Question

For the equilibrium, $2 H_2 O\rightleftharpoons H_3O^{+} + OH^{-}$, the value of $\Delta G^{\circ}$ at $298\, K$ is approximately :

• A. $ - 80 \; kJ \; mol^{-1}$
• B. $ - 100 \; kJ \; mol^{-1}$
• C. $ 100 \; kJ \; mol^{-1}$
• D. $ 80 \; kJ \; mol^{-1}$

Answer 1

Answer: (D)

$2 H _2 O = H _3 O ^{+}+ OH ^{-} K =10^{-14}$
$\Delta G ^{\circ}=- RT \ell nK$
$=\frac{-8.314}{1000} \times 298 \times \ell n 10^{-14}$
$=80\, KJ /$ Mole