Question

Determine the AHvap in kJ/mol using Clausius-Clapeyron equation given that the boiling at 1.5 atm of a substance is 400 K, and its normal boiling point is 348 K. (Only enter numericanswer to its first decimal place, do not enter unit kJ/mol) P2 In P1 -AH/1 R T2 "G. =)

Answer 1

Given:-

Pressure when boiling occurs at 400 K, P2 = 1.5 atm

Boiling point at 1.5 atm, T2 = 400 K

Normal boiling point, T1 = 348 K

Note:- At normal boiling point pressure is 1.0 atm.

Given Clausius-Clapeyron equation:-

-ΔΗ P2 in- P1 - (α) R Τ2 T1

Here,

P2 = pressure at temperature T2(400 K), atm = 1.5 atm

P1 =  pressure at temperature T1(348 K), atm = 1.0 atm ------(explained above in note)

ΔH = enthalpy of vaporization, J

R = universal gas constant, J/(mol K) = 8.3145 J/(mol K)

T2 = temperature at pressure P2(1.5 atm), K = 400 K

T1 = temperature at pressure P1(1.0 atm), K = 348 K

On putting all the values in equation(a), we get-

1.5atm in 1.0atm -AH 1 1 8.3145J/(molK) 400K 348K usingequation(a)

In(1.5) = -ΔΗ (348Κ - 400K) 8.3145J/(molK) (400K) * (348K)

0.405 -ΔΗ (-52K) 8.3145J/(molK) (139200K2)

0.405 : ΔΗ (0.00037K-1) 8.3145.J/(molK)

0.405 = ΔΗ 22257.27J/(mol)

ΔΗ = (0.405) * - (0.405) * (22257.27).J/mol

ΔΗ = 9024.54J/mol

ΔΗ = 9.02454kJ/mol - (1J = 0.001k:J

AH = 9.0kJ/mol - (rounded off upto first decimal place)

AH = 9.0 (without units as asked in question

Hence, the enthalpy of vaporization, i.e., ΔH(vap) is 9.0 kJ/mol or 9.0(only numeric answer without units as asked in question).