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1. (18 pts) The following kinetic data were obtained for the following "clock reaction" SzOs?-(aq) 3I-(aq) 7 2S042-(aq) + Is(aq):Initial concentration (mo...

Question

1. (18 pts) The following kinetic data were obtained for the following "clock reaction" SzOs?-(aq) 3I-(aq) 7 2S042-(aq) + Is(aq):Initial concentration (mol Le [S2082-J [I-lo 0.15 0.21 0.22 0.21 0.22 0.12 0.35 0.24Initial rate (mol L-les-I- 1.14 1.70 0.98Experiment2 3(a) Determine the rate law for the reaction (b) From the data, determine the value of the rate constant: (c) Use the data to predict the reaction rate for Experiment 4.

1. (18 pts) The following kinetic data were obtained for the following "clock reaction" SzOs?-(aq) 3I-(aq) 7 2S042-(aq) + Is(aq): Initial concentration (mol Le [S2082-J [I-lo 0.15 0.21 0.22 0.21 0.22 0.12 0.35 0.24 Initial rate (mol L-les-I- 1.14 1.70 0.98 Experiment 2 3 (a) Determine the rate law for the reaction (b) From the data, determine the value of the rate constant: (c) Use the data to predict the reaction rate for Experiment 4.



Answers

The following data were measured for the reaction $\mathrm{BF}_{3}(g)+\mathrm{NH}_{3}(g) \longrightarrow \mathrm{F}_{3} \mathrm{BNH}_{3}(g):$ $$ \begin{array}{lccc} \hline \text { Experiment } & {\left[\mathrm{BF}_{3}\right](M)} & {\left[\mathrm{NH}_{3}\right](M)} & \text { Initial Rate }(M / \mathrm{s}) \\ \hline 1 & 0.250 & 0.250 & 0.2130 \\ 2 & 0.250 & 0.125 & 0.1065 \\ 3 & 0.200 & 0.100 & 0.0682 \\ 4 & 0.350 & 0.100 & 0.1193 \\ 5 & 0.175 & 0.100 & 0.0596 \\ \hline \end{array} $$ (a) What is the rate law for the reaction? (b) What is the overall order of the reaction? (c) Calculate the rate constant with proper units? (d) What is the rate when $\left[\mathrm{BF}_{3}\right]=0.100 \mathrm{M}$ and $\left[\mathrm{NH}_{3}\right]=0.500 \mathrm{M} ?$

Okay. So for the first part, if you Yeah, doubling The concentration of image three. Why you? Yeah, Bs three is constant. Double storied and doubling. We have three concentration while mh three is constant. Also double 30. Trade. So, overall or like the rate of reaction would be Yeah, we have three. Chemistry. Okay. Overall, yes. Second order reaction. And then see part. We'll use The date Oklahoma Experiment one. They don't give me a call to. We need to work Concentration of PFC concentration of image three. Oh, This is the bombs conditional 4th of them. But I've noticed 3.4. 1 second. Okay. Mhm.

This is a method of initial rates with two reactant. To determine the order of the reaction with respect to each reactant, we need to treat each reactant separately, namely, if we look at experiments one and two, we see that the concentration of N. 02 is changing, but the concentration of F two is not. So let's just look at what a change to the concentration of N. 02 does to the rate. By looking at experiments one and 2 where F two stays constant. So with these two experiments a doubling of N 02 Only a doubling of eno to keeping F to constant, results in a doubling of the rate going from 20.262 point 51 This is indicative of a first order reaction with respect to N. 02 Now let's look at experiments two and three where the concentration of CO two is stained constant and the concentration of F two is being doubled from 20.12 point two. In this process we see that the rate is doubled from .512.103 Doubling the concentration resulting in doubling the rate is indicative of a first order reaction. So it's also first order with respect to F two. So the rate laws rate is equal to K, multiplied by each concentration raised to the first power. To solve for K will simply plug in the rate for any experiment and the two corresponding concentrations and we'll calculate a K value of 2.61 over Mueller seconds, one over Mueller seconds. Because the overall order is second order. First order with respect to each reactant. Or we could do this for all four experiments, Get four K values and then calculate the average. So the differential rate law is rate is equal to 2.61 over moller seconds, multiplied by each concentration raised to the first power giving us a second order overall

Calculate a slow over the first two rate constant slope AM is equal to Delta l and C L two or to buy by a Delta Day minus K is when a 16.93 to minus 16.68 37 Okay, I 517 meal as minus 1 72 mil s minus. K is minus 16.98 32 plus 16.6837 No, my 3. 45 mil s so minus care is a close to minus point to the bill 95 by 345 mills second. Okay, that was to 8.688 point 68 And pretend to the bar minus four meals. Second inverse. That is the value of the first order rate. Constant case, then the question. Yeah. After one house has passed that the state was to see how the concentration of sale Toto is half of the initial concentration off sailboat. So c l two or two is ah, half or cierto or do not. Any question toe is as follows L N C l do or two by c l two or two, not which is minus skating. So, Ellen Sale to talk and regional half sales tour to not by Cielito, Not which is minus Katie. Yeah. Uh huh. This This is canceled out. L in half is minus 0.693 Closed toe minus Katie After when? It's when it was considered out. So be half is back. You got ta half. It was okay by 0.693 Uh, this is a half life for the rest of the reaction. Now calculate the half life of the half life for the decomposition of the altar toe shown below right de half called the 0.693 My case substituting 816 stranger tenders par minus four music and in verse, four case, uh, could really have a ball. 2.693 Okay, s 6.8 point 68 into tend to the bar minus four meals 2nd and 1st, which is a 700 nineties, 8.38 muse. Second, which is approximately 798 music. And so this is half life for the decomposition. These are the halfway for the decomposition. Off the door toe. Yeah,

Yeah. To answer this question, we need to determine what happens to the rate when just one of the reactant concentrations change. If we look carefully at the data provided you'll notice that when only A is tripled, the rate is tripled. This is indicative of a first order reaction with respect to that reactant. If we look at experiments two and 3. When B is tripled, the rate actually increases 3 to the two or 9 fold. This is indicative of a second order reaction with respect to that reactant. So the rate law is going to be equal to rate equals K. Multiplied by A to the first power and be to the second power. Knowing the rate law allows us now to rearrange it and solve for the rate constant if we know the rate and the concentrations and we do three different times with three different experiments. So we can take the data the rate and the concentrations for any one of these three experiments and we will get a K Value of 10, 1 over Mueller one over Mueller squared seconds. Now knowing the K value, we could calculate the rate at different reactant concentrations. Rate is equal to K which we now know, multiplied by the concentrations provided .2 for a and .2 for B. Not, don't forget to square B. And we get a rate of .08, 0 molar per second


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