The binding of the hydrophobic amino acid tryptophan to a protein shows a pronounced hydrophobic effect. The thermodynamic data for the binding of is given below calculate ?

Cp; Calculate ?S and ?G at T=323K and 298K for this binding reaction assuming ?S=0 at T=287K Temp (K) ?H (kcal/mole) 298 -13.5 303 -15.5 313 -19 318 -21.5 323 -24 The protein lysozyme unfolds (melts) at 75.5C and at this equilibrium temperature the change in the enthalpy is 509kJ/mol. Calculate the change in the entropy at 75.5C Calculate the entropy change for unfolding lysozyme at 25C ( As with many other problems you have solved for finding thermodynamic property at an non-equilibrium temperature you need to assume the unfolding could occur at this temperature).

Given that the heat capacity of a dilute solution of lysozyme when folded is 21.2 kJ/Kmole and when unfolded it increases to 27.4kJ/K mole. Given that the unfolded protein exposes hydrophobic amino acids side chains to the water solvent explain the increase in the heat capacity based on the change in the structure of the solvent. A Carnot cycle uses 1 mole of an ideal gas as the working fluid. The initial conditions are P=4atm and T=400K. Step 1 is an isothermal expansion to P=1atm.

In step 2 the temperature is decreased to T=250K adiabatically. This is followed by an isothermal compression and then an adiabatic compression back to P=4atm and T=400K. Determine: q, w, ?U and ?S for each step and the overall cycle. Present the data in a table. Which values would change if instead step 2 was done first and then step1( in other words what values would change if first the temperature was changed from 400K to 250K and then the pressure changed from 4atm to 1 atm, but the overall cycle is still a Canot cycle). Comment on the reasons in one paragraph. Assume you have 1mole of liquid methane which you have held as the liquid while raising the temperature from -200°C to 0°C (note the warming of the liquid is not part of the...