Question

A $1\, M$ solution of glucose reaches dissociation equilibrium according to equation given below $6HCHO\rightleftharpoons C_{6}H_{12}O_{6}$. What is the concentration of HCHO at equilibrium if equilibrium constant is $6\times10^{22}$

• A. $1.6 \times 10^{-8}M$
• B. $32 \times 10^{-6}M$
• C. $3.2 \times 10^{-4} M$
• D. $1.6 \times 10^{-4}M$

Answer 1

Answer: (D)

$6HCHO \rightleftharpoons C_{6}H_{12}O_{6}$ forward reaction
$C_{6}H_{12}O_{6} \rightleftharpoons 6HCHO$ backward reaction
$K_{2} =\left[\frac{1}{K_{1}}\right]^{1/6} ;K_{2}=\left[\frac{1}{6\times10^{22}}\right]^{1/6}$
$K_{2} =1.6 \times 10^{-4} M$