Q. Formaldehyde polymerizes to form glucose according to the reaction, $6\ HCHO \rightleftharpoons C_{6}H_{12}O_{6}$ 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
Explanation
(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,
$\left\lbrack C_{6}H_{12}O_{6} \right\rbrack = 1\ M$
$K = \frac{\left\lbrack C_{6}H_{12}O_{6} \right\rbrack}{\lbrack HCHO\rbrack^{6}},\ \ \ \ \ \ \ \ \ i.e.,\ 6 \times 10^{22} = \frac{1}{\lbrack HCHO\rbrack^{6}}$
or $\lbrack HCHO\rbrack = \left( \frac{1}{6 \times 10^{22}} \right)^{\frac{1}{6}} = 1.6 \times 10^{- 4}$ M
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.