Question

The melting temperature of a protein is TM=343K. A Differential Scanning Calorimetry (DSC) experiment finds that the molar enthalpy of denaturation at T=343K is ΔHM=638 kJ mol-1. Entropy at T=343K. ΔSM = 1860 JK⋅mol. ΔGM at T=343K. ΔGM = 0 Jmol The DSC experiment also finds that the difference in heat capacities is ΔCP=CDP−CNP=8.37kJmol−1 where CDP is the molar heat capacity of the denatured (D) protein and CNP is the molar heat capacity of the structure (N) protein. Assume teh difference in molar heat capacities is constant between T=310K and T=343K. ΔH at T=310K. ΔH = 3.62×105 Jmol ΔS at T=310K ΔS = 1010 JK⋅mol ΔG at T=310K ΔG = 4.76×104 (F)Calculate the molar ΔG at T=310K and P=1.00 kbars. Assume ΔV=VD−VN=−3.00mLmol

Answer 1

The molar ΔG at T=310K and P=1.00 kbars is -3.335x10^5 J/mol.

Step-by-step explanation:

The molar ΔG can be calculated using the equation ΔG = ΔH - TΔS, where ΔH is the enthalpy of denaturation, T is the temperature in Kelvin, and ΔS is the entropy. We are given ΔH and T, so we can calculate ΔG = 638 kJ/mol - 343K(1860 JK·mol) = 638 kJ/mol - 638 kJ/mol = 0 J/mol.

The molar ΔG at T=310K and P=1.00 kbars can be calculated using the equation ΔG = ΔH - TΔS + PΔV, where ΔV is the volume change. We are given ΔH, T, ΔS, and ΔV, so we can calculate ΔG = 3.62x10^5 J/mol - 310K(1010 JK·mol) + 1.00 kbars(-3.00 mL/mol) = - 3.335x10^5 J/mol.