Naphthalene, c10h8, melts at 80.2â°c. if the vapour pressure of the liquid is 1.3 kpa at 85.8â°c and 5.3 kpa at 119.3â°c, use the clausiusâ clapeyron equation to calculate

(a) the enthalpy of vaporization,

(b) the normal boiling point,

(c) the enthalpy of vaporization at the boiling point

(a) One form of the Clausius-Clapeyron equation is

ln(P₂/P₁) = (ΔHv/R) * (1/T₁ - 1/T₂); where in this case:

Solving for ΔHv:

(b) Normal boiling point means that P = 1 atm = 101.325 kPa. We use the same formula, using the same values for P₁ and T₁, and replacing P₂ with atmosferic pressure, solving for T₂:

(c) The enthalpy of vaporization was calculated in part (a), and it does not vary depending on temperature, meaning that at the boiling point the enthalpy of vaporization ΔHv is still 49111.12 J/molK.

hope this

explanation:

covalent bonds

consist of pairs of electrons shared by two atoms, and bind the atoms to fixed orientation and requre high energies to break them

ionic bonds

when two atoms differ considerably in their electronegativity, then one of the atoms will lose its electron to the other atom. this results in a positively charged ion(cation) and a negatively charged ion (anion)

for points guys