lilsis76
  • lilsis76
this is a preexam for this tuesday. Need help solving and understand please. Benzene has a heat of vaporization of 30.72 kJ/mol, and a normal boiling point of 80.1 deg. Celsius. At what temperature does benzene boil when the external pressure is 745 torr? well since I have a pressure, and i have what seems to be 2 temperatures, 1 i gotta find. I have to use the ln (p2/p1)= - Delta H vap / R (1/t2 - 1/t1) right? @aaronq Im back I think I fixed it
Chemistry
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SOLVED
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katieb
  • katieb
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aaronq
  • aaronq
okay, so this: \(ln(\dfrac{P_2}{P_1})=\dfrac{-\Delta H_{vap}}{R}(\dfrac{1}{T_2}-\dfrac{1}{T_1})\) is called the "clausius-clapeyron equation", although it's name is not really important. you basically need to plug in the values given, and solve. Just keep track of what P2 and T2 are and P1 and T1 are. you have: heat of vaporization of 30.72 kJ/mol T= 80.1 deg. P=normal conditions (1 atm) they ask at what T? will it boil if the P= 745 torr can you try to set it up?
lilsis76
  • lilsis76
okay. well this is what I got, from information given.
lilsis76
  • lilsis76
\[\Delta H=30.72 \frac{ kj }{ mol }\] \[T_{1} 80.1 C +273=353.1 Kelvin\] \[P_{1} = 1atm_ normal boilig point = 760 torr\] \[P_{2} = 745 torr\]

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aaronq
  • aaronq
perfect
aaronq
  • aaronq
now all you do it plug it in and solve for T2
aaronq
  • aaronq
you don't have to write it here, i'll do it also and we'll compare answers, hopefully we have the same one.
aaronq
  • aaronq
don't forget to convert dHvap to match the units of R
lilsis76
  • lilsis76
YAY! im back on haha sorry. im back
lilsis76
  • lilsis76
dHvap? how hold on let me just try and ill show u what i have :)
lilsis76
  • lilsis76
okay, imma put what i have @aaronq
lilsis76
  • lilsis76
\[\ln \frac{ 760 }{ 745 } = \frac{ -30.72 \frac{ kJ }{ mol } }{ .082 \frac{ L * atm }{ mol * K } } (\frac{ 1 }{ T } - \frac{ 1 }{ 353 K })\]
lilsis76
  • lilsis76
@aaronq
aaronq
  • aaronq
you put the 353 in the wrong place
lilsis76
  • lilsis76
|dw:1378683096460:dw|
lilsis76
  • lilsis76
|dw:1378683170734:dw|
lilsis76
  • lilsis76
1/353 = .00283
lilsis76
  • lilsis76
Am i close? @aaronq
aaronq
  • aaronq
better now, whats your final answer? don't forget units!
lilsis76
  • lilsis76
1/T = .002776791 so now I have to multiply by 2, so i get ... .0055535 K
aaronq
  • aaronq
wait no sorry i didn't see your units for R, you're using 8.314
lilsis76
  • lilsis76
no...I think i did....oh shoot
aaronq
  • aaronq
not 0.8206
lilsis76
  • lilsis76
let me try again. im going to use.. .082
aaronq
  • aaronq
no use 8.314 J, convert dHvap to J because it's in kJ
lilsis76
  • lilsis76
okay hold on
lilsis76
  • lilsis76
|dw:1378683680317:dw| right?
lilsis76
  • lilsis76
@aaronq I dont remember the rest of the units for R.
lilsis76
  • lilsis76
|dw:1378683898602:dw|
aaronq
  • aaronq
R =J/K*mol but thats unimportant, whats your final answer?
lilsis76
  • lilsis76
.01993/-369.49= -.00005394 -53.94x10^6 = .00283K - 1/T
aaronq
  • aaronq
soo, which is your final answer?
