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 CH3CH(OH)2 (the hydrated form) at a concentration of 2.60 M. You have no information about how much, if any, of the anhydrous form (CH3CHO) is initially in the flask. 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?

Solution: Given the concentration of hydrated form =2.60M 2.0M CH3CHO is added to the flask Amount Change = 1.13M we are asked to find the final amount of CH3CHO.Answer: Final Amount = 1.79 Explanation: 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 = 0.92MHence 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

Answered: Acetaldehyde (CH3CHO) is an important…

Chemistry by OpenStax (2015-05-04)

1st Edition

ISBN:9781938168390

Author:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser

Publisher:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser

Chapter13: Fundamental Equilibrium Concepts

Section: Chapter Questions

Problem 3E: If a reaction is reversible, when can it be said to have reached equilibrium?

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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 CH₃CH(OH)₂ (the hydrated form) at a concentration of 2.60 M. You have no information about how much, if any, of the anhydrous form (CH₃CHO) is initially in the flask. 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?

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