Gibbs Free Energy & Gibbs- Helmholtz equation | Physical Chemistry PDF Download

Helmholtz free energy or Helmholtz function or work function.

The work function (A) is defined as

A = U – TS

Let an isothermal change at temperature T from the initial state indicated by subscript (1) to final state subscript (Z) so that
                 A₁ = U₁ – TS₁
&              A₂ =  – TS₂
                A₂ – A₁ = (U₂ – U₁) – T(S₂ – S₁)
                ΔA = ΔU – TΔS
or             dA = dU – TdS

                 dA = dU – TdS                                                         [∴ TdS = dq]
                 dA = dU – dq

From first law dq = dU – ω or  dU – dq = ω i.e.
dA = ω

Free energy function Gibb’s free energy function.
The free energy funct ion G is given by G = H – TS
Let us consider the change of system from (1) to (2) at constant temperature, we have

G₂ – G₁ = (H₂ – H₁) – T(S₂ – S₁)

ΔG = ΔH – TΔS
dG = dH – TdS

Variation of free energy with pressure:

We know that

dG = VdP – SdT

when process is isothermal then 

dT = 0

(dG)_T = V(dP)_T
⇒  

When pressure is constant the
dP = 0

(dG) _P = –S(dT)_P
⇒ 

Isothermal change in free energy for ideal gas
dG = VdP

Integrating wit h the limit s G₁ and G₂, P₁ and P₂ we get

we know that P₁V₁ = P₂V₂
then

Isothermal change in free energy for solid and liquid The volume occupied by solid and liquid is nearly independent of pressure:

dG = VdP
G₂ – G₁ = V(P₂ – P₁)

Problem.  Two mole of an ideal gas are allowed to expand reversibly and isothermally at 300 K from a pressure of 1 atm to a pressure of 0.1 atm.  What is the change is Gibb’s free energy.
Sol.

 
= 2.303 x 2 x 8.314 x 300 log 

ΔG = –11488.2 J = –11.4882 KJ

Gibbs Helmholtz equation. 

(A) Temperature depends of free energy we know that

         G = H – TS

&      dG = VdP – SdT
         dG = VdP – SdT

As pressure is constant, therefore

              (dG)P = –S(dT)P
∴            dG₁ = –S₁dT

S₁ & S₂ are entropy in the init ial and final states.

dG₂ = –S₂dT

Subtracting these equation we get
        dG₂ – dG₁ = –(S₂ – S₁)dT
   

 &   

We know that 

                      …(1)
This is Gibb’s Helmholtz equation.

ΔG = ΔH - T ΔS
⇒

Differentiating this equation w.r.t. T at constant P we get

 


                 ... (2)
 

(B)  Temperature dependence of Helmholtz function:

We know that      A = U – TS
                           ΔA = ΔU – TΔS
&                         dA = –PdV – SdT

As vo lume is constant, therefore

dA = –SdT dA₁ = –S₁dT

S₁ and S₂ are init ial and final entropy. dA₂ = –S₂dT Subtracting these equation we get
dA₂ – dA₁ = –(S₂ – S₁)dT
dΔA = –ΔSdT

⇒ 

This is Gibbs Helmholtz equation.

ΔA = ΔU – TΔS
⇒ 

Different iat ing this equation w.r.t. T at constant V we get

 


…(2)

Problem.  Prove that

Sol. (a) ΔG = ΔH – TΔS dividing by T we get

differentiation of by T at constant P, we get

 
or 

Since        therefore, we have

i.e. 

Gibbs Helmholtz equation.

(b)  Similarly

Gibbs Helmholtz equation.
 Problem.  For the reaction.

The value of enthalpy change and free energy change are –68.32 and –56.69 K cals respectively.  

Calculate the value of free energy change at 30°C.
 Sol. We know that 

T = 25° = 273 + 25 = 298 K
–56.69 = 68.32 298 
⇒ 
then at 30°C  

                         = –68.32 + 303 × 0.0390
                         ΔG = –56.495 kcal.