Answer:
[tex]\mathbf{\Delta H =97.815 \ kJ/mole}[/tex]
Explanation:
Given that:
The molecular formula = [tex]C_7H_{12}N[/tex]
The normal boiling point = 200°C
The enthalpy of the vapor = 200°C
The liquid temperature = 25°C.
By applying Kopp's rule:
[tex]C_p = 6 \times (0.012) + 12 (0.018)+1(0.033)[/tex]
[tex]C_p = 0.071+ 0.216+0.033[/tex]
[tex]C_p[/tex] = 0.32 kJ/mole
For Trouton's rule;
ΔHV = 0.088 Tb(t) for non polar liquid
ΔHV = 0.088(200)°C
ΔHV = 0.088(200+ 273.15)
ΔHV = 41.637 kJ/mol
ΔHV ≅ 41.637 kJ/mol
∴
[tex]\Delta H = \int \limits^ {250}_{25} C_p dT + \Delta HV (200^0C)[/tex]
Replacing the values from above; we have:
[tex]\Delta H = 0.321(200-25) \ kJ/mole+ 41.64 \ kJ/mole[/tex]
[tex]\Delta H = 0.321(175) \ kJ/mole+ 41.64 \ kJ/mole[/tex]
[tex]\Delta H =56.175 \ kJ/mole+ 41.64 \ kJ/mole[/tex]
[tex]\mathbf{\Delta H =97.815 \ kJ/mole}[/tex]
