A 1 , M solution of glucose reaches dissociat | vedclass

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(D) The given reaction is $C_6H_{12}O_6 ightleftharpoons 6HCHO$. The formation reaction is $6HCHO ightleftharpoons C_6H_{12}O_6$ with $K_1 = 6 	i

Vedclass · Accepted Answer

$1.6 	imes 10^{-4} \, M$

Vedclass · Answer

(D) The given reaction is $C_6H_{12}O_6 ightleftharpoons 6HCHO$. The formation reaction is $6HCHO ightleftharpoons C_6H_{12}O_6$ with $K_1 = 6 	imes 10^{22}$. The dissociation reaction is the reverse,so $K_2 = \frac{1}{K_1} = \frac{1}{6 	imes 10^{22}}$. For the reaction $C_6H_{12}O_6 ightleftharpoons 6HCHO$,the equilibrium constant $K = (K_2)^{1/6} = (\frac{1}{6 	imes 10^{22}})^{1/6} \approx 1.6 	imes 10^{-4}$. Since the initial concentration of glucose is $1 \, M$,at equilibrium $[C_6H_{12}O_6] \approx 1 \, M$. Thus,$[HCHO]^6 = K 	imes [C_6H_{12}O_6] = 1.6 	imes 10^{-4} 	imes 1$. $[HCHO] = (1.6 	imes 10^{-4})^{1/6} \approx 0.26 \, M$. Note: Given the options,the question implies calculating the equilibrium constant for the dissociation step directly as $K = (6 	imes 10^{22})^{-1/6} \approx 1.6 	imes 10^{-4}$.

$A$ $1 \, M$ solution of glucose reaches dissociation equilibrium according to the equation $C_6H_{12}O_6 \rightleftharpoons 6HCHO$. What is the concentration of $HCHO$ at equilibrium if the equilibrium constant $K_c$ for the formation of glucose from formaldehyde is $6 \times 10^{22}$?

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