Question

When 1-pentyne (A) is treated with 4N alcoholic KOH at 175°C, it is converted slowly into an equilibrium mixture of 1.3% 1-pentyne (A), 95.2% 2-pentyne (B) and 3.5% of 1, 2 pentadiene (C) the equilibrium was maintained at 175°C, calculate ΔG⁰ for the following equilibria.

\[\ce{B ⇌ A}\] `Δ"G"_1^0` = ?

\[\ce{B ⇌ C}\] `Δ"G"_2^0` = ?

Answer 1

T = 175°C = 175 + 273 = 448 K

Concentration of 1-pentyne [A] = 1.3%

Concentration of 2-pentyne [B] = 95.2%

Concentration of 1, 2-pentadiene [C] = 3.5%

At equiLibrium

\[\ce{B ⇌ A}\]

95.2% 1.3% = K₁ = `1.3/95.2` = 0.0136

\[\ce{B ⇌ C}\]

95.2% 3.5% = K₂ = `3.5/95.2` = 0.0367

`Δ"G"_1^0` = −2.303 RT log K₁

`Δ"G"_1^0` = – 2.303 × 8.314 × 448 × log 0.0136

`Δ"G"_1^0` = + 16010 J

`Δ"G"_1^0` = + 16 kJ

`Δ"G"_2^0` = – 2.303 RT log K₂

`Δ"G"_2^0` = – 2.303 × 8.314 × 448 × log 0.0367

`Δ"G"_2^0` = + 12312 J

`Δ"G"_2^0` = + 12.312 kJ