Asked by Jamie

At 57.7°C, the vapor pressure of CCl4 is 56.1 kPa, and its enthalpy of vaporization is 29.82 kJ/mol. Use this information to estimate the normal boiling point (in °C) for CCl4.

Answered by DrBob222
Isn't that the Arrhenius equation? You use p1 = vp given and use P2 as 101.325 kPa. Then T1 is kelvin at the vp given and T2 will be normal boiling point.
Answered by DrBob222
Sorry, not Arrhenius. Clausius-Clapeyron.
Answered by Jamie
yeah, I did that.

I've got

P1=65.1
P2=56.1
T1=330.85 K
T2=?
H=29,830 j/mol

Plugging this in, I got 40.6. But my HW system isn't accepting it.
Answered by DrBob222
Where did you get 65.1? That should be 1 atm.
Convert dH to joules. That is 29820 and not 29830
Convert kPa to atm. That is 56.1/101.325
Then T1 will come out in kelvin. Convert to C. Let me know how things turn out. Show your work and I'll find the error. BYW. R is 0.08205
Answered by DrBob222
Well, I goofed again, R is 8.314.
Answered by bobpursley
shouldn't pressures be in Pa?
Answered by DrBob222
Jamie, I just worked the problem and I got 40.65 C which I would round to 40.6. The problem with data bases sometimes is that we use the wrong number of significant figures OR the database is wrong. Hope this helps. Of course you don't need to convert kPa to anything. I usually use mm or atm but kPa works just as well.
Answered by Jamie
Okay, I wasn't converting kPa to atm. That was the problem. I got 76.8 C, and that was the right answer.

Thank you for your help!
Answered by DrBob222
Does it matter what units we use for P1 and P2 as long as they are the same? That is ln (mm/mm) or ln(kPa/kPa) or ln(Pa/Pa) should be the same I think.
Answered by Jamie
Huh... that's weird then.

I ended plugging this in:

ln(56.1/101.325)=(29820/8.314)((1/T2)-(1/380.85))

And solved

-0.59119=(29820/8.314)((1/T2)-(1/380.85)) to get 349.93. Then I subtracted 273.15 to get 76.78 C.

Answered by DrBob222
OK. I found my problem and I expect it's yours also for 40 C answer.. I reversed the 1/T2-1/T1 term. I didn't go through it but I'm sure that's the problem.
Answered by Jamie
Ah, Okay. Thank you so much for your help. I'll be sure to pay more attention to that in the future.
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