Free energy change and net useful work



Work other than pressure volume work (PΔV) is net useful work. So, total work of a process is,

W = Wnet + PΔV

Wnet = W – PΔV

From Gibb’s helmuntz equation,

           ΔG = H – TΔS      ……………. (i)

We have,

            H = E +   PΔV

            S = Q /T

            TΔS = Q

Putting the values of   H and TS  in eqn (i), we get

            ΔG = ΔE   + PΔV – Q

From first law of thermodynamics

            Q = ΔE + W

Putting the value of Q in above enq we get,

            G =  Δ E + PΔV – ΔE – W

            G = PΔV – W

            G = – (W – PΔV)

            G = – Wnet

            Wnet is – ΔG

Thus the net useful work is the amount of decrease in free energy of a system under constant temperature and pressure.

 

Relation between free energy change and cell potential

The net work done by galvanic cell in carrying charge is

W = charge × potential difference = nf Ecell

We have,

            ΔG = -Wnet

            ΔG = -nfEcell

 Relation between G and equilibrium constant

The variation of G is given by Nernst equation as,

            G =    ΔG + RT in Q

Where, Q = reaction quotient

            free energy

Where, the reaction is at equilibrium

            g = k (equilibrium constant)

            ΔG = O

            O = G°  + RT in k

            Δ = -RT InK

            ΔG = -2.303 RT logK

 The overall reaction for corrosion of iron by oxygen is,

4Fe (s) + 3O2 (g)            →        2Fe2O3 (s) Rust

Calculate the free energy change and equilibrium constant for this reaction at 250°c.

Given,

            Δ and    Δ  = -543 JK-1 and -1652 KJ mol-1

                        = -1652000 Jmol-1

           Δ   =   Δ   – TΔ

                = 1652000 – 298 × (-543)

               = -1490186 J

Again,

            Δ = -RTInK

              = -2.303 RTlogK

            1490186 = – 2.303 × 8.314 × 298 × log K

            logK = 261.17

            K = 10261.17

            K = 1.48 × 10261.17

 

 

 

Calculate the free energy at standard condition for galvanic cell having following cell reaction.

2Al (s) + 3 Cu ++ (aq)            →            2Al+++ (aq) + 3Cu (s)

E° Cu++/Cu = 0.34 V           →             E° Al+++/A1 = -1.66V

So, the standard cell potential,

            E°cell = E°cathode – E°anode

                        = E°cu++/cu – E°Al+++/A1

                        = 0.4 – (-1.66)

                        = 2V

Again,  Δ G = -nfEcell

                        = – 6 × 96500 × 2

 

                        = 1.158 × 106J

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