Question

As an approximation we can assume that proteins exist either in the native state and the denatured state. The standard molar enthalpy and entropy of the denaturation of a certain protein are 512 kj/mol and 1.60 kj/K mol.comment on the signs and magnitude of these quantities , and calculates the temperature at which the process favors the denatured state.

Answer 1

Answer:

Above 320 Kelvin temperature or 47°C , denaturation of protein will favored.

Explanation:

Standard molar enthalpy of denaturation of protein = ΔH = 512 kJ/mol

Standard molar entropy of denaturation of protein = ΔS = 1.60 kJ/mol

Gibbs free energy of the reaction = ΔG

$\Delta G = \Delta H-T\Delta S$

For reaction to be feasible ,  ΔG < 0 . So the value of ( ΔH-TΔS) shoulbe less than zero or negative.

Putting ΔG = 0

$0 = \Delta H-T\Delta S$

$\Delta H=T\Delta S$

$512 kJ/mol=T\times 1.60 kJ/mol$

T = 320 K

T = 320 K = 320 - 273°C = 47°C

Above 320 Kelvin temperature or 47°C , denaturation of protein will favored.