Dissociation of a gas A₂ takes place according to the following chemical reaction. At equilibrium, the total pressure is 1 bar at 300 K. A₂(g) ⇌ 2A(g) The standard Gibbs energy of formation of the involved substances has been provided below: Substance ΔG°f / kJ mol⁻¹ A₂ −100.00 A −50.832 The degree of dissociation of A₂(g) is given by (x × 10⁻²)¹ᐟ² where x = ____ . (Nearest integer) [ Given : R = 8 J mol⁻¹ K⁻¹ , log 2 = 0.3010 ] Assume degree of dissociation is not negligible.

Dissociation of a gas A₂ takes place according to the following chemical reaction. At equilibrium, the total pressure is 1 bar at 300 K.

Q. Dissociation of a gas A₂ takes place according to the following chemical reaction. At equilibrium, the total pressure is 1 bar at 300 K.

A₂(g) ⇌ 2A(g)

Substance    ΔG°f / kJ mol⁻¹
A₂    −100.00
A    −50.832

The degree of dissociation of A₂(g) is given by (x × 10⁻²)¹ᐟ² where x = ____ . (Nearest integer)

[ Given : R = 8 J mol⁻¹ K⁻¹ , log 2 = 0.3010 ]
Assume degree of dissociation is not negligible.

Correct Answer: 33

Step 1: Calculate ΔG° of Reaction

ΔG° = Σ ΔG°f (products) − Σ ΔG°f (reactants)

= [2 × (−50.832)] − [−100.00]

= −101.664 + 100

= −1.664 kJ mol⁻¹

Step 2: Convert into Joules

−1.664 kJ = −1664 J

Step 3: Use Relation ΔG° = −RT ln Kp

−1664 = − (8 × 300) ln Kp

−1664 = −2400 ln Kp

Step 4: Solve for ln Kp

ln Kp = 1664 / 2400

ln Kp = 0.693

Step 5: Convert into Kp

ln Kp = 0.693

Kp = e^0.693

Kp = 2

Step 6: Assume 1 mole A₂ Initially

Initial moles = 1

Dissociation = α

Step 7: Equilibrium Moles

A₂ = 1 − α

A = 2α

Total moles = 1 − α + 2α = 1 + α

Step 8: Expression of Kp

Kp = (P_A)^2 / P_A2

= [ (2α / 1+α)² ] / [ (1−α)/(1+α) ]

= 4α² / (1 − α²)

Step 9: Put Kp = 2

2 = 4α² / (1 − α²)

2(1 − α²) = 4α²

2 − 2α² = 4α²

2 = 6α²

α² = 1/3

Step 10: Compare with Given Form

α = (1/3)¹ᐟ²

= (33 × 10⁻²)¹ᐟ²

Therefore, x = 33