Formaldehyde polymerizes to form glucose according to the reaction 6 HCHOC 6 H 12 O 6 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 6 HCHOC 6 H 12 O 6 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 H C H O is very-very small. Hence, at equilibrium, we can take, [ C 6 H 12 O 6]=1 M K =([ C 6 H 12 O 6]/[ HCHO ]6) i.e, 6 × 1022=(1/[ HCHO ]6) or [ HCHO ]=((1/6 × 1022))1 / 6 =1.6 × 10-4 M

$1.6 \times 1 0^{- 2} M$

$1.6 \times 1 0^{- 4} M$

$1.6 \times 1 0^{- 6} M$

$1.6 \times 1 0^{- 8} M$

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​O6​]​ i.e, 6×1022=[HCHO]61​ or [HCHO]=(6×10221​)1/6 =1.6×10−4M

$1.6 \times 1 0^{- 2} M$

$1.6 \times 1 0^{- 4} M$

$1.6 \times 1 0^{- 6} M$

$1.6 \times 1 0^{- 8} M$