Question

A certain liquid has a vapor pressure of 92.0 Torr at 23.0 °C and 351.0 Torr at 45.0 °C. Calculate the value of ΔH°vap for this liquid.

Answer 1

there is a specific formula to use for these type of problems.

ln (P2/ P1)= Δvap/ R x (1/T1 - 1/T2)

R= 8.314
P1= 92.0 torr
T1= 23 C + 273= 296 K
P2= 351.0 torr
T2= 45.0 C + 273= 318 K

plug the values and solve for the unknown

ln( 351.0/ 92.0)= Δvap/ 8.314 x (1/296 - 1/318)

Δvap= 47630.6 joules

Answer 2

Answer : The value of
`\Delta H_(vap)` is 47.627 kJ/mol

Explanation :

To calculate
`\Delta H_(vap)` of the reaction, we use clausius claypron equation, which is:

`\ln((P_2)/(P_1))=(\Delta H_(vap))/(R)[(1)/(T_1)-(1)/(T_2)]`

where,

`P_1` = vapor pressure at temperature
`23^oC` = 92 torr

`P_2` = vapor pressure at temperature
`45^oC` = 351 torr

`\Delta H_(vap)` = Enthalpy of vaporization = ?

R = Gas constant = 8.314 J/mol K

`T_1` = initial temperature =
`23^oC=[23+2730]K=296K`

`T_2` = final temperature =
`45^oC=[45+2730]K=318K`

Putting values in above equation, we get:

`\ln((351torr)/(92torr))=(\Delta H_(vap))/(8.314J/mol.K)[(1)/(296)-(1)/(318)]\\\\\Delta H_(vap)=47627.347J/mol=47.627kJ/mol`

Therefore, the value of
`\Delta H_(vap)` is 47.627 kJ/mol