Question

Formaldehyde polymerizes to form glucose according to the reaction. $\sf HCHO\rightleftharpoons{C_6H_{12}O_6}$. Theoretically computed equilibrium constant for this reaction is found to be$\sf 6\times10^{22}$ . If 1 M solution of glucose dissociates according to the above equilibrium, the concentration of formaldehyde in the solution will be

Answer 1

$\large \pmb{ \sf{Question : }}$

Formaldehyde polymerizes to form glucose according to the reaction. $\sf HCHO\rightleftharpoons{C_6H_{12}O_6}$. Theoretically computed equilibrium constant for this reaction is found to be $\sf 6\times10^{22}$ . If 1 M solution of glucose dissociates according to the above equilibrium, the concentration of formaldehyde in the solution will be

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$\large \pmb{ \sf{Solution :}}$

$\sf HCHO\rightleftharpoons{C_6H_{12}O_6}$

As equilibrium constant for this reaction is very large, therfore for the reverse reaction involving dissociation of glucose, equilibrium constant is very very small . Hence , for the reverse reaction

$\sf{C_6H_{12}O_6 \rightleftharpoons6HCHO} \sf{\left(K = \frac{1}{6 \times {10}^{22} } \right)^{ \frac{1}{6} }}$

dissociation of glucose is negligible . Starting with 1 M, concentration at equilibrium at equilibrium $≅$1M

$\sf{[HCHO] = \frac{1}{6 \times {10}^{22} } }$

$\sf{[HCHO] = \left( \frac{1}{6 \times {10}^{22} } \right)^{ \frac{1}{6} } = \frac{ {(1)}^{ \frac{1}{6} } }{ {(6)}^{ \frac{1}{6} } \times {(10}^{22}) ^\frac{1}{6} } }$

$\sf{[HCHO] = \frac{1}{1.348 \times 4641.588} }$

$\sf{[HCHO] = \frac{1}{6256.86} = 0.000159} \\ \\ \pink{ \underline{\boxed{\large{ \sf{ = 1.6 \times {10}^{ - 4} M} }}}}$

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$\pmb{ \sf{Note : }}$

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Answer 2

EXPLANATION.

Formaldehyde polymerizes to form glucose.

⇒ HCHO ⇆ C₆H₁₂O₆.

Equilibrium constant for the reaction = 6 x 10²².

If 1M solution of glucose dissociates by above equilibrium.

As we know that,

First we balance the equation, we get.

6HCHO ⇆ C₆H₁₂O₆.

                    1M.

⇒ k = [C₆H₁₂O₆]/[HCHO]⁶.

Put the value in equation, we get.

⇒ 6 x 10²² = 1/[HCHO]⁶.

⇒ [HCHO]⁶ = 1/6 x 10²².

⇒ [HCHO] = (1)^1/6/(6)^1/6 x (10²²)^1/6.

⇒ [HCHO] = 1/1.348 x 4641.588.

⇒ [HCHO] = 1/6256.86.

⇒ [HCHO] = 0.000159.

⇒ [HCHO] = 1.6 x 10⁻⁴M.

                                                                                                                       

MORE INFORMATION.

Free energy changes (ΔG).

(1) = If ΔG = 0 Then reversible reaction would be in equilibrium.

(2) = If ΔG = (+)ve Then equilibrium will be displace in backward direction.

(3) = If ΔG = (-)ve Then equilibrium will displace in forward direction.