What is the value of the equilibrium constant K, for a reaction for which delta G(free energy) = -5.20kJ?

This is at 50 degrees Celsius
The answer is 6.93, I just need an explanation on how to get this value.

One Response to “What is the value of the equilibrium constant K, for a reaction for which delta G(free energy) = -5.20kJ?”

Dr.J Says:
2. September 2010 at 9:46 am

e^(-DeltaG^o/RT) = K

Plug in and solve. Be sure temp is in Kelvin, R is in units of J/mol-K, and Delta G is in units of J. You will get K=6.93.

e^(-DeltaG/RT)
e^{5200 J/[(8.31 J/molK)(273+50K)]} = 6.93

One Response to “What is the value of the equilibrium constant K, for a reaction for which delta G(free energy) = -5.20kJ?”