Question

ΔH = CpΔT ΔS = Cp ln T2/T1 ΔG = ΔH − TΔS

1. Consider a protein that unfolds at a transition temperature of 70.0ºC at a pressure of 1 bar, and the heat of unfolding (qu) is +638 kJ mol-1.

a. Calculate ΔuH, ΔuS, ΔuG and for the transition from folded to unfolded protein at 70.0ºC at 1 bar knowing that ΔCp = −8.37 kJ K-1 mol-1 and is independent of temperature.

b. Repeat the calculations from part (a) at 25ºC.

c. Has the reaction become more spontaneous at 25ºC compared to 70ºC? Is that what you expected?

d. Looking at the signs of ΔuH and ΔuS , speculate on what is happening at a molecular level to the protein upon unfolding at the two different temperatures.

Answer 1

Transition temperature = 70 oC = 343 K

Heat of unfolding (qu) = +638 kJ mol-1

(a). We know that,

uH = CpT

= −8.37 * (343 - 298)

= −8.37 * 45

uH = - 376.65 kJ/mol

uS = Cp ln (T2/T1)

= −8.37 * ln (343 / 298)

= −8.37 * 0.14

uS = - 1.172 kJ/mol-K

uG = uH - T * uS

= - 376.65 - 343 * - 1.172

= - 376.65 + 401.99

uG = 25.346 kJ/mol

(b). At 25 oC

T = 0

uH = 0 kJ/mol

uS = 0 kJ/mol-K

uG = 0 kJ/mol

(c). Yes, the reaction has become more spontaneous at 25ºC compared to 70ºC.