Consider the reaction:
2 NO( g) + O2( g) ∆ 2 NO2( g)
The followin...

Question

Consider the reaction: 2 NO( g) + O2( g) ∆ 2 NO2( g) The following data show the equilibrium constant for this reaction measured at several different temperatures. Use the data to find ΔH° rxn and ΔSrxn ° for the reaction.

Accepted Answer

hi everyone for this problem. It reads based on the given equilibrium constant at different temperatures. What are the standard entropy and standard entropy change for the reaction were given the reaction along with the following data. So what we want to do here is we want to find these two values. We want to find the value for our standard entropy change in our standard entropy change. And what we can do since we're given data is we can plot this data of our temperature versus our equilibrium constant. And when we plot that data, what we'll get is a regression line and that is translated to this. Okay, so when we plot it are linear form of our equation is going to be Ln. Of K is equal to the standard entropy change of our reaction over R gas constant R times one over temperature plus our standard entropy change for our reaction over r gas constant R. Okay, so the values that we're interested in here is our standard entropy change in our standard entropy change. Okay. And when we plot this data we're going to get a linear equation of Y. Is equal to M X plus B. Okay. And we have our temperature one over temperature represents R X and R L N. F K. Is what we're going to use as our K. P values. So we're going to need to plot this to get this line. And when we plot the data, the regression line we're going to get is why is equal to 24, . x plus negative 24.110239. Okay so that is the regression line we're going to get from our data. So like we said we're looking to solve for our standard entropy change and standard entropy change. So because we know that let's take a look at this part first. We know that our standard entropy change over our is going to equal this value because that is what is represented in our regression line. So if we set these two equal to each other, we'll be able to solve for our standard entropy change because we know what the value of R is. This is a value. We should have memorized. It is a constant. Okay so let's go ahead and solve for our standard entropy change for our reaction first. Okay so let's set those two equal to each other. So we set our standard and they'll be changed for our reaction over. R gas constant R is going to equal. There should be a negative in this equation. Okay so go ahead and put that negative there. Alright so we have negative. Okay, so now that we set this equal to each other. Our goal here is to solve for our standard entropy change. So we'll go ahead and multiply both sides of our equation by R. And when we do that are well cancel on the left. Okay and we can go ahead and plug in the value now. So our standard entropy change for our reaction is equal to And our our is 8.314 jewels over Mole Times Kelvin. OK so the value that we're going to get for this is negative 2.0 times 10 to the fifth jules. Okay so our standard entropy changes in jewels and we're going to want to convert this to kill a jules because that's the unit that it is normally in. So let's go ahead and do that conversion here and one killer jewel there is 1000 jewels will make sure our units cancel jewels cancels and will be left with an answer of negative 2.0 times 10 to the second killer jewels. Okay, so that is our answer to the first part are standard entropy change for our reaction is negative 2.0 times 10 to the second killer jewels. So we can go ahead and highlight that as our final answer. The next thing that we want to do now is calculate the standard entropy change for our reaction and we're going to do the same thing. Set it equal to what our regression line tells us that it equals. Okay, so we have our standard entropy change for our reaction over R is equal to -24.110239. So we're interested in our standard entropy change. So we'll go ahead and multiply both sides of our equation by our Okay. And so when we do that we'll get our standard entropy change for our reaction is equal to negative 24.110239 times 8.314 jewels over Mole Times Kelvin. Okay, so when we do that calculation we're going to get negative 200.45. Let me rewrite that out. Negative 200. jules. Now the unit that we normally have entropy in is killer jewels. So let's go ahead or the unit that we normally have this in is jules. So let's see here. Yeah, so we can just go ahead and simplify this out. So for a standard standard entropy change We'll have it in jewels so we can put this in exponent form. So when we put this in exponent form it is negative 2.0 times 10 to the second jewels. Okay so our our standard entropy change was in kila jewels and our standard entropy change is in jewels. So that is it for this problem. I hope this was helpful

Consider the reaction: 2 NO( g) + O2( g) ∆ 2 NO2( g) The following data show the equilibrium constant for this reaction measured at several different temperatures. Use the data to find ΔH° rxn and ΔSrxn ° for the reaction.

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