lilsis76
  • lilsis76
haha hold on im writing my work . okay so i divide both sides by .00283 = -19.06 x 10^-3 = - 1/T
lilsis76
  • lilsis76
Final answer is 38.12e-6
lilsis76
  • lilsis76
yes no??? :/ @aaronq
aaronq
  • aaronq
that can't be right, 0.00003812 K ..so it would boil almost at absolute zero? :P
lilsis76
  • lilsis76
haha what am I dont wrong? or am i calculating it wrong
aaronq
  • aaronq
you must've messed up somewhere in the process of plugging into the calculator
aaronq
  • aaronq
because i got a different answer
lilsis76
  • lilsis76
okay ill do it step by step of what i did u tell me where i messed up oka? :)
lilsis76
  • lilsis76
ln(760/745) = .01993 right?
lilsis76
  • lilsis76
-30.72e-6/8.314 = 36.95e-7
aaronq
  • aaronq
-30.72e-6 this is wrong
lilsis76
  • lilsis76
sorry. -36.95 e -7
aaronq
  • aaronq
-30.72 kJ = -30720 J
lilsis76
  • lilsis76
1/353 = .00283 K
lilsis76
  • lilsis76
and theres that - 1/T
lilsis76
  • lilsis76
so i get this>>>>|dw:1378685045886:dw| okay, now. is this right?
aaronq
  • aaronq
no, this:-30.72e-6/8.314 = 36.95e-7 is wrong because this: 30.72e-6 is wrong how did you convert dHvap kJ to J?
lilsis76
  • lilsis76
I multiplied by 1000
lilsis76
  • lilsis76
how would I do it?? if i didnt do it right?
aaronq
  • aaronq
you wrote: 30.72e-6 = 0.00003072 but, it really should be: 30.72 kJ = 30720 J
lilsis76
  • lilsis76
okay so then 30.72e6
lilsis76
  • lilsis76
.01993=
lilsis76
  • lilsis76
I dont put a negagive becuze of the -delta H?
aaronq
  • aaronq
no 30.72e6=30720000.0 way too many zeroes
lilsis76
  • lilsis76
okay then
aaronq
  • aaronq
you gotta work on your conversions ;P 30.72 kJ = 30720 J
lilsis76
  • lilsis76
|dw:1378685552337:dw| @aaronq okay, now is this right?? before i go further :/
aaronq
  • aaronq
that is correct !
lilsis76
  • lilsis76
okay now, do i...
lilsis76
  • lilsis76
|dw:1378685955801:dw| @aaronq okay this messes me up. what do i do first?
lilsis76
  • lilsis76
|dw:1378686212208:dw|
lilsis76
  • lilsis76
When I get to these parts I really dont know what to do. Could you help me
aaronq
  • aaronq
that was good so far, so: -0.0000053938157552083 -0.00283 =- \(\dfrac{1}{T}\) -0.0028353938157552083 =- \(\dfrac{1}{T}\) multiply both sides by -1 (to get rid of negative signs) 0.0028353938157552083 = \(\dfrac{1}{T}\) multiply both sides by T 0.0028353938157552083*T = \(\dfrac{1}{T}T\) divide both sides by 0.0028353938157552083 T = \(\dfrac{1}{0.0028353}\)= 352.4125348596419165 K
lilsis76
  • lilsis76
YAY! IM BACK! i hate this connection thingy. does it do it to you??
aaronq
  • aaronq
very rarely
lilsis76
  • lilsis76
okay, and i got 356.456 K
lilsis76
  • lilsis76
I rounded it to about the 2nd or third decimal place
lilsis76
  • lilsis76
would my answer be about right too?
lilsis76
  • lilsis76
@aaronq
lilsis76
  • lilsis76
i mean as long as its in the 300's Kelvin right?
aaronq
  • aaronq
i would leave it as 356.5 K because you have to use the least number of sig. figs given in the question (which was 80.1 deg. Celsius)
aaronq
  • aaronq
sorry 4 sig figs 357 K should be the answer
lilsis76
  • lilsis76
okay. 356.5 K it is then
aaronq
  • aaronq
3 sig figs, dang it
aaronq
  • aaronq
sorry i keep mistyping things
lilsis76
  • lilsis76
haha. Thank you for taking the time to explain how to do the steps. I feel like I know it, but when I try it Its completly wrong more than half the time.
lilsis76
  • lilsis76
Im closing this question. thank you @aaronq
aaronq
  • aaronq
haha no worries. all you need is practice. no problem !
lilsis76
  • lilsis76
thank you,

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