Question

Oxidising power of chlorine in aqueous solution can be determined by the parameters indicated below



The energy involved in the conversion of $\frac{1}{2}Cl_{2}\left(\right.g\left.\right)$ to $\left(C l\right)^{-}\left(\right.aq\left.\right)$ (Using the data)

$\text{ΔH}_{\text{diss}}\text{Cl}_{\text{2}}\text{=240} \, \text{kJ} \, \text{mol}^{- 1}$

$\text{ΔH}_{\text{eg}}\text{Cl= -349} \, \text{kJ} \, \text{mol}^{- 1}$

$\text{ΔH}_{\text{hyd}}\text{Cl= -381} \, \text{kJ} \, \text{mol}^{- 1}$

• A. $+152 \, kJ \, mol^{- 1}$
• B. $-610 \, kJ \, mol^{- 1}$
• C. $-850 \, kJ \, mol^{- 1}$
• D. $+120 \, kJ \, mol^{- 1}$

Answer 1

Answer: (B)

$\frac{1}{2}Cl_{2}\left(\right.g\left.\right) \rightarrow Cl^{-}\left(\right.aq\left.\right)$

$\Delta \text{H} = \frac{1}{2} \Delta \left(\text{H}\right)_{\text{diss}} \left(\right. \left(\text{Cl}\right)_{\text{2}} \left.\right) + \Delta \left(\text{H}\right)_{\text{eg}} \text{Cl} + \Delta \left(\text{H}\right)_{\text{hyd}} \left(\right. \left(\text{Cl}\right)^{-} \left.\right)$

$=\frac{240}{2}-349-381$

$=-610 \, kJ \, mol^{- 1}$