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

• A. $ 1.6\times {{10}^{-2}}M $
• B. $ 1.6\times {{10}^{-4}}M $
• C. $ 1.6\times {{10}^{-6}}M $
• D. $ 1.6\times {{10}^{-8}}M $

Answer 1

Answer: (B)

A very high value of $K$ for the given equilibrium shows that dissociation of glucose to form HCHO is very-very small. Hence, at equilibrium, we can take,
$\left[ C _{6} H _{12} O _{6}\right] =1 M$
$K =\frac{\left[ C _{6} H _{12} O _{ c }\right]}{[ HCHO ]^{6}} $
ie, $ 6 \times 10^{22}=\frac{1}{[ HCHO ]^{6}} $
or $[ HCHO ] =\left(\frac{1}{6 \times 10^{22}}\right)^{1 / 6}$
$=1.6 \times 10^{-4} M$