The standard Gibbs energy change for the reac | vedclass

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(A) The reaction is the auto-ionization of water: $H_2O(l) \rightleftharpoons H^+(aq) + OH^-(aq)$.The equilibrium constant $K_w$ for this reaction at

Vedclass · Accepted Answer

$+80$

Vedclass · Answer

(A) The reaction is the auto-ionization of water: $H_2O(l) ightleftharpoons H^+(aq) + OH^-(aq)$.The equilibrium constant $K_w$ for this reaction at $25 \, ^\circ C$ is $1.0 	imes 10^{-14}$.The standard Gibbs energy change is given by the formula: $\Delta G^\circ = -RT \ln K_w$.Substituting the values: $R = 8.314 \, J \cdot K^{-1} \cdot mol^{-1}$,$T = 298 \, K$,and $K_w = 10^{-14}$.$\Delta G^\circ = -(8.314 \, J \cdot K^{-1} \cdot mol^{-1}) 	imes (298 \, K) 	imes \ln(10^{-14})$.$\Delta G^\circ = -8.314 	imes 298 	imes (-14 	imes 2.303) \, J/mol$.$\Delta G^\circ \approx 80400 \, J/mol = 80.4 \, kJ/mol$.Among the given options,the closest value is $+80 \, kJ$.

The standard Gibbs energy change for the reaction $H_2O \rightleftharpoons H^{+} + OH^{-}$ at $25 \, ^\circ C$ is $... \ kJ$.

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