Question

An interaction between two subunits of a protein was determined to have a delta G of -57.05 kj/mol. What is the Keq for the reaction at 25 degrees celsius?

Answer 1

An assumption is made here that either the physiological temperature around the protein is `25^@ "C"`, or the protein was isolated outside of physiological conditions and placed into pseudo-physiological conditions at `25^@ "C"`.

So this may not be a realistic assumption, but we'll go with it so you get the "right" answer in accordance with your textbook.

Suppose `DeltaG^@ = -"57.05 kJ/mol"` and `"T = 298.15 K"` (room temperature).

Then, we have the following equation to consider:

`\mathbf(DeltaG = DeltaG^@ + RTlnQ)`

where:

• `DeltaG` is the Gibbs' free energy and `DeltaG^@` is the same thing, but at `25^@ "C"` and `"1 bar"`. Note that in nonstandard conditions, `DeltaG^@` describes exergonicity or endergonicity, not equilibrium or spontaneity.
• `R` is the universal gas constant, `8.314472xx10^(-3) "kJ/mol"cdot"K"`.
• `T` is the temperature in `"K"`.
• `Q` is the reaction quotient, i.e. the "not-yet-equilibrium" constant.

When we are at equilibrium and `"298.15 K"` like the problem asks, `DeltaG = 0`, and `Q = K`. Thus, we get:

`color(green)(DeltaG^@ = -RTlnK_(eq))`

So, to calculate `K_(eq)`, we divide `-RT` over and exponentiate.

`color(blue)(K_(eq)) = e^(-DeltaG^@"/RT")`

`= e^(-(-"57.05 kJ"cdot"mol"^(-1)"/"8.314472xx10^(-3) "kJ"cdot"mol"^(-1)"K"*"298.15 K")`

`= e^(23.01)`

`~~ color(blue)(9.879xx10^(9))`