The change in enthalpy [Delta H] in kJ mol^{- | vedclass

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Question

(D) The enthalpy change for the reaction $Mg + 2F \rightarrow MgF_2$ can be calculated by summing the energy changes involved in the ionization of $Mg

Vedclass · Accepted Answer

$1532$

Vedclass · Answer

(D) The enthalpy change for the reaction $Mg + 2F ightarrow MgF_2$ can be calculated by summing the energy changes involved in the ionization of $Mg$ and the electron gain of $F$.$Mg_{(g)} ightarrow Mg_{(g)}^{+2} + 2e^{-}$, $\Delta H_1 = IE_1 + IE_2 = 737 + 1451 = 2188\ kJ\ mol^{-1}$.$2F_{(g)} + 2e^{-} ightarrow 2F_{(g)}^{-}$, $\Delta H_2 = 2 	imes (-EA) = 2 	imes (-328) = -656\ kJ\ mol^{-1}$.Adding these steps, the total enthalpy change is $\Delta H = \Delta H_1 + \Delta H_2 = 2188 - 656 = 1532\ kJ\ mol^{-1}$.

The change in enthalpy $[\Delta H]$ in $kJ\ mol^{-1}$ for the reaction is $Mg + 2F \rightarrow MgF_2$. Given: $EA$ of $F = 328\ kJ\ mol^{-1}$, $IE_1$ of $Mg = 737\ kJ\ mol^{-1}$, $IE_2$ of $Mg = 1451\ kJ\ mol^{-1}$.

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