The enthalpy change for the conversion of fra | vedclass

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(D) The process involves three steps:$1. 	ext{Dissociation of } \frac{1}{2} Cl_{2(g)}$ $ightarrow Cl_{(g)}: \Delta H_1 = \frac{1}{2} 	imes 240 = 1

Vedclass · Accepted Answer

$610$

Vedclass · Answer

(D) The process involves three steps:$1. 	ext{Dissociation of } \frac{1}{2} Cl_{2(g)}$ $ightarrow Cl_{(g)}: \Delta H_1 = \frac{1}{2} 	imes 240 = 120 \, kJ \, mol^{-1}$$2. 	ext{Electron gain of } Cl_{(g)} ightarrow Cl^{-}_{(g)}: \Delta H_2 = -350 \, kJ \, mol^{-1}$$3. 	ext{Hydration of } Cl^{-}_{(g)} ightarrow Cl^{-}_{(aq)}: \Delta H_3 = -380 \, kJ \, mol^{-1}$Total enthalpy change $\Delta H^{\circ} = \Delta H_1 + \Delta H_2 + \Delta H_3$$\Delta H^{\circ} = 120 + (-350) + (-380) = -610 \, kJ \, mol^{-1}$The magnitude is $610$.

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