Question

For the given reaction, $H _{2}( g )+ Cl _{2}( g )->2 H ^{+}( aq )+2 Cl ^{-}( aq ) ; \Delta G ^{\circ}=--262 \cdot 4 kJ$ The value of free energy of formation $\left(\Delta G_{f}^{\circ}\right)$ for the ion $C l^{-1}(a q)$, therefore will be

• A. $- 131.2 \, kJ \, mol^{-1}$
• B. $ + 131.2 \, kJ \, mol^{-1}$
• C. $- 262.4 \, kJ \, mol^{-1}$
• D. $ + 262.4 \, kJ \, mol^{-1}$

Answer 1

Answer: (A)

$\left(\Delta G^{\circ}\right)$ reaction $= \Delta G^{0}_{f} $ (Products) $ - \Delta G^{0}_{f}$ (reactants)
$\therefore 264.4 = \left[2 \Delta G^{0}_{f} \left(H^{+}\right) + 2 \Delta G^{0}_{f} \left(C^{1-}\right)\right] $
or $264.4 = - \left[\Delta G_{f}^{0} \left(H_{2}\right) + \Delta G^{0}_{f} \left(Cl_{2}\right)\right] $
$ = \left[0+ 2\Delta G^{0}_{f} \left(Cl^{-}\right)\right] + \left[0 +0\right] $
or, $- 262.4 = 2\Delta G_{f}^{0} \left(Cl^{-}\right) $
or, $ \Delta G^{0}_{g}\left(Cl^{-}\right) = -131.2 \,kJ\,mol^{-1}$