Acetaldehyde (CH3CHO) is an important chemical both industrially and biologicall

Question

Acetaldehyde (CH3CHO) is an important chemical both industrially and biologically. For instance, it is a (somewhat toxic) intermediate in the body's metabolism of ethanol into acetic acid, and thus is possibly implicated in the "hungover" symptoms of someone who has had too much to drink the night before. In aqueous solution, it establishes an equilibrium with a hydrated form, shown below. CH3CHO (aq) + H2O (l) <-- --> CH3CH(OH)2 (aq) You start with an aqueous sample, already at equilibrium, with the concentration of CH3CH(OH)2 (the hydrated form) at a concentration of 2.60 M. You have no information about whether there is any of the anhydrous form (CH3CHO) initially in the flask, and if so, how much. If you add 2.0 M of CH3CHO to the reaction flask, and as the equilibrium is being restored the amount of CH3CH(OH)2 changes by 1.13 M, what is the final amount of CH3CHO?

Explanation / Answer

ANSWER:

The equilibrium is shown below.

CH3CHO (aq) + H2O (l) <-- --> CH3CH(OH)2 (aq)

Let equilibrium constant is K

K = [CH3CH(OH)2] / [CH3CHO] because water cencentration is taken as 1 as it is solvent and present in excess.

let at equilibrium xM is the concentration of CH3CHO.

K = 2 / x ============1)

When 2M of CH3CHO is added the equilibrium is shifted to right , i.e, more CH3CH(OH)2 is formed. The CH3CH(OH)2 formed is given in question and is equal to 1.13M. But equilibrium constant does not change. New concentration of CH3CHO is (x + 2)M. But out of (x + 2)M, 1.13M react to form CH3CH(OH)2. New equilibrium can be represented as

CH3CHO (aq) + H2O (l) <-- --> CH3CH(OH)2 (aq)

(x + 2 - 1.3)M (2.6 + 1.3)M

K =  (2.6 + 1.13)M / (x + 2 - 1.13)M ===========2)

equating 1) and 2)

(2.6 + 1.3)M / (x + 2 - 1.3)M = 2 /x

Or (2.6 + 1.3)M x = 2 (x + 2 - 1.13)M

3.9 x = 2x + 1.74

3.9x - 2x = 1.74

1.9x = 1.74

x = 1.74 / 1.9 = 0.92M

Hence the initial concentration of CH3CHO is 0.92M.

After adding 2M of CH3CHO the total concentration = 0.92 + 2 = 2.92M

OOut of 2.92M 1.13M react to attain new equilubrium, hence final amount of CH3CHO = 2.92M - 1.13M= 1.79M

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