Given the following and that R = 8.314 J/K ?mol, determine the equilibrium constant, K, at 298K for the following reaction:
AgBr(s) ? Ag⁺(aq) + Br-(aq)
$\begin{array}{ll}\text { Substance } & \underline{\Delta}_{\mathrm{f}} \mathrm{G}^{\circ}(\mathrm{kJ} / \mathrm{mol}) \text { at } 298 \mathrm{~K} \\\hline\mathrm{Br}-(\mathrm{aq}) & -104.0 \\\mathrm{Ag}^{+}(\mathrm{aq}) & 77.12 \\\mathrm{AgBr}(\mathrm{s}) & -96.9\end{array}$

Question

Given the following and that R = 8.314 J/K ?mol, determine the equilibrium constant, K, at 298K for the following reaction:
AgBr(s) ? Ag⁺(aq) + Br-(aq)

\begin{array}{ll}\text { Substance } & \underline{\Delta}_{\mathrm{f}} \mathrm{G}^{\circ}(\mathrm{kJ} / \mathrm{mol}) \text { at } 298 \mathrm{~K} \\\hline\mathrm{Br}-(\mathrm{aq}) & -104.0 \\\mathrm{Ag}^{+}(\mathrm{aq}) & 77.12 \\\mathrm{AgBr}(\mathrm{s}) & -96.9\end{array}

Quizplus · Accepted Answer

The equilibrium constant, K, can be calculated using the formula: $\Delta G^\circ = -RT \ln K$. First, calculate $\Delta G^\circ$ for the reaction: $\Delta G^\circ = (\Delta G^\circ_{\text{Ag}^+} + \Delta G^\circ_{\text{Br}^-}) - \Delta G^\circ_{\text{AgBr}}$. Substituting the given values: $\Delta G^\circ = (77.12 - 104.0) - (-96.9) = 70.02 \, \text{kJ/mol}$. Convert to J/mol: $70.02 \times 10^3 \, \text{J/mol}$. Then, solve for K: $70.02 \times 10^3 = -8.314 \times 298 \times \ln K$. Solving gives $K \approx 5.3 \times 10^{-13}$.

Given the Following and That R = 8 A) 5.3×10−135.3 \times 10 ^ { - 13 }5.3×10−13

Given the Following and That R = 8 A) $5.3 \times 10 ^ { - 13 }$