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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` = ?
Solution
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
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