Tamil Nadu Board of Secondary EducationHSC Science Class 11th

# 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) - Chemistry

Numerical

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

Concept: Gibbs Free Energy (G)
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#### APPEARS IN

Tamil Nadu Board Samacheer Kalvi Class 11th Chemistry Volume 1 and 2 Answers Guide
Chapter 7 Thermodynamics
Evaluation | Q II. 41. | Page 227
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