Question

For the reaction,
$\frac{1}{2}{H}_2\left(g\right)+\frac{1}{2}C{l}_2\left(g\right)\to H^{+}\left(aq\right)+C{l}^{-}\left(aq\right)$
$\Delta G_{recation}^{\circ }=-131.23kJmo{l}^{-1}$
The value of $\Delta G_{formation}^{\circ }$ of $A{g}^{+}(aq)$ shall be given by, (if $\Delta G_f^{\circ }\left(H^{+}aq=0\right)$

• A. $-54.12kJmo{l}^{-1}$
• B. $-131.23kJmo{l}^{-1}$
• C. $+77.11kJmo{l}^{-1}$
• D. $+54.12kJmo{l}^{-1}$

Answer 1

Answer: (B)

Given, $\Delta G_{\left(\text{Reaction}\right)}^{\circ }=-131.23KJmol$
and , $\Delta G_f^{\circ }\left[H^{+}aq\right]=0$
$\because \Delta G_{\text{reaction}}^{\circ }=\sum \left[G_{f\left(\text{product}\right)}^{\circ }\right]-\sum \left[G_{f\left(\text{reactant}\right)}^{\circ }\right]$
$=\left[G_f^{\circ }\left(H^{+}\right)+G_f^{\circ }\left(C{l}^{-}\right)\right]-\left[0\right]$
$\because G_f^{\circ }$ for elements in free state, i.e
$H_2$ and $O_2=$ zero
$\because A{g}^{+}$ ion will form $AgCl$ with $C{l}^{-}$ ions and cone, of $\left[A{g}^{+}\left(aq\right)\right]=\left[C{l}^{-}\left(aq\right)\right]$
$\therefore$ both are in $1:1$ ratio
$G_f^{\circ }\left(Cl\right)=G_f^{\circ }=\left(A{g}^{+}\right)-131.25=G_f^{\circ }\left[C{l}^{-}\left(aq\right)\right]$
$G_f\left[Ag+\left(aq\right)\right]=-131.25KJ/mo{l}^{-1}$