Question

The standard change in Gibbs energy for the reaction: $H_{2}O\rightleftharpoons H^{+}+O{H}^{-}$ at ${25}^{\circ }C$

• A. $100kJ$
• B. $-90kJ$
• C. $90kJ$
• D. $-100kJ$

Answer 1

The correct option is C $90kJ$

We know at equilibrium,
$\Delta G^{\circ }=-2.303RTlogK$

$Kfor{H}_{2}O=\frac{K_{w}at{25}^{o}C}{molarityofwater}$

$Kfor{H}_{2}O=\frac{{10}^{-14}}{55.6}$

Putting the values in formula for $\Delta G^{\circ }$,

$\Delta G^{\circ }=-2.303\times 8.314\times 298\times \text{log}\frac{{10}^{-14}}{55.6}=89.838$
$=90kJ$