For the phase transformation Sn ( white ) → Sn ( grey ) What will be Δ G ∘ at 300 K?

Sn ( w ) Sn ( g ) Δ S ∘ Δ H ∘ Δ G ∘ = ( 44 − 52 ) = − 8 J / k = − 2 kJ = − 2000 − 300 × ( − 8 ) = 400 J

Chemistry - Thermodynamics - II Question with Solution | TestHub

ChemistryThermodynamics - IIGibb's Free EnergyMedium2 minQB

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Passage / Comprehension

Various elements show allotropy; however, some of them are monotropic (one allotrope is more stable at all conditions) like carbon. Some are enantiotropic (different allotropes having different stability at different conditions of temperature and pressure) like sulfur. Based on this and the information given below, answer the question that follows.

$Δ H_{f}^{\circ} S n_{(w hi t e)} = 0$ , $Δ H_{f}^{\circ} S n_{(g rey)} = - 2 k J / m o l$

$S^{\circ} (w hi t e) = 52 J / K m o l$

$S^{\circ} (g r a y) = 44 J / K m o l$

The molar heat capacity ( $C_{p, m}$ ) of both allotropes is assumed to be the same.

All values are at 300 K.

For the phase transformation $Sn_{(white)} \to Sn_{(grey)}$

What will be $Δ G^{\circ}$ at 300 K?

Options:

Answer:

Solution:

$Sn_{(w)} Sn_{(g)}$

$Δ S^{\circ} Δ H^{\circ} Δ G^{\circ} = (44 - 52) = - 8 J / k = - 2 kJ = - 2000 - 300 \times (- 8) = 400 J$

Stream:JEESubject:ChemistryTopic:Thermodynamics - IISubtopic:Gibb's Free Energy

⏱ 2mℹ️ Source: QB

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