Question

For the following reaction
$A{g}_{\left(aq\right)}^{+}+C{l}_{\left(aq\right)}^{-}\to AgC{l}_{\left(s\right)}$
Given: $\Delta G_f^o$ , AgCl = - 112.44 kJ/mol; $\Delta G_f^o$ $C{l}^{-}$ = - 130 kJ/mol ; $\Delta G_f^{o}A{g}^{+}$ = 75 kJ/mol
Report your answer by rounding it upto nearest whole number. The $K_{sp}$ of AgCl is $n\times 10^{-10}$ .The value of 'n' is.

Answer 1

Answer: 1

$A{g}_{\left(aq\right)}^{+}+C{l}_{\left(aq\right)}^{-}\to AgC{l}_{\left(s\right)}$
$\Delta G_f^o$ , AgCl = - 112.44 kJ/mol; $\Delta G_f^{o}C{l}^{-}$ = - 130 kJ/mol ; $\Delta G_f^{o}A{g}^{+}$ = 75 kJ/mol
$\Delta G_{reaction}^o=\Delta G_f^o$ , AgCl- [ $\Delta G_f^{o}A{g}^{+}+\Delta G_f^{o}C{l}^{-}$ ]
= - 112.44 kJ-[75 kJ- 130 kJ]
= -57.44 kJ
-2.303RTlog $K_{sp}=\Delta G_{reaction}^o$
$log{K}_{sp}=\frac{57440}{2.303\times 8.314\times 298}\cong 10$
$K_{sp}=1\times 10^{-10}$
Hence n = 1