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

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SUMMARY

The discussion focuses on calculating the equilibrium constant (K) for the reaction between ethanol and acetic acid at 100°C. Given the initial concentrations of both reactants at 0.810 M and the equilibrium concentration of acetic acid at 0.748 M, the calculation involves applying the formula for pseudo first-order kinetics. The reaction is represented as C2H5OH (aq) + CH3COOH (aq) <=> CH3COOC2H5 (aq) + H2O (l). The value of K can be determined using the equation kt=ln(a/(a-x)), where t is the temperature in Kelvin.

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Orbit1212
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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.
 
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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.
 

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