Calculate Δ G ( kJ / mol ) for the above given reaction at 300 K . | TestHub

Passage / Comprehension

At $300 K$ , partial pressures of A₂, B₂, and AB are $1 0^{- 2} bar$ , $1 0^{- 2} bar$ , and $1 0^{- 4} bar$ respectively, for the reaction:

A₂(g) + B₂(g) ⇌ 2AB(g)

Given:

For AB(g): $Δ_{f} H = 180 kJ/mole$ ; $Δ_{f} S ° = 210 J/mole - K$

For A₂(g): $Δ_{f} H ° = 60 kJ/mole$ ; $Δ_{f} S ° = 190 J/mole - K$

For B₂(g): $Δ_{f} H ° = 29.5 kJ/mole$ ; $Δ_{f} S ° = 205 J/mole - K$

Use: $2.303 R \times 300 = 5750 J/mole$

Calculate $Δ G (kJ / mol)$ for the above given reaction at 300 K .

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Answer:

Solution:

$A_{2} (g) + B_{2} (g) ⇌ 2 AB (g)$

$Δ_{r} H^{\circ} Δ_{r} S^{\circ} Δ_{r} G^{\circ} Δ_{r} G^{\circ} Δ_{r} G Δ_{r} G = 2 \times 180 - (60 + 29.5) = 270.5 kJ / mole = 2 \times 210 - (190 + 205) = 25 J / mole - K = Δ_{r} H^{\circ} - T Δ_{r} S^{\circ} = 270.5 - 300 \times 25 \times 1 0^{- 3} = 263 kJ / mole = Δ_{rG}^{\circ} + 2.303 RT lo g Q_{P} = 263 + 5750 \times 1 0^{- 3} lo g \frac{( 1 0 ^{- 4} ) ^{2}}{1 0 ^{- 2} \times 1 0 ^{- 2}} = 263 + 5750 \times 1 0^{- 3} \times (- 4) = 240 kJ / mole$

Chemistry - Thermodynamics - II Question with Solution | TestHub

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