Question

write the equilibrium constant expression for this reaction: 2ch₃coo⁻(aq)+c₂h₆o₂(aq)→c₆h₁₀o₄(aq)+2oh⁻(aq)

Explanation:

Step1: Recall the formula for equilibrium constant

For a general reaction \(aA + bB
ightleftharpoons cC + dD\), the equilibrium constant \(K=\frac{[C]^c[D]^d}{[A]^a[B]^b}\), where \([X]\) represents the molar concentration of species \(X\).

Step2: Identify reactants and products

In the reaction \(2\mathrm{CH_3COO^-}(aq)+\mathrm{C_2H_6O_2}(aq)
ightleftharpoons\mathrm{C_6H_{10}O_4}(aq)+2\mathrm{OH^-}(aq)\), the reactants are \(\mathrm{CH_3COO^-}\) and \(\mathrm{C_2H_6O_2}\), and the products are \(\mathrm{C_6H_{10}O_4}\) and \(\mathrm{OH^-}\).

Step3: Apply the formula

Using the formula \(K=\frac{[C]^c[D]^d}{[A]^a[B]^b}\), with \(a = 2\) (for \(\mathrm{CH_3COO^-}\)), \(b=1\) (for \(\mathrm{C_2H_6O_2}\)), \(c = 1\) (for \(\mathrm{C_6H_{10}O_4}\)), and \(d = 2\) (for \(\mathrm{OH^-}\)), we get \(K=\frac{[\mathrm{C_6H_{10}O_4}][\mathrm{OH^-}]^2}{[\mathrm{CH_3COO^-}]^2[\mathrm{C_2H_6O_2}]}\).

Answer:

\(K=\frac{[\mathrm{C_6H_{10}O_4}][\mathrm{OH^-}]^2}{[\mathrm{CH_3COO^-}]^2[\mathrm{C_2H_6O_2}]}\)