The standard Gibbs energy ( Δ G 0 ) for the following reaction is A ( s ) + B 2 + ( aq ) → A 2 + ( aq ) + B ( s ) , K c = 1 0 12 at 2 5 ∘ C ( K c = equilibrium constant )

A + B 2 + ⇌ A 2 + + B G ∘ = − RT ln K c = − 8.314 × 298 × 2.303 × lo g 1 0 12 = 68.47 kJ / mol

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ChemistryThermodynamics - IIRandomness and Third law of TDEasy2 minPYQ_2024

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The standard Gibbs energy $(Δ G^{0})$ for the following reaction is $A (s) + B^{2 +} (aq) \to A^{2 +} (aq) + B (s), K_{c} = 1 0^{12} at 2 5^{\circ} C$ $(K_{c} =$ equilibrium constant $)$

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Answer:

Solution:

$A + B^{2 +} ⇌ A^{2 +} + B$ $G^{\circ} = - RT ln K_{c}$ $= - 8.314 \times 298 \times 2.303 \times lo g 1 0^{12} = 68.47 kJ /$ mol

Stream:BITSATSubject:ChemistryTopic:Thermodynamics - IISubtopic:Randomness and Third law of TD

⏱ 2mℹ️ Source: PYQ_2024

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