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poizon [28]
3 years ago
12

For double-helix formation, DG can be measured to be 2 54 kJ mol 1 ( 2 13 kcal mol 1 ) at pH 7.0 in 1 M NaCl at 25 8C (298 K). T

he heat released indicates an enthalpy change of 2 251 kJ mol 1 ( 2 60 kcal mol 1 ). For this process, calculate the entropy change for the system and the entropy change for the surroundings.
Chemistry
1 answer:
levacccp [35]3 years ago
8 0

Answer:

ΔS = -661.0J/mol is the entropy change for the system

ΔS = -842J/mol.K is the entropy change for the surroundings

Explanation:

From the relationship between ΔG, T, ΔH and ΔS,

Mathematically, ΔG = ΔH - TΔS

TΔS = ΔH - ΔS

ΔS = ΔH - ΔS / T

but ΔG = -54 kJ/mol, ΔH = -251 kJ/mol and T = 25 °C (298 K)

plugging into the equation,

ΔS = -251 kJ/mol - ( -54 kJ/mol) / 298

ΔS  = -0.6610KJ/mol or in J.mol

ΔS = -661.0J/mol is the entropy change for the system

  • For entropy change for the surroundings = ΔS = ΔH/T
  • ΔS  = -251 kJ/mol / 298K
  • ΔS = -0.84KJ/mol.K or -842J/mol.K is the entropy change for the surroundings
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From the equation; ΔTf = Kf × m
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\\ \sf\longmapsto CaCO_3

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