Gibbs energy-

Gibbs energy of a system is defined as the maximum amount of energy available to the system during a process that can be concentrated into a useful work.

                                                Or

The amount of energy i.e. converted into useful work is known as Gibbs free energy & it is denoted by G.

G = H – TS

G₁ = H₁ – TS₁

G₂ = H₂ – TS₂

ΔG = ΔH - TΔS → change in free energy & its units are KJ/mol or J/mol

Significance of Gibbs free enrgy

Gibbs free energy is used to measure the useful work obtained from the system

Work done is of types

W exp (PΔV) + W non exp (useful work done)

 ΔE = q –w

q = ΔE + W_exp + W_{non exp}

q = ΔE + PΔV + W_useful

 TΔs = ΔH + W_useful (Δs = (q/T), q = TΔs)

W_useful = ΔH - TΔs

W_useful = -ΔG

Decrease in free energy is equal to useful work done.

Standard free energy change-

It is the amount of energy change when 1mole of the substance is formed from the reactant in their standard state. & denoted by ΔG°

ΔG° = ∑ of G° Product - ∑ of G° Reactant

Question. Enthalpy & entropy changes of a reaction are 40.63 KJ/mol & 108.8 J/K respectively. Predict the feasibility of reaction at 27°C

40630 J/mol

ΔG = ΔH - TΔS

ΔG = 40630 - 300 x 108.8

ΔG = 40630 – 32640

ΔG = 7990

Reaction is not fisible.

Question. For the reaction at 298 K, 2A + B gives C

ΔH = 400 KJ/mol

ΔS = 2 KJ/K mol 

At what temperature the reaction become spontaneous.

At equilibrium, ΔG = 0

ΔG = ΔH - TΔS

TΔS = ΔH

T = (ΔH / ΔS)

  T = (40.0000/2000)

T = 2000 K.

Question. ΔH & ΔS for the reaction Ag₂O → 2Ag + ½ O₂ are 30.56 KJ & 6.6 J/K mol respectively Calculate the temperature at which ΔG is O. What will be direction of the reaction at this temp. at temp. below this temp

 ΔG = ΔH - TΔS

TΔS = ΔH

  T = ΔH/ΔS

  T = 305600/6.6

T = 4630.3

T = 4630 K

At this temperatire ΔG is 0.

If ΔG is negative, reaction is spontaneous

ΔG = - ve

∆H = + ve

TΔS = + ve

TΔS > ΔH                 Direction is forward

If ΔG is positive, reaction is non-spontaneous

ΔH = + ve

TΔS = + ve

ΔH > TΔS

ΔG = +ve                 Direction is backward

If ΔG is zero, reaction is at equilibrium

ΔH = + ve

TΔS = + ve

ΔH = TΔS

ΔG = 0      

There is no direction at 4630 K temp. bcoz reaction is at equilibrium

If temp. below this temp. then ΔH > TΔS, thus direction is backward as reactions is non-spontaneous

Question. At what temp. does the reaction of lead oxide to lead by carbon become spontaneous

PbO + C → Pb + CO

For this reaction, ΔH & ΔS are 108.4 KJ/mol & 190 J/K mol respectively.

ΔG = ΔH - TΔS

ΔH = TΔS

  T = ΔH/ΔS

  T = 108400/190

T = 570.5 K

At equilibrium

If temp. is greater than 570.5 K, reaction is spontaneous

ΔG = - ve