Question

For the equilibrium, $2H_{2}O\rightleftharpoons H_3{O}^{+}+O{H}^{-}$, the value of $\Delta G^o$ at 298 K is approximately :

• A. $-80kJmo{l}^{-1}$
• B. $-100kJmo{l}^{-1}$
• C. $100kJmo{l}^{-1}$
• D. $80kJmo{l}^{-1}$

Answer 1

Answer: (D)

$80kJmo{l}^{-1}$

$2H_{2}O=H_3{O}^{+}+O{H}^{-}$ $K={10}^{-14}$

$\Delta G^o=-RT\ell nK$

$=\frac{-8.314}{1000}\times 298\times \ell n{10}^{-14}$

$=80KJ/Mole$

Therefore, option is D.