Match standard free energy of the reaction with the corresponding Equilibrium constant


(i) ∆G0 > 0 - (d) K < 1

We know that,


∆G = ∆G° + RTln K… eq (1)


• ∆G = Gibbs free energy


• ∆G° = Standard Gibbs free energy (1atm pressure and 298k)


• T = temperature (in kelvin)


R = Gas constant


• K = Equilibrium constant


• At Equilibrium, ∆G = 0


• ∆G° + RTln K = 0


So, RTln K = -∆G°...eq (2)


From the eq (2) we get smaller the magnitude of -∆G°, the higher the rate constant K will be and the faster the reaction.


Explanation: when the value of ∆G° is positive (+) the value of K will become negative hence, reaction will proceed in reverse direction i.e. non spontaneous reaction.


(ii) ∆G0 < 0 (a) K > 1


Explanation: when the value of ∆G° is negative (-) the value of K will become positive hence, reaction will proceed in forward direction i.e. spontaneous reaction.


(iii) ∆G0 = 0 (b) K = 1


Explanation: we know that, ∆G = ∆G° + RT ln K


At Equilibrium, ∆G = 0


From eq (1), we can write:


∆G° + RTln K = 0


So, RTln K = -∆G°


( Log 1 = 0, so we get, RTln K = 0)


k = 1


Hence value of K will become 1.


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