# 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?

##### 1 Answer

An assumption is made here that either the physiological temperature around the protein is

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

Then, we have the following equation to consider:

#\mathbf(DeltaG = DeltaG^@ + RTlnQ)# where:

#DeltaG# is theGibbs' free energyand#DeltaG^@# is the same thing, but at#25^@ "C"# and#"1 bar"# . Note that in nonstandard conditions,#DeltaG^@# describes exergonicity or endergonicity,equilibrium or spontaneity.not#R# is theuniversal gas constant,#8.314472xx10^(-3) "kJ/mol"cdot"K"# .#T# is thetemperaturein#"K"# .#Q# is thereaction quotient, i.e. the "not-yet-equilibrium" constant.

When we *are* at equilibrium and

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

So, to calculate

#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))#