Question

For the reaction,
$ \frac{1}{2}H_{2}\left(g\right)+\frac{1}{2}Cl_{2}\left(g\right) \rightarrow H^{+}\left(aq\right)+Cl^{-}\left(aq\right) $
$ \Delta G^{\circ}_{recation}=-131.23 \, kJ \, mol^{-1} $
The value of $ \Delta G^{\circ}_{formation} $ of $ Ag^{+}(aq) $ shall be given by, (if $ \Delta G^{\circ}_{f} \left(H^{+}\,aq=0\right) $

• A. $ - 54.12\, kJ \,mol^{-1} $
• B. $ -131.23\, kJ \,mol^{-1} $
• C. $ + 77.11 \,kJ \,mol^{-1} $
• D. $ + 54.12\, kJ \,mol^{-1} $

Answer 1

Answer: (B)

Given, $\Delta G^{\circ}_{\left(\text{Reaction}\right)} =-131.23\,KJ\, mol$
and , $\Delta G_{f}^{\circ} \left[H^{+} aq\right]=0 $
$\because \Delta G_{\text{reaction}}^{\circ}=\sum\left[G^{\circ}_{f\left(\text{product}\right)}\right]-\sum\left[G^{\circ}_{f \left(\text{reactant}\right)}\right]$
$=\left[G_{f}^{\circ}\left(H^{+}\right)+G_{f}^{\circ}\left(Cl^{-}\right)\right]-\left[0\right]$
$\because G_{f}^{\circ} $ for elements in free state, i.e
$H_{2}$ and $O_{2}=$ zero
$\because Ag^{+}$ ion will form $AgCl$ with $Cl^{-}$ ions and cone, of $\left[Ag^{+}\left(aq\right)\right]=\left[Cl^{-}\left(aq\right)\right]$
$\therefore $ both are in $1 : 1$ ratio
$G_{f}^{\circ} \left(Cl\right)=G_{f}^{\circ}=\left(Ag^{+}\right)-131.25=G_{f}^{\circ}\left[Cl^{-}\left(aq\right)\right]$
$G_{f} \left[Ag+\left(aq\right)\right]=-131.25 \,KJ / mol^{-1}$