The standard molar enthalpy of vaporisation ( | vedclass

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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

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