The standard molar enthalpy of vaporisation ( | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The enthalpy of vaporisation is the amount of heat required to transform a liquid into a gas. Dipole-dipole interactions are intermolecular forces

Vedclass · Accepted Answer

$B > C > A$

Vedclass · Answer

(A) The enthalpy of vaporisation is the amount of heat required to transform a liquid into a gas. Dipole-dipole interactions are intermolecular forces of attraction between polar molecules. Stronger dipole-dipole interactions require higher heat energy to overcome these forces and vaporise the liquid. Given the molar enthalpies of vaporisation for liquids $A, B$ and $C$ are $23.3, 41$ and $29 \ kJ \ mol^{-1}$ respectively,the magnitude of intermolecular forces follows the order of their enthalpies. Therefore,the order of dipole-dipole attractive forces is $B > C > A$.

The standard molar enthalpy of vaporisation $(\Delta_{vap}H^{\circ})$ of $A, B$ and $C$ liquids is $23.3, 41$ and $29 \ kJ \ mol^{-1}$ respectively. The correct order of dipole-dipole attractive forces in these liquids is

Similar Questions

Why is the water molecule polar?

Explain why $BeH_{2}$ molecule has a zero dipole moment although the $Be-H$ bonds are polar.

If the dipole moment of chlorobenzene is $\mu$,what is the dipole moment of $1,3,5$-trichlorobenzene?

Carbon tetrachloride has no net dipole moment because of

Which pair of molecules are polar species?

Vedclass Test Series

Exam Paper Generator

Online Exam Module