Question

Using the following data, calculate equilibrium constant $K$ for the reaction,
$AgCl _{(s)} \rightleftharpoons Ag _{(a q)}^{+}+ Cl _{(a q)}^{-}$
$\Delta G_{f( AgCl )}^{\circ}=-109.7\, kJ$
$\Delta G_{f\left( Ag ^{+}\right)}^{\circ}=77.1\, kJ$
$\Delta G_{f\left( Cl ^{-}\right)}^{\circ}=-131.2 \,kJ$

• A. $55.6 \times 10^{3}$
• B. $2.95 \times 10^{-41}$
• C. $1.78 \times 10^{-10}$
• D. $9.75$

Answer 1

Answer: (C)

$\Delta G^{\circ}=\Sigma \Delta G_{f(\text { Product})}^{\circ}-\Sigma \Delta G_{f \text { (Reactant)}}^{\circ}$

$=77.1+(-131.2)-(-109.7)=55.6 kJ =55.6 \times 10^{3} J$

We know that $\Delta G^{\circ}=-2.303 R T \log K$

$\log K=-\frac{\Delta G^{\circ}}{2.303 R T}$

$=-\frac{55.6 \times 10^{3}}{2.303 \times 8.314 \times 298}=-9.75$

$K=1.78 \times 10^{-10}$