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vovangra [49]
3 years ago
10

The normal boiling point of a substance is defined to be the temperature at which the liquid phase of the substance is in equili

brium with the gas phase at 1 atm pressure. The normal boiling point of methanol is 80oC and ∆H vap = 38 kJ/mol. What is the ∆Scrap value
Chemistry
1 answer:
boyakko [2]3 years ago
5 0

Answer:

ΔSv = 0.1075 KJ/mol.K

Explanation:

Binary solution:

∴ a: solvent

∴ b: solute

in equilibrium:

  • μ*(g) = μ(l) = μ* +RTLnXa....chemical potential (μ)

⇒ Ln (1 - Xb) = ΔG/RT

∴ ΔG = ΔHv - TΔSv

⇒ Ln(1 -Xb) = ΔHv/RT - ΔSv/R

∴ Xb → 0:

⇒ Ln(1) = ΔHv/RT - ΔSv/R

∴ T = T*b....normal boiling point

⇒ 0 = ΔHv/RT*b - ΔSv/R

⇒ ΔSv = (R)(ΔHv/RT*b)

⇒ ΔSv = ΔHv/T*b

∴ T*b = 80°C ≅ 353 K

⇒ ΔSv = (38 KJ/mol)/(353 K)

⇒ ΔSv = 0.1075 KJ/mol.K

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What is the conjugate acid in the following equation:
yanalaym [24]

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5 0
3 years ago
I need some help with chemistry. Let me know if you are good at it!
tekilochka [14]

Answer:

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3 0
3 years ago
A reaction that occurs in the internal combustion engine is n2(g) + o2(g) ⇌ 2 no(g) (a) calculate δh o and δs o for the reaction
jekas [21]
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2) ΔS = 2mol·ΔS(NO) - (ΔS(O₂) + ΔS(N₂)).
ΔS = 2mol · 210.65 J/mol·K - (1mol · 205 J/mol·K + 1 mol · 191.5 J/K·mol).
ΔS = 24.8 J/K.
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2570°C: ΔG = 180.6 kJ - 2843.15 K · 24.8 J/K = 110.09 kJ.
3610°C: ΔG = 180.6 kJ - 3883.15 K · 24.8 J/K = 84.29 kJ.
7 0
3 years ago
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