At STP,the dissociation reaction of water is | vedclass

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(B) The dissociation constant of water at $STP$ $(298 \ K)$ is $K_w = [H^{+}][OH^{-}] = 10^{-7} 	imes 10^{-7} = 10^{-14}$.The relation between standa

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$19091 \ cal / mol$

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(B) The dissociation constant of water at $STP$ $(298 \ K)$ is $K_w = [H^{+}][OH^{-}] = 10^{-7} 	imes 10^{-7} = 10^{-14}$.The relation between standard free energy change and equilibrium constant is $\Delta G^{\circ} = -RT \ln K_w = -2.303 RT \log K_w$.Given $R = 1.987 \ cal \ K^{-1} mol^{-1}$,$T = 298 \ K$,and $K_w = 10^{-14}$.$\Delta G^{\circ} = -2.303 	imes 1.987 	imes 298 	imes \log(10^{-14})$.$\Delta G^{\circ} = -2.303 	imes 1.987 	imes 298 	imes (-14)$.$\Delta G^{\circ} = 2.303 	imes 1.987 	imes 298 	imes 14 \approx 19091 \ cal / mol$.

At $STP$,the dissociation reaction of water is $H_2O \rightleftharpoons H^{+}(aq.) + OH^{-}(aq.)$,and the $pH$ of water is $7.0$. The change of standard free energy $(\Delta G^{\circ})$ for the above dissociation process is given by

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