Equilibrium Constant for Alcohol and Acetic Acid Reaction at 25°C | Homework

[harambe]

Homework Statement

One mole of pure ethyl alcohol was treated with one mol of pure acetic acid at 25 °C.One third of acid changes into ester at equilibrium The equilibrium constant for the reaction will be

Homework Equations

C2H5OH+CH3COOH------>C2H5COOCH3[/B]
K={A}^m{B}n/{C}^x{D}^z where m,n, x, z are the concentrations of the reactants respectively.

The Attempt at a Solution

[/B]
At equilibrium ,I got the concenterations of alcohol and acetic acid to be 1-x and ester to be x where x is dissociation constant. Now I equated 1-x=3x as per the information in the question but got the wrong answer

 

[Charles Link]

Beginning with one mole of $C_2H_5 OH$ and one mole of $H C_2 H_3 O_2$, you need the volume of the final product.$\\$ I presume these are all in a homogeneous solution. The exponents on the concentrations are all equal to 1, but the equilibrium constant equation needs concentrations in the form of $x$ moles per $y$ liters, or $1-x$ moles per $y$ liters. $\\$ They give you $x=\frac{1}{3}$, but you first need to find the final volume $y$. $\\$ In the final product, you have 2/3 mole of each of the reactants remaining, and 1/3 mole of ester. You need to find the volume of these 3 components, and the total volume should be the sum of these 3. $\\$ And the reaction is in the form $A+C \rightarrow D$. Thereby $K=\frac{[D]}{ [A] [C]}$. $\\$ (I tried to use $B$ in the reaction, but $B$ inside the "[ ]" kept instructing the Latex to print bold, and it would not give a $B$ there. It must be some special command).

 

[Chestermiller]

Homework Statement

One mole of pure ethyl alcohol was treated with one mol of pure acetic acid at 25 °C.One third of acid changes into ester at equilibrium The equilibrium constant for the reaction will be

Homework Equations

C2H5OH+CH3COOH------>C2H5COOCH3[/B]
K={A}^m{B}n/{C}^x{D}^z where m,n, x, z are the concentrations of the reactants respectively.

The Attempt at a Solution

At equilibrium ,I got the concenterations of alcohol and acetic acid to be 1-x and ester to be x where x is dissociation constant. Now I equated 1-x=3x as per the information in the question but got the wrong answer [/B]

That reaction equation you wrote is not a balanced chemical reaction equation. You are missing a reaction product. No wonder you couldn't solve this.

 

[Charles Link]

That reaction equation you wrote is not a balanced chemical reaction equation. You are missing a reaction product. No wonder you couldn't solve this.

@Chestermiller I'm also interested in this one, but I only had a very little bit of organic chemistry. Question: When an alcohol is mixed with an acid to get an ester, is $H_2 O$ normally also one of the products? And from a google, that appears to be the case: https://sites.google.com/site/chemistryolp/formation-of-esters $\\$ Additional item: If the equilibrium constant takes the form $K=\frac{[C][D]}{[A][B']}$, the final volume will cancel out of the calculation.

 

[Chestermiller]

@Chestermiller I'm also interested in this one, but I only had a very little bit of organic chemistry. Question: When an alcohol is mixed with an acid to get an ester, is $H_2 O$ normally also one of the products? And from a google, that appears to be the case: https://sites.google.com/site/chemistryolp/formation-of-esters $\\$ Additional item: If the equilibrium constant takes the form $K=\frac{[C][D]}{[A][B']}$, the final volume will cancel out of the calculation.

Hi Charles. Yes, when an acid and an alcohol react, the products are an ester and water. Regarding volume, it seems that, in the absence of additional information, the only reasonable approximations is that the final- and the initial volumes are equal.

 

[Charles Link]

Hi Charles. Yes, when an acid and an alcohol react, the products are an ester and water. Regarding volume, it seems that, in the absence of additional information, the only reasonable approximations is that the final- and the initial volumes are equal.

In determining the equilibrium constant, if I'm not mistaken, all that is needed is the final volume. $\\$ $[A]$ is the final concentration, (using the final volume), and the same with B, C, and D. Thereby, algebraically the volume drops out of the expression. $\\$ Assuming I got the right answer, this problem, because the volumes cancel, becomes simple arithmetic.

 

[harambe]

Okay so the reaction is C2H5OH+Ch3COOH------>CH3COOC2H5 +H2O.

But still it gives me wrong answer . ...

 

[Charles Link]

Okay so the reaction is C2H5OH+Ch3COOH------>CH3COOC2H5 +H2O.

But still it gives me wrong answer . ...

What answer did you get?

 

[harambe]

1/9

 

[harambe]

Wait. . How is x=1/3 like you have quoted @Charles Link . Shouldn't the expression be 1/3rd of concentration if carboxylic acid equal to ester giving us

1-x/3 = x

Solci g this gives us x=1/4

 

[Charles Link]

1/9

$x =1/3$ moles of the ester are formed with 2/3 moles remaining of each of the reactants. And of course you get 1/3 mole of $H_2 O$. $\\$ Try computing the answer for $K$ one more time.

 

[harambe]

Okay. Upon reading the question carefully ,I think I got the mistake now .. . x=1/3 was the concenteration of ester at equilibrium No need for the weird calculation I did

Thanks for the help

 

[Charles Link]

Okay. Upon reading the question carefully ,I think I got the mistake now .. . x=1/3 was the concenteration of ester at equilibrium No need for the weird calculation I did

Thanks for the help

For completeness, please tell us the answer you got. We can at least verify it.

 

[Chestermiller]

Please tell us the answer you got. We can at least verify it.

Yes, and show us how you calculated K.

 

[harambe]

Okay.

Like we agreed the reaction is
CH3COOH +C2H5OH -----> CH3COOC2H5+ H2O (1-x) (1-x) (x) (x)

Snce x=1/3 therefore concenterations become

CH3COOH +C2H5OH -----> CH3COOC2H5+ H2O (2/3) (2/3) (1/3) (1/3)

Putting them in the expression of
$$K=\frac{[C][D]}{[A][ B]}$$ I get K=1/4

 

[epenguin]

(I tried to use $B$ in the reaction, but $B$ inside the "[ ]" kept instructing the Latex to print bold, and it would not give a $B$ there. It must be some special command).

This happens with several other capital letters inside square bracketswhich Latex interprets as instructions. Just put a space before the B. I think if you press reply you can see the code in the following:

$[A][ B][C]$