The theoretically computed equilibrium constant for the polymerization of formaldehyde to glucose in — Chemical Equilibrium Chemistry Question
Question
The theoretically computed equilibrium constant for the polymerization of formaldehyde to glucose in aqueous solution is 6.4 × 10¹⁹ M⁻⁵. If 1 M solution of glucose were taken, what would be the equilibrium concentration of formaldehyde, in milligram per litre?
💡 Solution & Explanation
Polymerization reaction: 6HCHO ⇌ C₆H₁₂O₆. K_c = [Glucose] / [HCHO]⁶ = 6.4 × 10¹⁹. Initial glucose = 1 M. Let x be the equilibrium concentration of HCHO. [Glucose] at equilibrium = 1 - x/6 ≈ 1 M (since K_c is very large, x is very small). 1 / x⁶ = 6.4 × 10¹⁹ = 64 × 10¹⁸. x⁶ = 1 / (64 × 10¹⁸) = (1/64) × 10⁻¹⁸. Taking the 6th root: x = (1/2) × 10⁻³ = 5 × 10⁻⁴ M. Molar mass of formaldehyde (HCHO) = 30 g/mol. Concentration in mg/L = 5 × 10⁻⁴ mol/L × 30 g/mol × 1000 mg/g = 15 mg/L. Therefore, correct answer is 0015.