Question

A protein has a binding site for a single ligand with a A.G-16.7 KJ/mol at 298 K. What is Keg for this reaction?

Answer 1

Step-by-step explanation:

The given data is as follows.

`\Delta_(r) G` = -16.7 kJ/mol =
`-16.7 * 10^(3)`, T = 298 K

R = 8.314 J/mol K,
`K_(eq)` = ?

Relation between
`\Delta_(r) G` and
`K_(eq)` is as follows.

`\Delta_(r) G` =
`-RT ln K_(eq)`

Hence, putting the values into the above equation as follows.

`\Delta_(r) G` =
`-RT ln K_(eq)`

`-16.7 * 10^(3) J/mol` =
`-8.314 J/mol K * 298 K ln K_(eq)`

`ln K_(eq)` =
`(-16.7 * 10^(3) J/mol)/(-8.314 J/mol K * 298 K)`

= 6.740

`K_(eq)` = antilog (6.740)

= 846

Thus, we can conclude that
`K_(eq)` for given values is 846.