Answer:
0.4774 KJ/K.mol
Explanation:
We are told that the transition at 25.0°C occurs in three steps. Steps i, ii and iii.
Thus;
the entropy of unfolding of lysozyme = ΔS_i + ΔS_ii + ΔS_iii
Now,
C_p,m(unfolded protein) = C_p,m(folded protein) + 6.28 kJ/K.mol
Now, for the first process, ΔS_i is given as;
ΔS_i = C_p,m × In(T2/T1)
We are given;
T1 = 25°C = 25 + 273.15K = 298.15 K
T2 = 75.5°C = 75.5 + 273.15 K=348.65 K
Thus;
ΔS_i = C_p,m × In(348.65/298.15)
Now, for the third process, ΔS_iii is given as;
ΔS_iii = (C_p,m + 6.28 kJ/K.mol) × In(T1/T2)
Thus;
ΔS_iii = (C_p,m + 6.28 kJ/K.mol) × In(298.15/348.65)
Now, we don't know C_pm. So, we have to find a way to eliminate it. We will do it by rewriting In(298.15/348.65) in such a way that when ΔS_iii is added to ΔS_i, C_p,m will cancel out. Thus;
In(298.15/348.65) can also be written as;
In(348.65/298.15)^(-1) or
- In(348.65/298.15)
Thus;
ΔS_iii = - [(C_p,m + 6.28 kJ/K.mol) × In(298.15/348.65)]
Now, let's add ΔS_iii to ΔS_i to get;
ΔS_i + ΔS_iii = [C_p,m × In(348.65/298.15)] + [(-C_p,m - 6.28 kJ/K.mol) × In(348.65/298.15)]
ΔS_i + ΔS_iii = [C_p,m × In(348.65/298.15)] - [C_p,m × In(348.65/298.15)] - [6.28In(348.65/298.15)]
First 2 terms will cancel out to give;
ΔS_i + ΔS_iii = -6.28In(348.65/298.15)
ΔS_i + ΔS_iii = -0.9826 KJ/K.mol
Now,for process ii;
ΔS_ii = standard enthalpy of transition/Transition Temperature
Thus;
ΔS_ii = (509 KJ/K.mol)/348.65
ΔS_ii = 1.46 KJ/K.mol
Thus;
the entropy of unfolding of lysozyme = ΔS_i + ΔS_ii + ΔS_iii = -0.9826 + 1.46 = 0.4774 KJ/K.mol
