Formaldehyde polymerizes to form glucose according to the reaction 6HCHOleftharpoons C6H12O6 The theoretically computed equilibrium constant for this reaction is found to be 6× 1022 If 1 M solution of glucose dissociates according to the above equilibrium, the concentration of formaldehyde in the solution will be

Question

Formaldehyde polymerizes to form glucose according to the reaction 6HCHOleftharpoons C6H12O6 The theoretically computed equilibrium constant for this reaction is found to be 6× 1022 If 1 M solution of glucose dissociates according to the above equilibrium, the concentration of formaldehyde in the solution will be (A)  1.6× 10-2M  (B)  1.6× 10-4M  (C)  1.6× 10-6M  (D)  1.6× 10-8M

Tardigrade Admin · Accepted Answer

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, [ C 6 H 12 O 6] =1 M K =([ C 6 H 12 O c ]/[ HCHO ]6) ie, 6 × 1022=(1/[ HCHO ]6) or [ HCHO ] =((1/6 × 1022))1 / 6 =1.6 × 10-4 M

$1.6 \times 10^{- 2} M$

$1.6 \times 10^{- 4} M$

$1.6 \times 10^{- 6} M$

$1.6 \times 10^{- 8} M$

1.6×10−2M

1.6×10−4M

1.6×10−6M

1.6×10−8M

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, [C6​H12​O6​]=1M K=[HCHO]6[C6​H12​Oc​]​ ie, 6×1022=[HCHO]61​ or [HCHO]=(6×10221​)1/6 =1.6×10−4M

$1.6 \times 10^{- 2} M$

$1.6 \times 10^{- 4} M$

$1.6 \times 10^{- 6} M$

$1.6 \times 10^{- 8} M$