Question

A certain liquid has a vapor pressure of 92.0 Torr at 23.0 °C and 323.0 Torr at 45.0 °C.

a.) Calculate the value of ΔH°vap for this liquid.

b.) Calculate the normal boiling point of this liquid.

Answer 1

For solving this, you will be needing clausius-clapyron eqaution:

the equation is:

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

in our case,
P1=92 torr

P2=323 torr

T1=(23+273 )= 296 K

T2=(45+273)= 318 K

deltaH = heat of vaporisation in cal/mole

T2 and T1 are temp in kelvin

R=1.99 cal/mole-k

now, plugging in the values,

log (323/92) =0.5454

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

1/T1-1/T2=0.0002337K

Therefore,

deltaH = 0.5454*4.5829/0.0002337 =10695.4 cal or 10695.4/239 kJ = 44.75 KJ

b)

suppose normal boiling point is T2

Corresponding vap pressure P2 would be atm pr =760 torr

log(760/92)=0.9170 , P1=92 torr

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

now,
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 given=47.9*239cal = 11448cal 1 kJ = 239.0057


Therefore, 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 degree celcius