91Ó°ÊÓ

Equilibrium constant(K_eq) of the reaction governs the equilibrium concentration of the reactants and products. For a general reaction of the type, aA,bB ⇌ cC,dD the equilibrium constant (K_eq)expression can be written as:

K_eq=[products]/[reactants]

=[C]^c[D]^d/[A]^a[B]^b

The position of equilibrium can be predicted form the value of equilibrium constant (K_eq) . The forward reaction is favored if K_eq>1 and backward reaction is favored if K_eq<1 .

The expression for Gibb’s free energy is ΔG= -RTInK_eq. This expression can also be written as ΔG= -2.303 RT(logK_eq)

The relationship between Gibb’s free energy(Δ G) , enthalpy(Δ H) and entropy(Δ S) is ΔG = ΔH - TΔ³§ .

As per given data,

K_eq=1

T = 1400K

ΔH= -137kJ/mol

Now,

ΔG = -2.303RT(logK_eq)

= -2.303RT * log(1)

= -2.303RT*0

=0

Again,

ΔG = ΔH - TΔ³§

0= ΔH - TΔ³§

Δ³§ =ΔH / T

Δ³§= -137kJ/mol/1400K

Δ³§= -0.09785kJ/mol.K

Δ³§= -97.85 J/mol.K

Δ³§= -98J/mol.K

Hence, the value of Δ³§ for the reaction is found to be -98J/mol.K.