Choose any two temperatures and corresponding vapor pressures in ...

Question

Choose any two temperatures and corresponding vapor pressures in the table given in Problem 11.30, and use those values to calculate ΔHvap for dichloromethane in kJ/mol. How does the value you calculated compare to the value you read from your plot in Problem 11.32?

Accepted Answer

hi everyone for this problem. It reads using the first two temperatures and vapor pressures in the table, calculate the entropy of vaporization for benzene and killing joules per mole, plot the values in the table, then determine the entropy of vaporization for benzene and kill a joules per mole and compare the values obtained using the two methods. So we have three things that we want to answer here. And the first one is calculating the entropy of vaporization for benzene in the units of joules per mole. The second part is to plot the values okay, in the table and then determine the entropy of vaporization for benzene in the units of joules per mole. Okay. And so let's go ahead and do this problem. And the equation that we're going to want to use to solve this problem is the two point form of the classiest clip Aaron equation and that equation is Ellen of P two over P one which is pressure to overpressure one is equal to the negative entropy of vaporization over gas constant. R times this should be an R over gas constant our times one over T two, which is temperature to minus one over T one. And so because we're solving for the entropy of vaporization, what we want to essentially solve for is this value right here and in the problem we're told to use the first two temperatures and vapor pressures. Okay, so what that means then is we're looking at the first two And we're gonna go ahead and define the variables here. So what this means is T1 is equal to 300 Kelvin. T two is equal to 310 Kelvin, P one is equal to 662 mm of mercury and P two is equal to 1028 0. millimeters of mercury. So these are pressures and temperatures are is a gas constant and this value is 8. jewels over mole times kelvin. Okay, so we have everything we need. So let's go ahead and plug these values into our Closius clip Aaron equation so that we can solve for the entropy of vaporization. So when we plug these values in we're going to get L. N. Of P two over P one. So this is going to be Ln of 28.9 millimeters of mercury over p 1 millimeters of mercury equals negative entropy of vaporization over r where r is 8. jewels over mole times kelvin. This is going to be multiplied one by one over T two. Where T two is 310 kelvin minus one over T one, which is 300 kelvin. Okay, so we've plugged in all of our values so let's go ahead and simplify this. Okay, so for the Left side are Ln of 1028.9 over 662. We're gonna get Ln of 1. 4-3. When we simplify the division. Okay, and then this is going to equal the entropy of vaporization over 8.314 jewels over mole times kelvin. And let's simplify our one over 310 kelvin minus one over 300 kelvin. When we simplify that, that becomes -1.075 times 10 to the -4 Calvin inverse. Okay, and so now we're gonna simplify this more by dividing both sides. Well, let's go ahead and do the Ln function. So we're going to take the L n f 1.55423. And what this becomes is 0. 098. Okay, so this is going to equal the entropy of vaporization over 8.314 Jewels over Mole, Kelvin times negative 1.75 times 10 to the negative four kelvin's inverse. Okay, so what we're gonna do is we're going to divide both sides by this negative 1.75 times 10 to the negative four kelvin's inverse. We're going to divide both sides of our equation by this. Okay, so let's go ahead and write that in red. So we're going to divide both sides By - 075 times 10 to the -4 Kelvin's inverse. Okay, divided by negative 1.075 times to the -4 Calvin's inverse. So when we do that, we get negative 4102. Calvin is equal to. So this is going to cancel here. So that's going to equal the entropy of vaporization over 8. jewels over Mole Times Kelvin. So now we're going to go ahead and multiply both sides by 8.314 jewels over Mole Times Kelvin. Okay, so we're gonna multiply both sides by 8.314 jules over more times kelvin. And we're gonna get so this is going to now cancel and we're gonna get the negative entropy of vaporization is equal to negative three negative 34,105 jewels per mole. So we can divide both sides by one and we're going to get a positive number. So this should be negative entropy of vaporization. So we'll divide both sides by negative one. So we'll get our entropy of vaporization is equal to positive 34, jules Permal. Okay, and the question asks us to report it and kill a jules Permal. So we need to go from Jules Permal to kill a jules Permal. So we'll go ahead and do that conversion and one kill a jewel there is 1000 jewels Our units of jewels cancel. And now as you can see we're left with the units of joules per mole, which is what the question asked us to report this number in. So let's do this calculation. And when we do that, we get 34.1 kila