Question

The enthalpy of vaporization of Substance X is 28.0kJmol and its normal boiling point is 144.°C. Calculate the vapor pressure of X at 102.°C

Round your answer to 2 significant digits.

Answer 1

Answer

0.405 atm

Explanation

According to classius clapeyron equation is

ln(P₁/P₂) = ∆H_vap/R ( 1/T₂ - 1/T₁)

where,

P₁ = vapor pressure 1 , 1.00 atm ( a substance boils when vapor pressure of substance reaches 1atm)

P₂ = vapor pressure 2, ?

T₁ = Temperature 1, 144℃ = 417.15K

T₂ = Temperature 2, 102℃ = 375.15K

∆H_vap = 28.0kJ/mol

R = gas constant, 0.008314 kJ/mol K

substituting the values

ln(1.00atm/P₂) =( 28.0kJ/mol / 0.008314kJ/mol K )(1/375.15K - 1/417.15K)

2.303log(1.00atm/P₂) = ( 28.0kJ/mol/0.008314kJ/mol K)(1/375.15K - 1/417.15K)

2.303log(1.00atm/P₂) = 3367.81K( 0.0002684K^-1)

log(1.00atm/P₂) = 0.3925

1atm/P₂ = 2.469

P₂ = 0.405 atm