The value of Delta H_{O-H} is 109 kcal mol^ | vedclass

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(A) The reaction for the formation of one mole of $H_2O_{(g)}$ from its gaseous atoms is: $2H_{(g)} + O_{(g)} \longrightarrow H_2O_{(g)}$ In this proc

Vedclass · Accepted Answer

Release of $218 \ kcal$ of energy

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(A) The reaction for the formation of one mole of $H_2O_{(g)}$ from its gaseous atoms is: $2H_{(g)} + O_{(g)} \longrightarrow H_2O_{(g)}$ In this process,two $O-H$ bonds are formed. The energy released during the formation of one mole of bonds is equal to the bond dissociation energy. Since the bond dissociation energy of one $O-H$ bond is $109 \ kcal \ mol^{-1}$,the energy released for the formation of two $O-H$ bonds is: $\Delta H = -2 	imes 109 \ kcal \ mol^{-1} = -218 \ kcal \ mol^{-1}$. The negative sign indicates the release of energy. Therefore,the formation is accompanied by the release of $218 \ kcal$ of energy.

The value of $\Delta H_{O-H}$ is $109 \ kcal \ mol^{-1}$. Then,the formation of one mole of water in the gaseous state from $H_{(g)}$ and $O_{(g)}$ atoms is accompanied by:

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