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Reaction Order ResultsTmitoi DopEach Chemical[stock solutton drup of (In well) Final concentratlon (ulul UFODS well HCI rate order (show all work for each question)...

Question

Reaction Order ResultsTmitoi DopEach Chemical[stock solutton drup of (In well) Final concentratlon (ulul UFODS well HCI rate order (show all work for each question) CIIDI were made Dy adding A1/C together t0 get total volume of 2Odrops delermine Ine concentration caused by diluting OM HCI and 0,30M Na_S_O1CZiDZ were made by adding AZICZ together to get total volume of ZOdrops determine the concentration caused by diluting 1,OM HCI and 0,30M Na_Sz01C3/D3 were made Dy adding A3/C3 together - t0 g

Reaction Order Results Tmitoi Dop Each Chemical [stock solutton drup of (In well) Final concentratlon (ulul UFODS well HCI rate order (show all work for each question) CIIDI were made Dy adding A1/C together t0 get total volume of 2Odrops delermine Ine concentration caused by diluting OM HCI and 0,30M Na_S_O1 CZiDZ were made by adding AZICZ together to get total volume of ZOdrops determine the concentration caused by diluting 1,OM HCI and 0,30M Na_Sz01 C3/D3 were made Dy adding A3/C3 together - t0 get = total volume of 2Odrops delermlne Ihe concentration caused by dlluling OM HCI and 30M Na 5-01



Answers

Upon dissolving $\operatorname{In} \mathrm{Cl}(s)$ in $\mathrm{HCl}, \operatorname{In}^{+}(a q)$ undergoes a disproportionation reaction according to the following unbalanced equation:
$$
\operatorname{In}^{+}(a q) \longrightarrow \operatorname{In}(s)+\operatorname{In}^{3+}(a q)
$$
This disproportionation follows first-order kinetics with a half-life of 667 s. What is the concentration of $\operatorname{In}^{+}(a q)$ after 1.25 $\mathrm{h}$ if the initial solution of $\operatorname{In}^{+}(a q)$ was prepared by dis- solving 2.38 $\mathrm{g} \operatorname{InCl}(s)$ in dilute $\mathrm{HCl}$ to make $5.00 \times 10^{2} \mathrm{mL}$ of solution? What mass of In $(s)$ is formed after 1.25 $\mathrm{h}$ ?

In this question, we have a reaction of CH three cl Reacting with three moles of cl two producing CCL four and three moles of hcl. We are given for data for four experiments where concentration of the reactant are provided and the initial rate is provided and were asked to determine the order of the reaction with respect to each reactant so that we can write the rate law for the reaction and then calculate the K value. To do this, we need to determine, we need to determine the order with respect to each reactant We noticed in experiments one and two, the concentration of cl two is unchanging or staying constant, But the concentration of CH three cl is being doubled. If we double the concentration of CH three cl, we see that the rate doubles from 30.142 point 0 to 9. It's not an exact doubling, but it's probably because of rounding issues. So, a doubling of the concentration, resulting in a doubling in the rate is indicative of first order. So the reaction is first order with respect to CH three cl But then if we look at experiments two and three Where the CH three cl concentration is staying constant And we are doubling the cl two concentration, We're going from .052.1. We have something that is not double nor quadruple the rate we go from .0292.041. If it were first order doubling, the concentration would double the rate And the rate would be more like .058. If it were second order, then doubling the concentration would quadruple the rate that didn't happen. If it was zero order, then doubling the concentration would result in the rate staying constant. So it's not one of these very common orders. 0 1st or second, it must be a fraction order to calculate the order. That is a fraction. We need to set up an algebraic expression where we will take the rate law and plug in the data for experiment to where the rate is 0.0 to nine. Set that equal decay multiplied by the concentration of CH three cl raised to the first power because we determined that up here multiplied by the concentration of cl two raised to an unknown power. Well then divide that by the differential rate law with the data. For experiment too. Where the rate is .041. Set that equal to K. Multiplied by the concentration of CH three cl again raised at the first power, Multiplied by the concentration of Cl two raised to an unknown power. We set up a ratio of rate laws so that the case will cancel. You'll notice that the .1 concentration staying constant also cancel. So we'll take the a ratio of these two numbers which gives us .707. And then the ratio of these two numbers gives us five. But remember it's raised to the X. So we can now take the log of both sides. We take the log of both sides. The exponent now becomes a multiplier. We can then divide both sides by little log of 50 And X. is the log of .77 divided by the log of .50 which is .5. So the order with respect to the L two is .5 or 1/2 order. The overall order would then be first order plus 1/2 order or 1.5 order. So to calculate the K value. We can take data for any experiment, plug it into a single rate law. I'm choosing the experiment one where the rate is .014. Okay. I will set that equal to K. Which I don't know. The concentration of both of them is 0.5. So it'll be five. Race to the first power for CH three cl and then five raised to the one half power for all to I can then solve for K. and I get 1.25 With units of one over polarity to the one half seconds because the overall order is 1.5. A better K value could be calculated by doing the same thing for all four experiments, Summing them up and divided by 4 to get an average K value. I'm just going to go with the K. Value. I got for experiment one because it should be pretty close to the true value. So then I can say that the rate law is simply equal to the rate constant, which I calculated to be 1.251 over, Mueller to the one half seconds, multiplied by the concentration of CH three cl raised to the first power, multiplied by the concentration of cl two raised to the one half.

