The electron gain enthalpy Delta_{eg}H of Cl_ | vedclass

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(C) The process of electron gain is represented as: $Cl_{(g)} + e^{-} \xrightarrow{\Delta_{eg}H} Cl^{-}_{(g)}$Energy of the product $Cl^{-}_{(g)}$ is

Vedclass · Accepted Answer

$x-349$

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(C) The process of electron gain is represented as: $Cl_{(g)} + e^{-} \xrightarrow{\Delta_{eg}H} Cl^{-}_{(g)}$Energy of the product $Cl^{-}_{(g)}$ is equal to the sum of the energy of the reactant $Cl_{(g)}$ and the enthalpy change $\Delta_{eg}H$ of the reaction.$	ext{Energy of } Cl^{-}_{(g)} = 	ext{Energy of } Cl_{(g)} + \Delta_{eg}H$Given that the energy of $Cl_{(g)}$ is $x \ kJ \ mol^{-1}$ and $\Delta_{eg}H = -349 \ kJ \ mol^{-1}$.$	ext{Energy of } Cl^{-}_{(g)} = x + (-349) = x - 349 \ kJ \ mol^{-1}$.

The electron gain enthalpy $\Delta_{eg}H$ of $Cl_{(g)}$ is $-349 \ kJ \ mol^{-1}$. If the ground state energy of $Cl_{(g)}$ is $x \ kJ \ mol^{-1}$,the ground state energy (in $kJ \ mol^{-1}$) of $Cl^{-}_{(g)}$ is

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