jewels per mole. So our final answer for the entropy of vaporization is equal to 34.1 killer jewels per mole. So that is going to be part one of this problem. Okay. And part two of this problem asks us to plot the values in the table and then determine the heat of vaporization for benzene and kill a jules Permal. So what we're going to want to do for part two of this problem is we're going to want to write the linear form of the classiest clap, Aaron equation. Okay, so we can go ahead and right here the linear form of this equation is Ln of P is equal to the negative entropy of vaporization over are Times one over temperature plus C. And when we write this in linear form, what this becomes is Y equals M X plus b. Ok, actually, excuse me, this should be over R. And this one over T. Let's do this correction negative heat of vaporization over R. And this is going to be times one over T. So this is gonna be Times one over T. So as you can see why equals M X plus B. So that is our linear form for this equation. And so in order for us to plot this, we're going to need to calculate the values for one over T and L N. Of P. So let's go ahead and go back up to the top so that we can do that. So we're going to add two rows to our table. That was given. We'll move these values over to the side because we don't need them at this moment. So move this over here. And so what we want to calculate so that we can plot this, we're going to add two rows to this table. We're going to add a row for one over T. And L. N. Of P. Because we have temperature and pressure, we can calculate these values for each row in our table. Okay, so when we calculate this for each row on our table, let's go ahead and start off with row one. So number this row 123456. So for row 11 over T. So one over 300 Calvin becomes 0.3333. And Ellen of p. R p is 662 millimeters of mercury. So Ellen of P. R. Pressure is 6. 52 64. So we're going to do this for each row in our table. So let's go ahead and do that and plug in these values. So for row two, R one over T. Is 0.32 to six and R L. N F p is 6. row three. R one over T is 0.3125. And Ellen of P is 7. row 41 over T. Is 0.303. And Ellen of P Is 7. 148 row 51 over T. Is 0.2941. And L N. Of P is 8.103702. And lastly row six R one over T is 0.2857. And LNFP is 8.448367. So this is the data that we're going to need to plot in order to find the heat of vaporization. So when we plot this data, the line that we're going to get is y equals negative 4101. X plus 20.167. Okay. And what this equates to is our slope M. M is our slope. So M is this number here? And we said that in our line below. Okay. So y equals mx plus B. Right here. What M equates to? Is the negative heat of vaporization over our Okay, so let's go ahead and write that up here. So M equals negative he of vaporization or the negative entropy change of vaporization over our All right, so now what we can do is we can use this to solve for our we can use this to solve for the entropy of vaporization. Alright, so will rearrange this equation by multiplying both sides by our so when we rearrange this equation to isolate our entropy of vaporization, what we get is it's equal to m. Times are where M is -4101. Calvin. Okay, times are is our constant 8. jewels over Mole Times, Kelvin. Okay, so as you can see here our so this is the entropy of vaporization. We're going to make sure our units cancel properly, so kelvin cancels and we're left in units of joules per mole. So let's go ahead and do this calculation. And when we do this calculation we're going to get the entropy of vaporization is 34, 99. jules per mole. Alright, so let's go ahead and write that down below. So our entropy a vaporization equals 34, 0.871 jules per mole. But the first one that we calculated is in units of kilo jewels per mole. So let's go ahead and convert this from jules per mole, tequila, joules per mole. Okay. And one kg jewel, there is 1000 jewels. Our units of jewels cancel and we're left with units of killer jewels. So when we do this we get 34.1 kg jewels per mole. Alright, so the heat of vaporization Is equal to 34.1 kg jewels per mole. So as you can see the heat, the entropy of vaporization is the same, but the values that we obtained, we use two different methods, so that is the part three of this problem, compare the values obtained using the two methods, so the values are the same. But for the first one we used the two point form of the classiest club Tehran equation, and for the second one we plotted the values in the table to determine the entropy of vaporization, so the entropy of vaporization are the same. We just used two different methods to solve that. Is it for this problem? I hope this was helpful.

Choose any two temperatures and corresponding vapor pressures in the table given in Problem 11.30, and use those values to calculate ΔHvap for dichloromethane in kJ/mol. How does the value you calculated compare to the value you read from your plot in Problem 11.32?

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