Question

The theoretically computed equilibrium constant for the polymerisation of formaldehyde to glucose in aqueous solution:
$6HCHO\rightleftharpoons C_6{H}_{12}{O}_6$
is $1.0\times {10}^{24}$. If $1$ M-solution of glucose was taken, what should be the equilibrium concentration of formaldehyde?

• A. $6.0\times {10}^{-4}$M
• B. $1.0\times {10}^{-4}$M
• C. $1.0\times {10}^4$M
• D. ${\left(\frac{1}{6}\right)}^{1/6}\times {10}^{-4}$M

Answer 1

The correct option is B $1.0\times {10}^{-4}$M

The given reaction is :-

$6HCHO\rightleftharpoons C_6{H}_{12}{O}_6{K}_C=1.0\times {10}^{24}$

Now, for $C_6{H}_{12}{O}_6\rightleftharpoons 6HCHO;{K_C}^{{}^{\prime }}=\frac{1}{K_C}={10}^{-24}$

Now, ${K_C}^{{}^{\prime }}=\frac{{\left[HCHO\right]}^6}{\left[C_6{H}_{12}{O}_6\right]}$

$\Rightarrow {10}^{-24}=\frac{{\left[HCHO\right]}^6}{1}$

$\Rightarrow \left[HC{H}_O\right]={\left({10}^{-24}\right)}^{1/6}={10}^{-4}M$