Calculate K at 100 C for Reaction of Ethanol and Acetic Acid

[Orbit1212]

I have this problem to solve and I am having dificulity it doing so:

An aqueous solution of ethanol and acetic acid, each with a concentration of 0.810 M is heated to 100 C. At equilibrium, the acetic acid concentration is 0.748 M. Calculate K at 100 C for the reaction.

C2H5OH (aq) + CH3COOH (aq) <=> CH3COOC2H5 (aq) + H2O (l)

This isn't for homework or anything like that. I am just going back through old college textbooks and came across this one.

Thank you for help in advance.

 

[chaoseverlasting]

I think this reaction follows pseudo first order kinetics, so $$kt=ln\frac{a}{a-x}$$. You have t=373k, a=.810 a-x=0.748, solve for k.

 

[Blueshift5]

To calculate K at 100 C for this reaction, we first need to write out the equilibrium expression:

K = [CH3COOC2H5]/[C2H5OH][CH3COOH]

Next, we plug in the given concentrations at equilibrium:

K = (0.748 M)/(0.810 M)(0.810 M) = 0.921

Therefore, the equilibrium constant at 100 C for this reaction is 0.921. This means that at 100 C, the reaction favors the products slightly more than the reactants.