Question

You are writing energy balances for a compound for which you cannot find heat capacity or latent heat data. All you know about the material are its molecular formula (C7H12N) and that it is a liquid at room temperature and has a normal boiling point of 200°C. Use this information to estimate the enthalpy of the vapor of this substance at 200°C relative to the liquid at 25°C.

Answer 1

`\mathbf{\Delta H =97.815 \ kJ/mole}`

Step-by-step explanation:

Given that:

The molecular formula =
`C_7H_(12)N`

The normal boiling point = 200°C

The enthalpy of the vapor = 200°C

The liquid temperature = 25°C.

By applying Kopp's rule:

`C_p = 6 * (0.012) + 12 (0.018)+1(0.033)`

`C_p = 0.071+ 0.216+0.033`

`C_p` = 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

∴

`\Delta H = \int \limits^ {250}_(25) C_p dT + \Delta HV (200^0C)`

Replacing the values from above; we have:

`\Delta H = 0.321(200-25) \ kJ/mole+ 41.64 \ kJ/mole`

`\Delta H = 0.321(175) \ kJ/mole+ 41.64 \ kJ/mole`

`\Delta H =56.175 \ kJ/mole+ 41.64 \ kJ/mole`

`\mathbf{\Delta H =97.815 \ kJ/mole}`