Calculate the entropy change for the following reversible process- 1 mol at 1atm - -Tin - 1 mol at 1 atm - -Tin at 13oC - H trans -2090J mol -1

Δ S _ trans =Δ H _ trans T = 2090 286 =7.3076J K ^ 1 mol ^ 1 ...

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Standard XII

Chemistry

Entropy

Calculate the...

Question

Calculate the entropy change for the following reversible process:
$α - T i n 1 m o l a t 1 a t m ⇌ β - T i n 1 m o l a t 1 a t m$ at $13^{o} C$
$(Δ H_{t r a n s} = 2090 J {m o l}^{- 1})$

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Solution

$Δ S_{t r a n s} = \frac{Δ H_{t r a n s}}{T} = \frac{2090}{286} = 7.3076 J K^{- 1} {m o l}^{- 1}$

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Calculate the entropy change for the following reversible process:α−Tin1molat1atm⇌β−Tin1molat1atm at 13oC(ΔHtrans=2090Jmol−1)

ΔStrans=ΔHtransT=2090286=7.3076JK−1mol−1

Calculate the entropy change for the following reversible process:
$α - T i n 1 m o l a t 1 a t m ⇌ β - T i n 1 m o l a t 1 a t m$ at $13^{o} C$
$(Δ H_{t r a n s} = 2090 J {m o l}^{- 1})$

$Δ S_{t r a n s} = \frac{Δ H_{t r a n s}}{T} = \frac{2090}{286} = 7.3076 J K^{- 1} {m o l}^{- 1}$