The equilibrium constant, Kc, for the reaction to form ethyl ethanoate from ethanol and ethanoic acid, C2H5OH + CH3CO2H ⇋ CH3CO2C2H5 + H2O, at 60oC is 4.00. When 1.00 mol each of ethanol and ethanoic acid are allowed to reach equilibrium at 60oC, what is the number of moles of ethyl ethanoate formed?
Correct answer: B. 2/3
- A. 1/3
- B. 2/3
- C. 1/4
- D. 3/4
Explanation
Note: Ignore volume in this case as it will cancel out in the Kc expression.Let the no. of moles of ethyl ethanoate at equilibrium = xInitial moles of ethanol and ethanoic acid = 1Moles of ethanol and ethanoic acid at equilibrium = 1-xKc= (x)(x) / (1-x)(1-x)(x/1-x)2 = 4x/1-x = 4x/1-x = 2X = 2-2x3x = 2X = 2/3
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Reversible reactions reach dynamic equilibrium when forward and reverse rates become equal, and Le Chatelier's principle predicts the effect of changing concentration, pressure or temperature. The chapter also covers solubility product, the common ion effect, buffer action and the conditions used in Haber's process, with equilibrium shifts distinguished from changes in the equilibrium constant.
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