Question

At 25 ∘C , the equilibrium partial pressures for the reaction 3A(g)+3B(g)↽−−⇀C(g)+3D(g) were found to be ?A=4.55 atm, ?B=4.49 atm, ?C=5.77 atm, and ?D=5.62 atm. What is the standard change in Gibbs free energy of this reaction at 25 ∘C ? (Answer in kj/mol)

Answer 1

The equation is given as:

3A (g) + 3B (g) <---> C(g) + 3D (g)

The partial pressures of the components are given:

P_{​​​​​​A} = 4.55 atm

P_{​​​​​​B} = 4.49 atm

P_{​​​​​​C} = 5.77 atm

P_{​​​​​​D} = 5.62 atm

The standard change in Gibb's free energy (∆G°) = ?

We know,

∆G = ∆G° + RT lnQ ........(1)

Where, ∆G = Gibb's free energy

R= universal gas constant

T = temperature

Q = reaction quotient

As the reaction is at equilibrium,

∆G = 0

lnQ = ln Keq , where Keq = equilibrium constant of the reaction

Thus, equation (1) becomes:

∆G° = -RT lnKeq ....... (2)

Given, T = 25° C = (25 + 273) K = 298 K

We know ,R = 8.314 J/ K .mol

Now, using the given partial pressures, we must find out Keq.

Keq = (P_{​​​​​​C} × P_{​​​​​​D}^​​​3 ) / (P_​​A ³ × P_{​​​​​​B} ³ )

==> Keq = [ (5.77) × (5.62)³ ] / [ (4.55)³ × (4.49)³ ]

==> Keq = 0.12 atm^{​​​​-2}

Substituting the values in equation (2),

∆G° = -RT lnKeq

==> ∆G° = - (8.314 J/K.mol) × (298 K) × ln(0.12)

==> ∆G° = 5253.11 J/mol = 5.25311 kJ/mol

Rounding up to 2 decimal digits,

∆G° = 5.25 kJ/mol