Question

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). The 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.

Answer 1

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