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?

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))#