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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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 $^{- 1}$