Question

The activation energy for proline isomerization of a peptide depends on the identity of the preceding residue and obeys Arrhenius rate behavior. Experiments are conducted on the isomerization of an alanine- proline peptide. At 25°C (298 K) the observed rate constant is 0.05 sec–1 and the value of EA is calculated to be 60 kJ•mol–1. Similar measurements are performed on a phenylalanine-proline peptide at 25°C, with a measured rate constant of 0.005 sec–1. Assuming an identical preexponential factor as the alanine-proline peptide, what is the activation energy for this peptide (kJ/mol)?

Answer 1

Answer:

Activation energy of phenylalanine-proline peptide is 66 kJ/mol.

Explanation:

According to Arrhenius equation-     $k=Ae^{\frac{-E_{a}}{RT}}$    , where k is rate constant, A is pre-exponential factor, $E_{a}$ is activation energy, R is gas constant and T is temperature in kelvin scale.

As A is identical for both peptide therefore-

                                   $\frac{k_{ala-pro}}{k_{phe-pro}}=e^\frac{[E_{a}^{phe-pro}-E_{a}^{ala-pro}]}{RT}$

Here $\frac{k_{ala-pro}}{k_{phe-pro}}=\frac{0.05}{0.005}$ , T = 298 K , R = 8.314 J/(mol.K) and $E_{a}^{ala-pro}=60kJ/mol$

So, $\frac{0.05}{0.005}=e^{\frac{[E_{a}^{phe-pro}-(60000J/mol)]}{8.314J.mol^{-1}.K^{-1}\times 298K}}$

   $\Rightarrow$ $E_{a}^{phe-pro}=65705J/mol=66kJ/mol$ (rounded off to two significant digit)

So, activation energy of phenylalanine-proline peptide is 66 kJ/mol