So I have been doing this for about 30 years, and I don't think I've ever solved a problem with Indian before. But that is not a problem. I'm not worried about what? I don't know. Let's go with what we do know. So it's a first order reaction. We've got the half life, and so first thing that we can do is calculate the rate constant. Okay. And the T 1/2 are related to each other, and so we can calculate the value of the rate constant per second. And then, um, ordinarily, we worked this in units of polarity. So first, let's convert the grams of Indian chloride two moles and then dividing by the volume convert to leaders. So now we know the polarity of the starting solution. Um, and now we want to know How much do we have left after? Ah, an hour and 1/4. 75 minutes. Uh, so, um yeah, we've been working per second. We're gonna have to take that our and converted two seconds here, but, uh, anyway, ah, the integrated great law for a first order reaction allows us to calculate later amounts after knowing the initial amount And so since we've got K and T, let's leave the log term alone. Invite out what Katie is. And so here's decay that we calculated and the time converted two seconds. And so Katie is a minus 4.68 So that's the log of the ratio of the concentrations. And the way that we clear that log is to take the exponential of both sides. And that then reduces Thelancet side just to the ratio of the concentrations and e to the minus 4.68 is 0.9 and then multiplying that by the initial concentration tells us how much what the mill arat E will be. Let's see, what is the concentration of the Indian? Plus I on There it is. That's the concentration of the Indian. Plus I on after an hour and 1/4 uh, then we were asked, Well, what? Ah, amount of indium metal or solid is produced, and they gave us an unbalanced equation. Plus one on the left, plus three on the right, one indium on the left to Indians on the right. Ah, the way to balance this. I guess you could try it. I guess you could do it by trial and error. But the, uh the way you have not you learned yet probably is to put the appropriate number of electrons on your side. So I've got a positive charge on the left. So I have to look at the conversion of the Indian Plus I am to indium neutral and in a separate half reaction, look at the conversion of the Indian plus ion to Indian three plus. And so, uh, I've got one plus on the left and neutral on the right in this half reaction. And so it's one Elect Tron here. And then I've got one positive and on the left and three on the right here. So two electrons on the right. And so the way we balance thes half reactions and combine them is that the electrons have to all cancel. So I've gotta multiply that 1st 1 by two. So there's two Indian plus plus another one here is three, and we're going to get to indium neutral and one Indian three plus. And so that tells me that I'm gonna get to Indian plus for every three indium ions that were consumed. So how to calculate how much India my on was consumed. This was my initial concentration, converted to decimal. And then this is Thea Mount that I have left. And so subtracting it tells me how much was consumed. If that is three, decrease in the polarity of the India my on. Ah, then that is multiply by the volume to find out how many moles of India my on were consumed. And then it's a 2 to 3 ratio two Indian neutral for every three Indian plus and then finally multiplying that that up to this point gets me the moles of India. And then times the molar mass finally gets May that you'll make 1.2 grams of indium in this way.

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.

Generally rate off a reaction increases with increasing concentration off the reactant variation for the high concentration hydrochloric. It's it. It's much faster because it dissolves to think much faster, which is what the reaction is. That's increasing the increasing. The concentration off the concentration off hydrochloric a seared increases the craziest e rate off the reaction mhm rate off the reaction. Then, as we double the concentration off in two or three, the rate also doubles, which means the reaction is first order. Reaction is forest order with respect, fit, respect toe the reactant So increasing the concentration but always in Christie rate, thus increasing the initial concentration, initial concentration off and tow, or three the increased E increase the initial reaction rate initial reaction, right?


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