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

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To calculate the equilibrium constant (K) for the reaction between ethanol and acetic acid at 100°C, the initial concentrations of both reactants are 0.810 M, and the equilibrium concentration of acetic acid is 0.748 M. The change in concentration (x) can be determined by subtracting the equilibrium concentration from the initial concentration. The reaction follows pseudo first-order kinetics, and the equation kt = ln(a/(a-x)) can be applied, where t is the temperature in Kelvin (373 K), a is the initial concentration, and a-x is the equilibrium concentration. By substituting these values into the equation, K can be calculated for the reaction at the specified temperature.
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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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