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.