For the equilibrium, 2 H2 Oleftharpoons H3O+ + OH-, the value of Δ G° at 298 K is approximately :

Question

For the equilibrium, 2 H2 Oleftharpoons H3O+ + OH-, the value of Δ G° 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

Tardigrade Admin · Accepted Answer

2 H 2 O = H 3 O ++ OH - K =10-14 Δ G °=- RT ℓ nK =(-8.314/1000) × 298 × ℓ n 10-14 =80 KJ / Mole

Q. For the equilibrium, $2 H_{2} O ⇌ H_{3} O^{+} + O H^{-}$ , the value of $Δ G^{\circ}$ at $298 K$ is approximately :

$- 80 k J m o l^{- 1}$

$- 100 k J m o l^{- 1}$

$100 k J m o l^{- 1}$

$80 k J m o l^{- 1}$

$2 H_{2} O = H_{3} O^{+} + O H^{-} K = 1 0^{- 14}$
$Δ G^{\circ} = - RT ℓ n K$
$= \frac{- 8.314}{1000} \times 298 \times ℓ n 1 0^{- 14}$
$= 80 K J /$ Mole

Q. For the equilibrium, 2H2​O⇌H3​O++OH−, the value of ΔG∘ at 298K is approximately :

−80kJmol−1

−100kJmol−1

100kJmol−1

80kJmol−1