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

Explanation:

Step1: Recall equilibrium - constant formula

For a reaction $aA + bB\rightleftharpoons cC + dD$, the equilibrium - constant expression $K=\frac{[C]^c[D]^d}{[A]^a[B]^b}$, where [ ] represents the molar concentration of the species and a, b, c, d are the stoichiometric coefficients.

Step2: Identify reactants and products and their coefficients

In the reaction $2CH_3COO^-(aq)+C_2H_6O_2(aq)\rightleftharpoons C_6H_{10}O_4(aq)+2OH^-(aq)$, the reactants are $CH_3COO^-$ with coefficient $a = 2$ and $C_2H_6O_2$ with coefficient $b = 1$, and the products are $C_6H_{10}O_4$ with coefficient $c = 1$ and $OH^-$ with coefficient $d = 2$.

Step3: Write the equilibrium - constant expression

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

Answer:

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