Calculate the entropyJ K-1 mol-1 change for the following reversible process-Tin - -Tin at 13- C1 mol at 1 atm - Htrans-2288 J mol-1

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Standard XIII

Chemistry

Entropy for a Reversible Process

Calculate the...

Question

Calculate the entropy $(J K^{- 1}$ mol $^{- 1})$ change for the following reversible process:
$α - T i n ⇌ β - T i n$ at $13^{\circ} C$
1 mol at 1 atm
$Δ H_{trans} = 2288 J$ mol $^{- 1}$

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Solution

The correct option is A 8
We know the relation , at equilibrium;
$Δ S_{trans} = \frac{Δ H_{trans}}{T}$

$Δ S_{t r a n s} represents the change in entropy$
$Δ H_{t r a n s} represents the change in enthalpy$

$= \frac{2288}{286} = 8 J K^{- 1}$ mol $^{- 1}$

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Q. Calculate the entropy

(J K^{- 1}

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α - T i n ⇌ β - T i n

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1 mol at 1 atm

Δ H_{trans} = 2288 J

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Q. Calculate the entropy change for the following reversible process:

α - T i n 1 m o l a t 1 a t m ⇌ β - T i n 1 m o l a t 1 a t m

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(Δ H_{t r a n s} = 2090 J {m o l}^{- 1})

The entropy change for the conversion of 1 mole of $α$ tin (at 13 $^{0}$ C, 1 atm) to 1 mole of $β$ tin (13 $^{0}$ C, 1 atm), if the enthalpy of transition is 2.095 KJ mol $^{- 1}$ , is :

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Latent heat of fusion of ice = $1440 c a l m o l^{- 1}$ at $0^{\circ} C$

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Entropy for a Reversible Process

Standard XIII Chemistry

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Calculate the entropy(JK−1 mol−1) change for the following reversible process: α−Tin ⇌ β−Tin at 13∘C 1 mol at 1 atm ΔHtrans=2288J mol−1

The correct option is A 8We know the relation , at equilibrium; ΔStrans=ΔHtransT ΔStrans represents the change in entropy ΔHtrans represents the change in enthalpy =2288286=8 JK−1 mol−1

Calculate the entropy $(J K^{- 1}$ mol $^{- 1})$ change for the following reversible process:
$α - T i n ⇌ β - T i n$ at $13^{\circ} C$
1 mol at 1 atm
$Δ H_{trans} = 2288 J$ mol $^{- 1}$

The correct option is A 8
We know the relation , at equilibrium;
$Δ S_{trans} = \frac{Δ H_{trans}}{T}$

$Δ S_{t r a n s} represents the change in entropy$
$Δ H_{t r a n s} represents the change in enthalpy$

$= \frac{2288}{286} = 8 J K^{- 1}$ mol $^{- 1}$