What is the equilibrium constant at #25^@"C"# for a reaction with #DeltaG^@ = -"19.600 kJ/mol"#?

1 Answer
F · Truong-Son N. · Ernest Z.
Nov 5, 2017

#K_(eq) = 2.716*10^3#

Explanation:

At equilibrium, #DeltaG = 0#, so

#cancel(DeltaG)^(0) = DeltaG^@ + RTlncancel(Q)^(K_(eq))#

Therefore,

#ΔG^@ = -RTlnK_(eq)#

#=> K_(eq) = e^(-(ΔG^@)/(RT))#

#K_(eq)= e^-((-19600 J/(mol))/(8.314J/(molK)(25+273.15)K))#

#K_(eq) = 2.716*10^3#

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