Consider the following concentration–time data for the
reaction ...

Question

Consider the following concentration–time data for the reaction of iodide ion and hypochlorite ion 1OCl-2. The products are chloride ion and hypoiodite ion 1OI-2. (d) Propose a mechanism that is consistent with the rate law, and express the rate constant in terms of the rate constants for the elementary steps in your mechanism. (Hint: Transfer of an H+ ion between H2O and OCl- is a rapid reversible reaction.)

Accepted Answer

Hello. And this problem we're told that the concentration time data for the reaction of hyper chloride ion and iodine are given in the table below the products are hyper bromine and chlorine dine. The fallen mechanism is proposed for this reaction where asked is the proposed mechanism consistent with the rate law and to express the rate constant in terms of the rate constants for the elementary steps. In the proposed mechanism. Looking at the proposed mechanism, we see that we have water as a reactant going into the reaction in the first step and it's on the product side in the last step, something that goes into and out of the reaction unchanged is a catalyst going back to our first step. We see that we have hipaa chorus acid being produced in the first step and then consumed in this next step. Something that is produced and then consumed is referred to as an intermediate. We have hydroxide ion which are also being produced in the first step and consumed in the third step. And then we have hyper and being produced in the second step and consumed in the third step. So we have three intermediates within this reaction mechanism, we'll make use of the second step, which is slow and rate determining to write our reaction rate law. So we have the rate then is equal to the reaction rate constant. We'll put subscript too. Since this is for the second step in the reaction mechanism times the concentration of hipaa chorus acid times the concentration of roman ruins. So generally an intermediate does not appear in our reaction rate law. So we will make you So the first step which is fast and reversible. To substitute in for our concentration of our intermediate which is hipAA Cloris acid. So given that the first step is fast and reversible, that then tells us that it will come to a state of chemical equilibrium. So we have the right then of the forward reaction is equal to the rate of the reverse. So our rate for the forward reaction will be equal to actually constant for the forward reaction. We call that subscript one. Since this is the first step in the mechanism times the concentration of hydrochloric dine. This will be cool to the reaction rate constant for the reverse we'll call that K subscript -1. And as times the concentration of hydrochloric acid times the concentration of hydroxide ions. We'll now move everything over to the left hand side, isolating concentration of our intermediate. And we will now substitute this into our initial reaction rate law. So we now have the rate is equal to the reaction rate constant for the first step and going into four direction times were actually constant for the second step provided by the reaction rate constant for the first step, going in the reverse direction, times the concentration of hip chord, iron, times the concentration of bromide iodine all over the concentration of hydroxide in so this is the rate law based on the proposed mechanism we're now going to determine the rate law based on the experimental data that were provided. In order to do this, we need to find then the rate of the reaction for each of the experiments. So we can write the rate then in terms of the change of our reactant, hypochlorite over changing time and this is negative because it's a reactant so it's disappearing over time. So for the first experiment then we have this is equal to our concentration after 20 seconds, which is 1.80 Times 10 to - -2.00 times 10 to the -3. That would be the concentration initially all divided by our change in time, which is 20 seconds. So our rate for the first experiment then is equal to 1.0 times 10 to minus five clarity for a second. For the second experiment we will again calculate the negative of the change in the concentration of hypochlorite over change in time. The time 20 seconds, it's 2.60 times 10 -3 moller. Initially it was four Times 10 to -3 Moller Our change in time is 20 seconds rate for the second experiment then is to Times 10 to -5, similarity, similarity per second. To find the rate for the third experiment in the same way, Concentration after 20 seconds is 1.9 times 10 to - Mueller. Initially it was to Mueller And our change in time is 20 seconds. This works out to five times 10 96. Hilarity per second. And then our last experiment We calculated in the same way as the previous three. After 20 seconds, the concentration is 1.9, 0 times 10 -3. Initially it was to Times 10 : -3 Mueller. all over changing time of 20 seconds. This works out to five times 10 : -6. Liberty per second. Now we'll write a general form of our reaction rate law. So we have, the rate then is equal to reaction rate, constant times the concentration of hip Coraline to some order Y. Times concentration of bromide, iron, some order X. And the concentration of hydroxide in to some order Z. So to find why. Then we're going to compare experiment one and to find X. We're going to compare experiments one and three and two finds the we're gonna compare experiments one and four. So beginning with finding why? Then we have the rate for Experiment two divided by that. For experiment one, if we have a concentration it's being held constant, then it will be eliminated from the rate law expression. So the concentration that is changing. Is that related to hipaa chloride? So we have the concentration of hyper quiet for the second reaction is four times 10 to minus three, this is at time zero. And for the first experiment our initial concentration for hip accordion is two times 10 -3. and our rate for experiment two, There's two times 10 : -5. And that for the first experiment is one Times 10 : -5. So our units have been removed for our rates, the units were canceled. They're the same for both the Right to and Write one. This then works out to to to the Y is equal to two. So that tells us then that Y is equal to one, find X. We're now going to compare The rate for experiment 12, that for experiment three in this case then we are changing the concentration of bromide and the rest are being held constant. So we have the initial concentration of in for Experiment one is four times 10 to minus three and the initial concentration bromide ion for the third experiment is two times 10 to the minus three. And our rate for experiment one Is one times 10 : -5. And that for experiment three is five times 10 to minus six. So we get then to the X. Is equal to two. That tells us then X is equal to one and now we'll compare the rate for experiment one to that. For experiment four, in comparing one and four, then the concentration of the hydroxide ion is changing and the rest are being held constant. So we have the concentration of hydroxide in initially, for the first experiment is 1.50 the Z. And the initial concentration of hydroxide ions. For experiment four is three. The rate for experiment one is 1 times 10 to the -5. And the rate for the 4th experiment is five times 10 -6. So we get then 1/2 to the sea Is equal to two. This tells us then that Z is equal to -1. So we can write in. Our rate law is equal to the reaction rate constant times the concentration of accordion to the X. Which is one times the concentration. Or sorry, two Y. Which is one times the concentration of bromide to the X. Which is equal to one times the concentration of hydroxide ions, which is to the minus one. So we can rewrite this with the hydroxide in and the denominator. So we have the rate is equal to the reaction rate constant, times the concentration of hip accordion, times the concentration of bromide in all over. The concentration of hydroxide in this then matches the rate law. The expression that we got from the reaction mechanism and our reaction rate constant then would be equal to reaction rate constant for the second step in the mechanism times that for the first step in the mechanism going in for direction divided by the reaction rate constant. For the first step going in the reverse direction. And so this rate law then is consistent or the mechanism is consistent with the rate law and we can write the reaction rate constant again in terms of the reaction rate constant for the elementary steps within this mechanism. Thanks for watching. Hope this help.

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