Here is the top portion of Table 2-2 from Voet, Voet, & Pratt. Note these values
ID: 141240 • Letter: H
Question
Here is the top portion of Table 2-2 from Voet, Voet, & Pratt. Note these values are for 25'C TABLE 2-2 Thermodynamia Changes for Tranaferring s. from Water to Nonpolar Solvents at 25 C -TAS kJ mol ) -22.6 -22.6 -25.1 18.8 J mol-') -10.9 -12.1 -15.9 -12.1 -8.0 Procas Cl, in H,OCH in CHs CH in HyOCH4 in CCl 11.7 10.5 9.2 GH in HoCH,in bennene CaHg in HyOCHa in benneme 6.7 a) What is the 4G for the transfer of methane from water to benzene at 5'C? (Show your work.) b) Why are all of the AH values >0? Which molecular interaction(s) are responsible for this? c) Why are all of the -TAS valuesExplanation / Answer
a) For the reaction of Methane from water to benzene, we are given all the values at T=25°C
we need to calculate dS and dH at T=5°C.
Assuming that dH and dS remain constant at given range of time
So dH1 = dH2 =11.7
similarly dS1 = dS2 = (22.6/298) =0.0758KJ/K
We know that G =H -TS
So, G2 = H2 - T2S2
G2 = 11.7 - (278*22.6)/298 kJ/mol
= -9.38 kJ
b) All the values of H>0 because for the reaction Hydrogen bonds are needed to be broken
So, energy is needed to break these strong bonds, hence Endothermic.
c) All the values of -TS<0 because Like dissolves like. So, the spontaneity of the reaction is much more which makes Entropy positive and hence, -TS negative.
d) The reason H decreases in going from C2H6 to C2H4 to C2H2 while -TS increases is because of steric hindrance and spontaneity is decreasing
e. Gibbs-Helmholtz equation, which is used to predict denaturation, is as follows:
G(T) = Hm (1-T/Tm) – Cp[(Tm-T) + Tln(T/Tm)
Cold denaturation is a result of changes in the interaction between water and hydrophobic groups. It leads to partial unfolding of the polypeptide chain. With a decrease in temperature, the free energy cost for hydrophobic effect decreases and as a result, hydration increases.
In cold water, the protein is surrounded by ordered water molecules in the hydration shell. Molecules in the shell are highly ordered and have low energy. In this condition, hydrogen bonds involving interfacial water are more favorable than that in the bulk water. This is the cause of cold denaturation.