What is the equilibrium constant at 25^@"C" for a reaction with DeltaG^@ = -"19.600 kJ/mol"?
1 Answer
Nov 5, 2017
Explanation:
At equilibrium,
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