Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: noiseordinance on May 23, 2009, 02:01:32 PM
-
Good morning (or afternoon depending on where you are). We did an experiment a couple days ago that involved heating ethanol to determine its heat of vaporization. We collected data for both temperature (in Celsius) as well as pressure (in kPa). We monitored between 85 degrees C to 35 degrees C. We ended up with a couple values that I'm having a hard time stringing together. From the CRC, I know that the Heat of Vaporization for Ethanol is:
38.56 kJ/mol @ 78.3 degrees C
42.32 kJ/mol @ 25.0 degrees C
As for our data from our first trial, we go:
m (slope) : -4044
b (y-intercept) : 16.13
Correlation : -0.9986
RMSE : 0.02940
I realize that Correlation and RMSE are both values for accuracy that just simply need to be reported. However, I'm having a hard time fitting m (slope) and b (y-intercept) into the Clausius-Clapeyron equation:
ln P = - (:delta:Hvap/RT) + C
I don't know which values convert to which. It looks like "C" in the Clausius-Clapeyron equation is related to b, the y-intercept. I'd also imagine that R is the gas constant, 8.314 J mol−1K−1. What I'm not sure is where I'm supposed to fit in the slope. I am also guessing that since our data was originally collected in C, we'll need to convert to K since we're using the gas constant, but I don't know which data to convert to K. Argh.
Anyways, from this data, I need to know how to report relationships between vapor pressure and temp, the significance of the slope, and I'm also supposed to be able to calculate the heat of vaporization for ethanol and the boiling point.
Could anyone lend some pointers?
-
I guess to ask more specific a question, do I have enough data to determine the boiling point and heat of vaporization or am I mission something?
From what it sounds like, my slope and intercept alone are supposed to be enough to determine both values. Afterwards, I'm supposed to compare my boiling point and heat of vaporization to the CRC values.
So I'm attempting to start with the empty equation:
ln P = - :delta: Hvap / RT + C
Then I fill in what I know how to fill in...
ln P = - :delta: Hvap / (8.314 J mol−1K−1) T + 16.13
This leaves me with ln P, :delta: Hvap and temperature. Problem is, I only have my slope value, which I haven't filled in cause I don't know where it's supposed to go, or even what it means. Perhaps it's my :delta: Hvap value in joules? If this is the case, I could plug in T in Kelvin and end up with pressure, and plug in pressure and get T... but how does that tell me the heat of vaporization? Or am I all messed up?
-
I'm going to guess that this thread totally doesn't make sense... I could use any pointers anyone may have to offer.
-
ln P = - :delta: Hvap / RT + C
ln P = - ( :delta: Hvap / R) T-1 + C = m T-1 + b
However, you don't know what you are doing, yet you have fitted something to something. What to what?
-
ln P = - ( :delta: Hvap / RT ) + C is the equation that my teacher gave us to use. Are you saying that this is not a valid equation?
-
This is the same equation, just written in different form.
-
I realize that. I wanted to make sure that the original equation that I provided was written in the exact way it was specified by the instructor. Yes, this is the same equation as I originally wrote. That said, is it a valid equation?
-
http://en.wikipedia.org/wiki/Clausius-Clapeyron
-
Thanks. Unfortunately I've looked at all the Clausius-Clapeyron equations I could find on the internet, as well as the material in my text book. Unfortunately, none that I could find involved integrating the experimental slopes and intercepts to determine heats of vaporization and boiling points...
-
Wiki page lists the C-C equation in the form identical to the one you were given.
Find out what m&b stand for just by comparing coefficients side by side in the equation I have listed earlier.
-
Gosh, I still really don't know what you mean. I'd like to think I'm not 100% dense though. Is there a more plain english way that this can be explained?
-
ln P = - ( :delta: Hvap / R) T-1 + C = m T-1 + b
Can you see what m & b must be equal to for the equation to be correct?
I suppose you are making it harder than it is in reality. It is pretty obvious.
-
I don't know what that means... Considering I was top of the class for the majority of my math classes, I'm not feeling too dumb about this. It really isn't that obvious to me. So ln P = m T-1 + b, then ln P = -4044 T-1 + 16.13... that still leaves two unknown variables, which in algebra, can be difficult to figure out. I really don't know. You're right, I'm making this too hard, and with the holiday weekend, there hasn't been any tutor center, and the instructor has not emailed me back. After fighting this problem for 3 days, I'm really starting to get discouraged.
-
As far as I can guess P & T are not unknowns.
From the equations you wrote so far C=16.13... can you tell what :delta: Hvap/R is equal to?
Trick is, you use m&b values from SOME fitting - but it is not clear what you have fitted to what. Seems to me you did it not having a slightest idea about the reason behind the operation.
-
I sorta thought that the point of this experiment is to make a curve of pressure as a function of temperature. So the temperature would be the x-axis and the kPa would be the y-axis, and since the y-intercept is 16.13, that would mean maybe that at 0 degrees C, the pressure is 16.13 kPa? I dunno, I could have this all wrong.
So maybe I'm totally wrong, and I'll probably be totally wrong asking the next question... Is :delta: Hvap the slope value, -4044? Ie, can I substitute :delta: Hvap with -4044?
I appreciate your help.
-
I sorta thought that the point of this experiment is to make a curve of pressure as a function of temperature
only to
to determine its heat of vaporization
your own words :)
Is :delta: Hvap the slope value, -4044? Ie, can I substitute :delta: Hvap with -4044?
Twice wrong, but you are very close. Just look at the equation again.
-
Haha... umm, crap. Ok, we'll what about -4044 / 8.314 as the slope... ie, -486.4?
-
That would be the correct approach - that is, assuming you fitted data correctly.
-
So could you verify if I'm on the right track still? Using the following equation:
ln P = - ( delta Hvap / R) T + C
Then I can fill in what I know...
ln P = - (-4044 / 8.314) T + 16.13
ln P = 486.4 T + 16.13
Would this basically be the equation used for solving for pressure or temperature?
-
Not T!
-
Ok, so let me ask this then. If the boiling point for fluids is 760 mmHg, then I could plug in 760 mmHg for ln P in this equation and get the boiling point? (I'd check on my calculator but I haven't one since I'm sneaking on at work)... that sound right?
-
That's how it is supposed to work (assuming you do it right).
-
Gotcha. Thanks. You're right, I was making that hard. I just really didn't understand what value my slope value was supposed to replace, if that makes sense. Thanks for your help and patience!