Question

For double-helix formation, change in Gibbs free energy, ΔG, can be measured to be −54 kJ⋅mol−1 (−13 kcal⋅mol−1) at pH 7.0 in 1 M NaCl at 25 °C (298 K). The heat released indicates an enthalpy change of -251 kJ/mol (-60 kcal/mol). For this process, calculate the entropy change for the system and the entropy change for the surroundings.

Answer 1

Step-by-step explanation:

Entropy change in the system : --

ΔG = −54 kJ⋅mol−1 (−13 kcal⋅mol−1) = −54 kJ⋅mol−1 (−13 x 4.2 kJ⋅mol−1)

= - 108.6 KJ / mol

ΔH = -251 kJ/mol (-60 kcal/mol) = -251 kJ/mol (-60 x 4.2 kJ/mol)

= - 503 KJ / mol

ΔG = ΔH - TΔS

ΔS = ( ΔH - ΔG ) / T

= - 503 + 108.6 / ( 273 + 25 ) KJ / mol k⁻¹

= - 1323.48 J / mol k⁻¹

Entropy change in the surrounding

+ 1323.48 J / mol k⁻¹