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

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

Step-by-step explanation:

According to Arrhenius equation-
`k=Ae^{(-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-

`(k_(ala-pro))/(k_(phe-pro))=e^([E_(a)^(phe-pro)-E_(a)^(ala-pro)])/(RT)`

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

So,
`(0.05)/(0.005)=e^{([E_(a)^(phe-pro)-(60000J/mol)])/(8.314J.mol^(-1).K^(-1)* 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