The enthalpy change for the conversion of $$\dfrac{1}{2}$$Cl$$_2$$(g) to Cl$$^-$$(aq) is (-) ______ kJmol$$^{-1}$$ (Nearest integer)
Given: $$\Delta_{dis}H^0_{Cl_2(g)} = 240$$ kJmol$$^{-1}$$
$$\Delta_{eg}H^o_{Cl(g)} = -350$$ kJmol$$^{-1}$$
$$\Delta_{hyd}H^o_{Cl^-(g)} = -380$$ kJmol$$^{-1}$$

The conversion $$\frac{1}{2}Cl_2(g) \rightarrow Cl^-(aq)$$ can be broken into three steps:

First, dissociation of $$\frac{1}{2}Cl_2(g)$$ to $$Cl(g)$$:.

$$ \Delta H_1 = \frac{1}{2} \times \Delta_{dis}H^0 = \frac{1}{2} \times 240 = 120 \text{ kJ/mol} $$

Next, electron gain by $$Cl(g)$$ to form $$Cl^-(g)$$:.

$$ \Delta H_2 = \Delta_{eg}H^0 = -350 \text{ kJ/mol} $$

Now, hydration of $$Cl^-(g)$$ to form $$Cl^-(aq)$$:.

$$ \Delta H_3 = \Delta_{hyd}H^0 = -380 \text{ kJ/mol} $$

Total enthalpy change (by Hess's law):

$$ \Delta H = \Delta H_1 + \Delta H_2 + \Delta H_3 = 120 + (-350) + (-380) = -610 \text{ kJ/mol} $$

The enthalpy change is $$(-) 610$$ kJ/mol.