Question

When ethene is treated in a calorimeter with H₂and a Ptcatalyst, the heat of reaction is found to be -137kJ/mol(-32.7kcal/mol), and the reaction goes to completion. When the reaction takes place at 1400⁰K, the equilibrium is found to be evenly balanced, with K_eq=1. Compute the value of ΔSfor this reaction.

Short answer

ΔS = -98J/mol.K

Step-by-step solution

Equilibrium constant (Keq)

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

Position of equilibrium

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 .

Expression for Gibb’s free energy (ΔG)

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

Relationship between ,ΔG  ΔH , and ΔS.

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

Calculation

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 - TΔS

0= ΔH - TΔS

ΔS =ΔH / T

ΔS= -137kJ/mol/1400K

ΔS= -0.09785kJ/mol.K

ΔS= -97.85 J/mol.K

ΔS= -98J/mol.K

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