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The reactionI- (aq) + OC1- (aq) IO- (aq) + Cl- (aq)tlsstudied, and the following data were obtained:[I~Jo (moVL) [OC1-Jo (moVL) Initial Rate (moVL 0.20 0.16 7.70 x ...

Question

The reactionI- (aq) + OC1- (aq) IO- (aq) + Cl- (aq)tlsstudied, and the following data were obtained:[I~Jo (moVL) [OC1-Jo (moVL) Initial Rate (moVL 0.20 0.16 7.70 x 10-20.100.163.90 x 10-2 9.60 * 10-30.0500800.400807.70 x 10-2What is the rale law?(Use k for the rale constant; )RateCalculate the value of the rate constant:k-L/molCalculate the initial rate for an experiment where bothand OCl are initially present at 0.14 molLRatemolLSubmit AnswerTry Another Version10 item attempts remaining

The reaction I- (aq) + OC1- (aq) IO- (aq) + Cl- (aq) tls studied, and the following data were obtained: [I~Jo (moVL) [OC1-Jo (moVL) Initial Rate (moVL 0.20 0.16 7.70 x 10-2 0.10 0.16 3.90 x 10-2 9.60 * 10-3 0.050 080 0.40 080 7.70 x 10-2 What is the rale law? (Use k for the rale constant; ) Rate Calculate the value of the rate constant: k- L/mol Calculate the initial rate for an experiment where both and OCl are initially present at 0.14 molL Rate molL Submit Answer Try Another Version 10 item attempts remaining



Answers

The reaction
$$
\mathrm{I}^{-}(a q)+\mathrm{OCl}^{-}(a q) \longrightarrow \mathrm{IO}^{-}(a q)+\mathrm{Cl}^{-}(a q)
$$
was studied, and the following data were obtained:
a. What is the rate law?
b. Calculate the value of the rate constant.
c. Calculate the initial rate for an experiment where both I'" and $\mathrm{OCl}^{-}$ are initially present at 0.15 $\mathrm{mol} / \mathrm{L}$ .

So we have here an unusual experiment of this type. Uh, when we look in the data table for the different concentrations and rates, we see that the Hypo chloride ion, the OOCL minus, is held constant and experiments one and two and then again in three and four. It's a 40.90 in three and four, so we can pick either pair. Experiments wanted to or experiments three and four toe. Look at the effect of changing the iodide. But then, when we look at the iodide concentration, there are no experiments where it's helped. Constance. We'll see what we need to do in this situation. Uh, what we need to do first is what we can do. Which is, uh, I picked Teoh. Compare experiments three and four. As I say, we could have done one or two or three and four, so I've set it up. The rate ratio and three and four is equal to the iodide concentration ratio and three and four, where the Hypo chloride is 0.90 molar in both three and four, and so have plugged in the rates for experiments three and four and the ah iodide concentrations in three and four. And basically those are both ratios of eight. And so that tells me that the, uh, reaction is first order in iodide. Once I know that, then I can pick any two of the other experiments. Even though I can't hold the iodide constant, I can take account of its effect because I know the order with respect to it. So now when I write the ratio, I picked experiments three and two. Um, I include the iodide concentration ratio in those two experiments, as well as thehyperfix chloride concentration ratio. And so, uh, plugging in the rates in experiments two and three, I find that the ratio of the rates is four and that both of the reactant is doubled in concentration in those two experiments. And so that gives me a known for the iodide to the first powers to and dividing both sides by two. Then I have That, too, is to to the Z Z being the letter I chose for the experiment on the hypochlorite. And so the reaction is first order with respect to both reactive. Now we need the rate constant here. And so what we do is, we write the great law. Uh, and so for K divide both sides by the iodide. Yeah, and the hyper correct. And so picking. Experiment 17.9 times 10 to the minus two. And that is Moeller per second. Our units canceled everywhere else. Units were the same in all of those ratios. But in calculating K, we want to take account of the units, uh, and the concentrations and experiment one or 0.12 Moeller and she report 18 Mueller. And so I can calculate the value of K, and we get 3.66 Eso won power of Moeller and two powers of Muller per muller per second. And so that then is theory rate constant. And for part c, the rate it's going to be I'm going to take a plug in given concentrations. And so, uh, both art. 0.15 Moeller very much like trial number one. We're going to get a rate. It's about the same as we had in trial number one. So notice Moeller and Moeller Moeller squared. Uh, the units on K need to combine with those on the concentration. So is to give a rate in units of polarity per second. And so, without rounding off as I did to write this down, I'm just going to go ahead and multiply 8.2 four times 10 to the negative, too. Muller per second. That is how you get an initial rate with given concentrations. You've gotta have the rate law with the dependence on concentrations, work up the value of the rate constant so that you can plug that into the rape law and then plug in any concentrations and you can calculate the right.

