Calculating the enthalpy of vapourisation (heeeeeeeeeelp meeeee)

Heya! I'm trying to calculate the enthalpy of vapourisation of hydrogen sulphide given that it boils at 212K at 760mmHg. I'm thinking that you're supposed to change mmHg into JK^ -1 mol ^ -1 and then multiply that with 212 but everywhere I search down the internet, there's always a different answer on how to convert mmHg into JK^ -1 mol ^ -1. Can anyone pleeeeeeeeeeeeaaaaaaase tell me whats the actual number for how many JK^ -1 mol ^ -1 are there per mmHg pleeeeeeeeeeeeeeeeease. Also, I've been asked what the value of the enthalpy of vapourisation of hydrogen sulphide tell me about the attractive forces between hydrogen sulphide molecules relative to those between H2O molecules given that the the enthalpy of vapourisation of H2O is 40.6kJmol-1 ..... all I could think about is that H20 has hydrogen bonds so more energy is required to overcome the intermolecular forces of attraction. Do you think that I should talk about entropy?

Reply 1

oceane

Are you not given any more information? The eqtn I know for calculating enthalpies of vaporisation is:

dlnP/ dT = delta Hvap/ RT^2

which requires knowing the boiling temp at another value of pressure.

Reply 2

Da Mouse

I'm going to try and look.... its not written anywhere on my sheet.... pleeeeeeeeeeeaaaaaaaaaaaaaaaaassssssssssse will you be here tomorrow????

Reply 3

Da Mouse

right - all I found was that that equation can be intergrated into ln p = -( delta vap H)/(RT) + c where p is the vapour pressure, R is the gas constant 8.314Jmol^-1K^-1, T is the temperature in K and c is a constant, which depends on the liquid under study..... I'm sure you already knew that already...... so now I'm screwed :bawling:

Reply 4

oceane

Yes, when you're integrating you need to integrate between 2 limits so you would need at least another vapour pressure and temperature to solve:

P1 DHvap 1 1
ln (---) = ---- (--- - ---)
P2 R T2 T1

Unless there's another eqtn other than this Clausius-Clapeyron one.

I wldn't worry about it, maybe the question's just incomplete.

okey dokey - cheers!

I don't know any online chem databases but here, from a google search:

Vapour pressure of H2S at 293K is 18.75 * 10^(5) Pascals.

from:
http://www.inchem.org/documents/ehc/ehc/ehc019.htm

Reply 7

charco

Study Forum Helper

Enthalpy of vaporisation can be calculated from Gibbs free energy change: delta G = delta H - Tdelta S

at the boiling point deltaG = 0
therefore deltaH = TdeltaS

if you know the absolute entropy of H2S gas and liquid then you can calculate delta H (the enthalpy of vaporisation) from the boiling point.

Reply 8

ChemistBoy

charco

if you know the absolute entropy of H2S gas and liquid then you can calculate delta H (the enthalpy of vaporisation) from the boiling point.

You should be able to find it tabulated somewhere. Unfortunately it isn't a liquid at STP so you can't use trouton's rule.

got it! thanx guys!

j/km cannot be converted into mmhg.because mmhg is pressure unit and j/km is unit of enthalpy.so if u want to calculate heat of vaporization at a given temp,u should understand following example
for example u want to calculate heat of vaporization of any substance at normal boilig point?
now first of all u should know clausious equation which is given
d(lnP)/dT=deltaH/RT^2
where
lnP=vapour pressure as a function of temperature
deltaH=heat of vaporization
R=ideal gas constant
T=temperature

to calculate lnP we use
lnP=lnA(boiling point at normal pressure)-deltaH(at normal temperature)/R*T(t is normal temperature)
after putting values in above equation we get lnP
then we differentiate lnP and we get
d(lnP)/dT
then we calclate deltaH from given clausius equation at a given temperature. :smile: