Using the Gibbs energy change,Delta G^{o} = + | vedclass

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(B) The relationship between standard Gibbs energy change and the equilibrium constant is given by $\Delta G^{o} = -RT \ln K_{sp}$ or $\Delta G^{o} =

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$8.0 	imes 10^{-12}$

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(B) The relationship between standard Gibbs energy change and the equilibrium constant is given by $\Delta G^{o} = -RT \ln K_{sp}$ or $\Delta G^{o} = -2.303 \ RT \log K_{sp}$.Given $\Delta G^{o} = +63.3 \ kJ \ mol^{-1} = 63.3 	imes 10^{3} \ J \ mol^{-1}$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$,and $T = 25 + 273 = 298 \ K$.Substituting these values into the equation:$63.3 	imes 10^{3} = -2.303 	imes 8.314 	imes 298 	imes \log K_{sp}$$63.3 	imes 10^{3} = -5705.84 	imes \log K_{sp}$$\log K_{sp} = -\frac{63300}{5705.84} \approx -11.094$$K_{sp} = 10^{-11.094} = 8.05 	imes 10^{-12} \approx 8.0 	imes 10^{-12}$.

Using the Gibbs energy change,$\Delta G^{o} = +63.3 \ kJ \ mol^{-1}$,for the following reaction in water at $25 \ ^{\circ}C$,calculate the solubility product constant $(K_{sp})$:
$Ag_{2}CO_{3(s)} \rightleftharpoons 2Ag^{+}_{(aq)} + C{O_{3}}^{2-}_{(aq)}$
$(R = 8.314 \ J \ K^{-1} \ mol^{-1})$

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Using the Gibbs energy change,$\Delta G^{o} = +63.3 \ kJ \ mol^{-1}$,for the following reaction in water at $25 \ ^{\circ}C$,calculate the solubility product constant $(K_{sp})$:$Ag_{2}CO_{3(s)} \rightleftharpoons 2Ag^{+}_{(aq)} + C{O_{3}}^{2-}_{(aq)}$$(R = 8.314 \ J \ K^{-1} \ mol^{-1})$