Question

A certain liquid has a vapor pressure of 92.0 Torr at 23.0 °C and 313.0 Torr at 45.0 °C. Calculate the value of ΔH°vap for this liquid

Answer 1

Calculate the normal boiling point of this liquid.

Answer 2

Do you know the formula?

Answer 3

yea, its the Clausius-Clapeyron equation

Answer 4

i know how to get the first part, its the temperature i dont know how to get

Answer 5

Well show me what you know and we'll go from there.

Answer 6

ln(p2)-ln(p1)+(Hvap/R)(1/T1-1/T2)

Answer 7

rearrange for Hvap

Answer 8

ln(p2/p1)-(1

Answer 9

ugh

Answer 10

ln(p2/p1)-(1/T1-1/T2)/R=Hvap

Answer 11

I think thats wrong

Answer 12

I get 43.55 as H vap

Answer 13

i re- arranged it wrong

Answer 14

Here:
logP2/P1= delta H/(2.303*R) * (1/T1-1/T2)

P1=92 torr P2=372, T1=296k, T2=318

R=1.99 cal/mole-k

log372/92=0.6067

2.303*R=2.303*1.99=4.5829cal/mole-k ...

1/T1-1/T2=0.0002337K

deltaH = 0.6067x4.5829/0.0002337 =11897 cal

11897/239kj = 49.78kj

I believe my math is right.This is the first half of the problem

Answer 15

ok

Answer 16

Do you understand so far?

Answer 17

yep

Answer 18

Do you know what to do next?

Answer 19

um nope. thats the part i got stuck on

Answer 20

log760/92=0.9170 , P1=92 torr

2.303*R=2.303*1.99=4.5829cal/mole-k ...

1/T1-1/T2= (1/296-1/T2) , T1=273+23=296k

1/T1-1/T2=( 0.003378 - 1/T2)

Heat of vaporisation = 47.9kj

47.9*239cal = 11448cal

0.9170*4.5829/11448=0.003378-1/T2

0.0003671=0.003378-1/T2

1/T2=0.003378-0.0003871=0.0030109 ...

T2=332K or 59 degc

I hope you understand!

Answer 21

it says its wrong

Answer 22

What are your anwser choices, if you have any?

Answer 23

i dont have any

Answer 24

ry my H value

Answer 25

try*

Answer 26

could u try it with my H value...i try it myself and i get a very big number

Answer 27

could u help with another question

Answer 28

Sorry I had my own school work to take care of.