Magnitude of Δ G ∘ of the reaction in KJ mol − 1 is

Balanced chemical equation is 3 I 2 ( s ) + 6 OH − ( aq ) ⇌ 5 I − ( aq ) + IO − 3 ( aq ) + 3 H 2 O ( ℓ ) Δ G ∘ = 5 × ( − 50 ) + ( − 123.5 ) + 3 × ( − 233 ) − 0 − 6 × ( − 150 ) = − 172.5 kJ / mol

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Chemistry - Thermodynamics - II Question with Solution | TestHub

ChemistryThermodynamics - IIFree Energy and EquilibriumMedium2 minQB

ChemistryMediumnumerical

Passage / Comprehension

The following reaction is at equilibrium in basic medium at 300 K:

I₂(s) ⇌ I⁻(aq) + IO₃⁻(aq)

Equilibrium concentrations at 300 K are: [I⁻] = 0.1 M, [IO₃⁻] = 0.1 M.

Given:

$Δ G_{f}^{\circ}$ (I⁻, aq) = –50 kJ/mol

$Δ G_{f}^{\circ}$ (IO₃⁻, aq) = –123.5 kJ/mol

$Δ G_{f}^{\circ}$ (H₂O, ℓ) = –233 kJ/mol

$Δ G_{f}^{\circ}$ (OH⁻, aq) = –150 kJ/mol

R = 25/3 J mol⁻¹ K⁻¹, log₁₀e = 2.3

Magnitude of $Δ G^{\circ}$ of the reaction in $KJ mol^{- 1}$ is

Answer:

172.50

Solution:

Balanced chemical equation is

$3 I_{2} (s) + 6 OH^{-} (aq) ⇌ 5 I^{-} (aq) + IO^{-}_{3} (aq) + 3 H_{2} O (ℓ)$

$Δ G^{\circ} = 5 \times (- 50) + (- 123.5) + 3 \times (- 233) - 0 - 6 \times (- 150) = - 172.5 kJ / mol$

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

⏱ 2mℹ️ Source: QB

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