Question

Question: How does the bond energy of $\mathrm{H}\mathrm{C}\mathrm{l}(\mathrm{g})$ differ from the standard enthalpy of formation of $\mathrm{H}\mathrm{C}\mathrm{l}(\mathrm{g})$?

Short answer

The bond energy differs from standard enthalpy for $\mathrm{H}\mathrm{C}\mathrm{l}$ as bond energy and enthalpy breaks $\mathrm{H}\mathrm{C}\mathrm{l}$ into $\mathrm{H}$ and $\mathrm{C}\mathrm{l}$ atoms and, ${\mathrm{H}}_2$ and $\mathrm{C}\mathrm{l}$, molecules respectively.

Step-by-step solution

Concept Introduction

Bond energy gives the measure of the strength of bond in a chemical reaction.

Difference in bond energy and standard enthalpy of Hydrochloric Acid

The enthalpy change is when a mole of a substance is created from its elements in their standard states under standard conditions is called enthalpy of formation.

$\mathrm{H}\mathrm{C}\mathrm{l}(\mathrm{g})\to {\mathrm{H}}_2(\mathrm{g})+\mathrm{C}\mathrm{l}(\mathrm{g})$

Bond enthalpy is quite different and is a measure of the strength of an individual bond.

$\mathrm{H}\mathrm{C}\mathrm{l}(\mathrm{g})\to \mathrm{H}(\mathrm{g})+\mathrm{C}\mathrm{l}(\mathrm{g})$

Bond energy involves breaking of $\mathrm{H}\mathrm{C}\mathrm{l}$ molecule into $\mathrm{H}$ and $\mathrm{C}\mathrm{l}$ atoms whereas enthalpy of formation involves making of $\mathrm{H}\mathrm{C}\mathrm{l}$ from ${\mathrm{H}}_2$ and $\mathrm{C}\mathrm{l}$, molecules.

Therefore, bond energy of $\mathrm{H}\mathrm{C}\mathrm{l}$ differ from the standard enthalpy of formation of $\mathrm{H}\mathrm{C}\mathrm{l}$.