem 27 you've plotted everything, you will get the rate constant for each of them from the slope. So Experiment one is first order its rate constants point to one experiment to zero order. Its rate constant is 00.1 an experiment three of second order and its rate constant is essentially 30.1 also so to determine the concentration at 110 seconds for the zero order reaction. Well, it goes to zero at 100 seconds. So if we're at 110 it's still nothing for B consulates. For first order, we use the first order integrated rate law, natural log of concentration at 110 seconds over, natural over a natural log of concentration at 110 seconds, divided by the concentration at time zero will be equal to negative K multiplied by t 110 seconds. Doing a little bit of algebra will find out that the natural log of concentration is negative 1.1 or 0.33 Moller. And that should make sense because here we see that at 100 seconds it's 1000.37 So it should be just a little bit less than that. At 100. And 10 0.33 is reasonable for the second order. Reaction will use the second order Integrated rate law one over. Concentration at time, zero minus one over concentration at time T equals positive. I'm sorry. Negative, Katie. I'm doing a little bit of algebra. Will see that one over. Concentration 110 seconds equals 2.9. Take the inverse of 2.9 and we get 0.48 Moller. And at 100 seconds for the second order we see it's 0.5. So it should be just a little bit less than that. 2.48 makes sense at 110 seconds.

To answer this question, we need to prepare graph so you can go into Excel and prepare graph something similar to what I've got here and then we'll draw a tangent Right at 60 seconds and then we'll do the change in more clarity over the change in time for this tangent. And it appears we've got 1,5 5 & zero is the change in why? And then the change in X is going to be zero and about 180 because the CH three ch oh is being consumed than the instantaneous rate of reaction of the CH three ch oh is going to be negative. If however they want to know just the instantaneous rate of the reaction itself, then the rate of the reaction is always positive. So I'm not quite sure exactly what they want, whether it's going to be negative or positive, but nonetheless it's about 86 or nine times 10 to the negative. Four Mueller per second. Then when we go to 120 seconds, a slope is less steep, and we're looking at a difference of about 0.12 And then a difference here of about 265. And so we get negative 45 times 10 to the negative four, about one half what we had at time 60 seconds.

To correctly answer this equation. The best way is to use Excel and plot the data. You need to plot the concentration as a function of time. One over the concentration as a function of time and the natural log of the concentration as a function of time. If the concentration is a function of time gives you a straight line Then it is zero order with respect to that reactant. If one over the concentration as a function of time gives you a straight line then its second order with respect to that reactant. And if natural log of concentration as a function of time gives you a straight line, then its first order with respect to that reactant. So the chemical reaction that we are considering is C four H eight goes to two C two H four. I went into Excel I plotted the data. I took the data that was provided in the problem for our the concentration as a function of time. I then in Excel put in equals one over and then referenced this cell and it calculated for me, one over the concentration and then I put equals natural log and then I referenced this cell and it calculated the natural log for me. I then plotted concentration is a function of time and got this graph here. I plotted one over. The concentration is a function of time, got this graph and then the natural log of the concentration is a function of time. And got this graph. They all look pretty close to being straight lines. You'll notice a little curvature here and a little curvature here to determine which one is the best straight line in Excel. You can ask it to show the equation of the line and the r squared value. The graph that has the r squared value closest to one is going to be the best straight line which is the natural log of concentration. So because the natural log of concentration is a function of time gives us the best straight line. This is indicative of first order for a first order reaction. When we plot natural log of concentration as a function of time, the negative of the negative slope gives us the positive rate constant. Great constants are always positive. So we get a rate constant 2.0112. The units, because its first order would be won over time, time is seconds. The last part of the question asks you now, knowing the rate constant to determine the rate of the reaction. When the concentration is .25 Mueller to calculate the rate of the reaction, we're going to use the differential rate law or sometimes just called the rate law. Where rate is equal, decay multiplied by concentration. So we'll take rate equals the K value, which we determine 0.112 multiplied by the concentration in question 0.25 Mueller, And we get a rate of .0028 moller per second.


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