Formaldehyde polymerises to form glucose according to the reaction 6 HCHO hArr C_(6) H_(12)O_(6) ltbr gt The theoretically computed equiibrium constant for this reaction is found to be6 xx 10^(22).If 1 M solution of glucose dissociates according to the above equilibrium , what will be the concentration of formaldehyde in the solution ?

Formaldehyde polymerises to form glucose according to the reaction 6 H

1.	Formaldehyde polymerises to form glucose according to the reaction 6 HCHO hArr C_(6) H_(12)O_(6) ltbr gt The theoretically computed equiibrium constant for this reaction is found to be6 xx 10^(22).If 1 M solution of glucose dissociates according to the above equilibrium , what will be the concentration of formaldehyde in the solution ?
Answer» Solution :` 6 HCHO hArr C_(6) H_(12)O_(6)` As equilibrium constant for this reaction is very large, therfore for the reverse reaction involving dissociation of glucose, equilibrium constant is very very small . HENCE , for the reverse reaction ` C_(6) H_(12)O_(6) hArr6 HCHO"" (K= 1/(6 xx 10^(22)))^(1//6)` dissociation of glucose is negligible . Starting with 1 M, concentration at equilibrium at equilibrium `cong 1 M` ` K= ([HCHO]^(6))/([C_(6)H_(12)O_(6) )` ` 1/(6 xx 10^(22)) = ([HCHO]^(6))/1 or [HCHO] = (1/(6xx10^(22)))^(1//6)` `log [HCHO] = 1/6 [-log (6xx10^(22)] = 1/6 [-22* 778 ] = -3* 7963 = bar 4* 2137` ` :.[HCHO] = 1* 636 xx 10^(-4) "M"`