Question

Formaldehyde polymerizes to form glucose according to the reaction
$6HCHO\rightleftharpoons {{C}_{6}}{{H}_{12}}{{O}_{6}}$ . The theoretically computed equilibrium constant for this reaction is found to be $6\times {{10}^{22}}$ . If 1 M solution of glucose dissociates according to the above equilibrium, the concentration of formaldehyde in the solution will be

Answer 1

Hint: The dissociation constant or equilibrium constant is the ratio of products to original chemical (reactant). It is denoted as Ka. The formula to represent the equilibrium constant of a reaction is as follows.
\[Ka=\dfrac{concentration\text{ }of\text{ }the\text{ }products}{concentration\text{ }of\text{ }the\text{ }reac\tan ts}\]

Complete step by step answer:
- In the question it is given that 6 moles of formaldehyde polymerizes and forms one mole of glucose.
- The representation of the above statement is as follows.
$6HCHO\rightleftharpoons {{C}_{6}}{{H}_{12}}{{O}_{6}}$
- The equilibrium constant of the above reaction is given as $6\times {{10}^{22}}$ (Ka).
- Now we have to find the concentration of the formaldehyde solution, if the 1 M solution of glucose dissociates according to the given equilibrium.
- By using the below formula we can calculate the concentration of the formaldehyde in the solution.
\[Ka=\dfrac{concentration\text{ }of\text{ }the\text{ }products}{concentration\text{ }of\text{ }the\text{ }reac\tan ts}\]
Here Ka = $6\times {{10}^{22}}$
Concentration of the reactants = 1 M
Concentration of the product (formaldehyde) = ?
- Substitute the known values in the above formula to get the concentration of the formaldehyde in the solution.
\[\begin{align}
  & Ka=\dfrac{concentration\text{ }of\text{ }the\text{ }products}{concentration\text{ }of\text{ }the\text{ }reac\tan ts} \\
 & 6\times {{10}^{22}}=\dfrac{1}{{{[HCHO]}^{6}}} \\
 & {{[HCHO]}^{6}}=\dfrac{1}{6\times {{10}^{22}}} \\
 & [HCHO]={{\left( \dfrac{1}{6\times {{10}^{22}}} \right)}^{\dfrac{1}{6}}} \\
 & [HCHO]=\dfrac{1}{{{\left( 6 \right)}^{\dfrac{1}{6}}}\times {{\left( {{10}^{22}} \right)}^{\dfrac{1}{6}}}} \\
 & [HCHO]=\dfrac{1}{6256.86} \\
 & [HCHO]=0.000159\approx 1.6\times {{10}^{-4}}M \\
\end{align}\]
Therefore the concentration of the formaldehyde in the solution is $1.6\times {{10}^{-4}}M$.

Note: Formaldehyde and glucose both are organic molecules and soluble in water. In the polymerization reaction, formaldehyde associates or polymerizes and forms glucose as the product. The reverse reaction is called dissociation reaction and glucose converts back into formaldehyde.