Solveeit Logo
Login

Question

Chemistry Question on Equilibrium

Formaldehyde polymerizes to form glucose according to the reaction 6HCHO⇌C6H12O66HCHO\rightleftharpoons {{C}_{6}}{{H}_{12}}{{O}_{6}} The theoretically computed equilibrium constant for this reaction is found to be 6×10226\times {{10}^{22}} If 1 M1\, M solution of glucose dissociates according to the above equilibrium, the concentration of formaldehyde in the solution will be

A

1.6×10−2M1.6\times {{10}^{-2}}M

B

1.6×10−4M1.6\times {{10}^{-4}}M

C

1.6×10−6M1.6\times {{10}^{-6}}M

D

1.6×10−8M1.6\times {{10}^{-8}}M

Answer

1.6×10−4M1.6\times {{10}^{-4}}M

Explanation

Solution

A very high value of KK for the given equilibrium shows that dissociation of glucose to form HCHO is very-very small. Hence, at equilibrium, we can take, [C6H12O6]=1M\left[ C _{6} H _{12} O _{6}\right] =1 M K=[C6H12Oc][HCHO]6K =\frac{\left[ C _{6} H _{12} O _{ c }\right]}{[ HCHO ]^{6}} ie, 6×1022=1[HCHO]66 \times 10^{22}=\frac{1}{[ HCHO ]^{6}} or [HCHO]=(16×1022)1/6[ HCHO ] =\left(\frac{1}{6 \times 10^{22}}\right)^{1 / 6} =1.6×10−4M=1.6 \times 10^{-4} M