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

1 Answer

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