The tabulated data were collected for this reaction at 500 °C:
CH...

Question

The tabulated data were collected for this reaction at 500 °C:
CH3CN(g)¡CH3NC( g)
Time (h) [Ch3Cn] (M)
0.0 1.000
5.0 0.794
10.0 0.631
15.0 0.501
20.0 0.398
25.0 0.316
a. Determine the order of the reaction and the value of the rate
constant at this temperature.

Accepted Answer

hey everyone in this example, we're asked to find the order of the reaction and the rate constant for the reaction based on the gathered data at 250 degrees Celsius. So are given reaction is bioscience ion which produces and chlor oh susana meid as an an ion. And below were given the data set where we have time corresponding to the concentration of our assign it an ion in units of polarity. So step one is going to be finding our order of the reaction. And what we would do is recall the following plots where first for a zero order reaction we would plot the concentration of the reactant being the assign an ion versus time according to the data set. Given now for the first order reaction plot we are going to take our Ln of our concentration of the assign it an ion at a given time and were plotting that versus time. And then lastly for a second order reaction, we want to plot one over our concentration of the synod an ion at a given time versus time. Now, what we're going to do is analyze each plot for the equation of the line that they generate as well as the correlation coefficient. So looking at the plot for the zero order reaction according to our data set for our correlation coefficient, which we recall is r squared. We're gonna get a value equal to 0.98848. And because we have a value being 0.988 which is very close to one. We would say that therefore the reaction is going to be zeroth order most likely. Now moving on to the second plot, we have a plot for a first order reaction. And according to the correlation coefficient we would generate in our calculators. According to this plot we're gonna get a value equal to 0.99999. And because this is very close to one, we would say that therefore this is likely also going to be the order of the reaction. So we would likely have a first order reaction. And then lastly we have for our plot for a second order reaction, we're going to generate a correlation coefficient value equal to 0.98789. Now here we have 0.987 which is not as close to one as our zero order reaction correlation coefficient. And so we would therefore not consider the reaction to be second order. And sorry, that should say reaction is not second order again because this value here is not as close to one as our other two plot values were for the correlation coefficient. So now we have determined that our order of our reaction is likely going to be zero order or first order. And these are our first two answers. But now we want to find for part to our rate constant. And so now we want to recall our integrated rate law for a zero order reaction where we would take the concentration of our reactant being theo sign it equal to our rate constant which is negative K. Which is what we're solving for times time t added to our concentration of co sign initially. And so for our zeroth order integrated rate law we can plug in what we know from the data set. So I'm sorry this should say at a given time. So for our concentration of assigning we're going to pick any time value from the plot. Let's just go with five hours. So we can say that according to our data set the concentration for theo sign is going to be 0.861 at the time being five hours. And we're setting that equal to negative one times our rate constant K. The time here is going to be five because we're plugging that in from the data set and then this is added to our initial concentration of co. Sign it. Which according to the plot at zero hours or zero seconds is a value of one molar. So we can plug that in as one. So to simplify this so that we can isolate for the rate constant K. We're gonna subtract one from both sides. And when we simplify this is going to give us negative 0.139 set equal to negative one times the rate constant K. Multiplied by the time being five. And now we're going to isolate for K. Further by dividing both sides by five. And what we would get here is that our rate constant K. Is equal to 0.278. Which we can round 2366. That we just have 0.3. And recall that the rate constant does not have units. So now we want to go ahead and do the same thing by recalling our integrated rate law for a first order reaction. Since we determined that this reaction could also potentially be first order where we would take the natural log of our concentration of our diocesan and ion at a given time and we're setting that equal to negative one times the rate constant K. Multiplied by the time and I'm just going to make some more room here and then this is going to be added to the natural log of our initial concentration of assigning and sorry just so everything is visible, we'll just make this a bit smaller. So this should have also said the initial concentration of co sign it for the zeroth order integrated rate law as well. But now we're going to do the same thing and plug in from our data set. The chosen values at the time being five hours for the concentration of co sign it. So we would say that the natural log of our concentration at five hours being negative zero or sorry that's positive. So at five hours we have positive 0.861 Moeller. We're setting this equal to negative one times our rate constant K. Multiplied by the time. Which at this concentration would be five hours then added to the Ln of our initial concentration of co sign. Which according to the time being at zero hours we have a concentration of one molar. And just so everything's clear. Let's just move this over Since these are two separate calculations. So moving into our next line, we want to further isolate for the rate constant K. So when we take the Ln of one we would get a value equal to zero. And so we can just simplify this so that we have the Ln of our concentration of co sign an ion at five hours is equal to negative K. Times five. And now we're just going to divide both sides by five yet again. And what we're going to do and sorry? So the Ln of our value above is equal to a value of negative 0.149661. So we can just fill that in there and now we can just divide by five. So what we can simplify too. Once we divide by five is a value of negative 0.29932. And this is equal to negative one times the rate constant K. And so we still have that negative one attached. So we're gonna divide both sides by -1. And what this is going to give us is our rate constant K equal to a value of 0.0299. Which we can also round to about three or sorry, round to about zero 0.3 again with no units. So because we have the same value for the rate constant being equal to 0.3, we can say that therefore our rate constant K based on the reaction being both zero order or first order is going to be 0.3. So Everything highlighted in yellow represents our final answers. We determined our order of our reaction either being zero or first order as well as our rate constant, which is equal to 0.03. So I hope that everything I reviewed was clear. But if you have any questions just leave them down below and I will see everyone in the next practice video.

The tabulated data were collected for this reaction at 500 °C: CH3CN(g)¡CH3NC( g) Time (h) [Ch3Cn] (M) 0.0 1.000 5.0 0.794 10.0 0.631 15.0 0.501 20.0 0.398 25.0 0.316 a. Determine the order of the reaction and the value of the rate constant at this temperature.

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