Calculate the entropy J K -1 mol -1 change for the following reversible process- Tin - Tinat 13- C1 mol at 1 atm - H trans -2288 J mol -1A- 8B- 5C- 0D- 10

Name: NEET - Grade 12 - Chemistry - Stereochemistry-1 - Session 03 - W10
Uploaded: 2022-07-04T10:36:42+05:30
Description: The correct option is A 8We know the relation , at equilibrium; Δ S_trans=Δ H_transT Δ S_trans represents the change in entropy Δ H_trans represents the chang ...

The correct option is A 8We know the relation , at equilibrium; Δ S_trans=Δ H_transT Δ S_trans represents the change in entropy Δ H_trans represents the chang ...

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Standard XII

Chemistry

Third Law of Thermodynamics

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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Third Law of Thermodynamics

Standard XII 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 $^{- 1}$