Protein denaturations are usually irreversible but may be reversible under a narrow range of conditions. At pH 2.0, at temperatures ranging from about 40o C to 50o C, there is an equilibrium between the active form P and the deactivated form D of the enzyme trypsin: P ¡ê D. Thermodynamic values are ¥ÄH¥è =283 kJ/mol and ¥ÄS¥è =891 J/Kmol. Assume these values to be temperature independent over this narrow range. At what temperature will there be equal concentrations of P and D?
To find the temperature at which there will be equal concentrations of P and D (i.e., the equilibrium point), we can use the van 't Hoff equation and the Gibbs free energy change (ΔG).
The van 't Hoff equation is given by:
ΔG = ΔH - TΔS
In this equation:
- ΔG is the change in Gibbs free energy
- ΔH is the change in enthalpy
- T is the temperature in Kelvin
- ΔS is the change in entropy
We need to rearrange the equation to solve for the temperature (T):
ΔG = ΔH - TΔS
TΔS = ΔH - ΔG
T = (ΔH - ΔG) / ΔS
Using the given values:
ΔH = 283 kJ/mol
ΔS = 891 J/Kmol
Note that the units of ΔH and ΔS should be consistent, so we need to convert ΔH to J/mol:
ΔH = 283 kJ/mol * 1000 J/kJ = 283,000 J/mol
Now substitute the values into the equation:
T = (ΔH - ΔG) / ΔS
T = (283,000 J/mol - 0) / 891 J/Kmol
The value of ΔG is not given, but since there is an equilibrium between P and D, ΔG must be equal to zero (ΔG = 0) at the equilibrium.
Substituting ΔG = 0 into the equation:
T = (283,000 J/mol - 0) / 891 J/Kmol
T = 283,000 J/mol / 891 J/Kmol
T ≈ 318 K
Therefore, at approximately 318 Kelvin, there will be equal concentrations of P and D.