In this problem, a reaction is given and the mechanism for this reaction is also given since the reaction is taking place in the echo solution. So the concentration of water remains constant. The read for the forward reaction that is the step one is given. No, we have to write the rate off the reaction that is rate law for the reaction. So from this low step off the reaction that is Step number two, we can write the rate as que two in two concentration off I negative multiplied by concentration off H o seal. Let this equation be number one. No, We can also write the create for the fast equilibrium step that is for step one. It can be written as given into concentration off O C. L Negative is equals toe K minus one into concentration off H O seal multiplied by concentration off O h Negative. Rearranging this equation we can write the concentration off h o C l equals two given into concentration off OOCL negative divided by K minus one into concentration off O h Negative. Now, when we substitute this concentration off h oocl in equation number one, we can read the read that is this can be further written is okay and to concentration off negative into concentration off all seal negative, divided by concentration off O H negative. This is the required rate law for the reaction.

This question. We also have the generic reaction of a single reactant. A going to products. You're provided data showing the concentration and the corresponding initial rate to determine the order of the reaction. It's a little bit more difficult because we don't just have a simple doubling of the concentration if you double the concentration and you double the rate, its first order. If you double the concentration, nothing happens to the rate at zero order. If you double the concentration and you quadruple the rate, it's second order. So to determine the order without simply identifying the change in the rate due to the doubling of concentration. Because we don't have a doubling of concentration, We need to set up an algebraic expression which will be a ratio of the differential rate laws for two experiments. I'm choosing experiments one and 2. Where the experiment where the rate for experiment one is .0078. That's equal decay, multiplied by the concentration .1 to raise to some power, which we don't know. I'll divide that whole thing by the data. For experiment too. Um incorporated into the differential rate law where the rate is .0104. That's equal decay. And then the concentration is .16 raised to some power. We don't know the rate constants will cancel. And we'll get this ratio here being .75. That will be equal to this ratio here which is .75. Race to the X. And it's hopefully obvious now that X. is one. So this is a first order reaction. So the rate will be equal to K. Multiplied by the concentration raised to one. We can calculate the rate constant by taking the rate and the concentration from any experiment. I'm choosing experiment one where the rate is .0078 and the concentration is .12 And I get a rate constant 2.065, 1 over seconds to get a more accurate K value. You could do this three times for the three experiments and calculate an average. So the rate law is, rate is equal to my rate constant 0.651 over seconds, multiplied by the concentration raised to the first power.

Here we have the general reaction of A. Goes to products with concentration of A. And initial rates. And were asked to first determine the order of the reaction. To do this will take a ratio of the differential rate laws. I'm choosing experiments one and 2. Where the rate is 3.89 times 10. The negative four with the concentration of 40.1 raised to some power, which we don't know, and then multiplied by a rate constant, which we don't know, divided by. Again the rate. This is now for experiment to set equal to the rate constant, which we don't know, and the corresponding concentration that gave us that rate raised to a power. We don't know. We do this ratio of differential rate laws so the case will cancel. And all we have left is the unknown for the order Dividing these two numbers, we get .444, Dividing these two numbers, we get 666. We raise that to the X. We can then multiply well, we can then take the log of both sides of the equation in order to bring that X down as a multiplier. Then divide the two log terms to get eggs and we get second order. So the differential rate laws rate is equal to the rate constant, multiply by the concentration squared. Because its second order, we can solve for the rate constant by taking any rate in its corresponding concentration. Now, knowing that second order and solve for K. Or we could do it for all three experiments, calculate three K values and then determine the average. I get a K value of 30.0 to 7 one over moller seconds because these are the units for a second order reaction. So the differential rate laws rate is equal to my rate constant .0271 over moller seconds multiplied by concentration squared.


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