What is Gibbs Free Energy?

In the previous post of thermodynamics we have seen how the change in entropy decides the spontaneity of any reaction. But entropy alone cannot decide spontaneity, enthalpy of the reaction and temperature at which the reaction is being carried out also plays an important role. To sum up all these factors we have another thermodynamic property known as Gibbs free energy “G”.

G = H – TS

‘G’ is an extensive property. It is a state function as it doesn’t depend on path. Units of Gibbs energy is J.

Let’s find change in Gibbs free energy of the system at constant temperature:

ΔG_sys = ΔH_sys – TΔS_sys    ......(1)

This is one of the important equations of thermodynamics. Difference in G decides the spontaneity of any reaction, let’s find out how.

We know that total entropy change for a spontaneous reaction must be more than zero. Total entropy means the entropy of the system and surrounding.

ΔS_total = ΔS_sys  + ΔS_surr       ......(2)

For an exothermic reaction ΔH heat is released from the system to the surrounding. It means when -ΔH_sys heat is lost from the system, it is added to the surrounding as ΔH_sur.

ΔS_sys = -ΔH_sys / T

Then,

ΔS_sur = ΔH_sur / T

Since ΔH_sur = - ΔH_sys

ΔS_sur = ΔH_sur / T = - ΔH_sys / T

On putting the values of ΔS_sur in equation 2 we get,

ΔS_total  = ΔS_sys + (-ΔH_sys / T)

TΔS_total  = TΔS_sys  - ΔH_sys

TΔS_total  =  - (ΔH_sys  - TΔS_sys)

For spontaneity ΔS_total must be greater than zero, so

- (ΔH_sys  - TΔS_sys) > 0

On comparing it with equation 1 we can say that for spontaneity,

-ΔG_sys > 0

- ΔG_sys can be more than zero only when ΔG_sys is negative,

- (-ΔG_sys)  > 0

Or

ΔG_sys  < 0

So, for spontaneity ΔG_sys must be negative.  
 

Gibbs Free Energy and Spontaneity

We know that ΔG_sys = ΔH_sys – TΔS_sys, where ΔH_sys is the enthalpy of the system and TΔS_sys is the energy required to be released or absorbed by the system and by subtracting them you will get the remaining energy ΔG which can be used to do any work. That’s why ΔG is known as free energy which is available to do useful work. For any spontaneous reaction ΔG is negative which means all energy has been used to carry the reaction forward (spontaneous reactions are irreversible). Reactions for which ΔG is negative are known as ‘exergonic reactions’. ‘Endergonic reactions’ have positive ΔG.

Let’s see how we can predict the spontaneity of any reaction by using equation of ΔG:

A + B ⟶ C + D ...... ΔH = (+ve) and ΔS = (+ve)

On applying Gibbs equation

 ΔG_sys = ΔH_sys – TΔS_sys

At lower temperature ΔG will be positive so that above reaction will be non-spontaneous at lower temperature. But if temperature is higher enough to overrule the ΔH, then it can be spontaneous.

Now we understand that if ΔG < 0 then reaction will be spontaneous, and if ΔG > 0 then it will be non-spontaneous. So what happens if ΔG = 0? Reaction will be at equilibrium if ΔG = 0.

ΔG = ΔG^ө + RT lnQ

At equilibrium reaction quotient Q becomes equal to equilibrium constant K and ΔG = 0

ΔG = ΔG^ө + RT lnK = 0

ΔG^ө =- RT lnK

Or

 ΔG^ө =- 2.303RT logK

lnK = -ΔG^ө/ RT

Taking antilog of both side,

K = e ^(-ΔGө/RT)

where ΔG^ө is the standard gibbs energy of the reaction.

When ΔG^ө < 0 then  (-ΔG^ө/ RT) will be positive and e ^{(-ΔGө /RT)} will be more than zero. That means K > 1 which means forward reaction is favoured and equilibrium shifts forward. Here product concentration exceeds reactant concentration.

When ΔG^ө > 0 then  (-ΔG^ө/ RT) will be negative and e ^{(-ΔGө /RT)} will be less than zero. That means K < 1 which means backward reaction is favoured and equilibrium shifts backward and product formation is suppressed.

I hope these posts have been helpful in understanding the thermodynamic concepts. The next post of thermodynamics will focus on the applicability of these concepts where we will try to apply these concepts to solve some numerical problems.

We have discussed all about feasibility and spontaneity of any reaction but how do we decide the speed of the reaction? On which factors it depends? Can we accelerate its speed? We will find the way to solve this mystery in the coming posts of chemical kinetics.

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