For the equilibrium 2 H 2 O ⇌ H 3 O + + OH − ; the value of Δ G ∘ at 298 K is approximately:

Δ G = Δ G ∘ + RT ln Q At equilibrium; Δ G = 0 and Q = K eq ⇒ Δ G ∘ = − 2.303 RT lo g K w = − 2.303 × 8.314 × 298 × lo g 1 0 − 14 = 79.9 kJ / mol ≈ 80 kJ / mol

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Chemistry - Thermodynamics - II Question with Solution | TestHub

ChemistryThermodynamics - IIFree Energy and EquilibriumMedium2 minPYQ_2019

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

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Solution:

$Δ G = Δ G^{\circ} + RT ln Q$ At equilibrium; $Δ G = 0$ and $Q = K_{eq}$ $\Rightarrow Δ G^{\circ} = - 2.303 RT lo g K_{w} = - 2.303 \times 8.314 \times 298 \times lo g 1 0^{- 14}$ $= 79.9 kJ / mol \approx 80 kJ / mol$

Stream:JEESubject:ChemistryTopic:Thermodynamics - IISubtopic:Free Energy and Equilibrium

⏱ 2mℹ️ Source: PYQ_2019

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