The enthalpy change for the conversion of (1 / 2) Cl_2( g) to Cl^-(aq) is (-) ___________ kJ mol^-1 (Nearest integer) Given : Delta_ dis H_ Cl_2( g)^theta^ominus=240 kJ mol^-1, Delta_ eg H_ Cl_(g)^ominus=-350 kJ mol^-1, Delta _hydH_Cl_(g)^ - ^Theta = - 380 kJ mol^-1

(1 / 2) Cl_2( g) -> Cl_ (aq) ^- Delta H=? \ Delta H =(1 / 2) Delta_ diss H_ Cl_2^°+Delta_ eg Delta H_ Cl( g)^°+Delta_ hyd H_ Cl_( g)^-^° \ =(1 / 2) x 240+(-350)+(-380) \ =-610 kJ mol^-1

The enthalpy change for the conversion of frac 12 Cl2g ... | (Chemistry)

The enthalpy change for the conversion of $\frac{1}{2} \mathrm{Cl}_{2}(\mathrm{~g})$ to $\mathrm{Cl}^{-}$ (aq) is ( $-$ ) ___________ $\mathrm{kJ} \mathrm{mol}^{-1}$ (Nearest integer)

Given : $\Delta_{\mathrm{dis}} \mathrm{H}_{\mathrm{Cl}_{2(\mathrm{~g})}^{\theta}}^{\ominus}=240 \mathrm{~kJ} \mathrm{~mol}^{-1}, \Delta_{\mathrm{eg}} \mathrm{H}_{\mathrm{Cl_{(g)}}}^{\ominus}=-350 \mathrm{~kJ} \mathrm{~mol}^{-1}$ ,

${\mathrm{\Delta _{hyd}}H_{Cl_{(g)}^ - }^\Theta = - 380}$ $\mathrm{kJ~mol^{-1}}$

Explanation

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