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Asked in UHS MDCAT 2017 2017

Consider the following reversible reaction:Initial concentrations:CH3-CH2-OH) = 1 mol.dm-3(CH3-COOH)= 1 mol.dm-3(CH3-CH2-O-CO-CH3)= 0 mol.dm-3(H2O)= 0 mol.dm-3Equilibrium concentrations:(CH3-CH2-OH) = 0.33mol.dm-3(CH3-COOH)= 0.33 mol.dm-3(CH3-CH2-O-CO-CH3)= 0.66 mol.dm-3(H2O)= 0.66 mol.dm-3Kc= 4 at temperature 100 CWhat are the new equilibrium concentrations of all species if 1 mol.dm-3 of CH3-CH2-OH and CH3-COOH are added to this equilibrium mixture? (Apply Le-Chatelier's principle)(Temperature and Kc remain constant)

Correct answer: C. (CH3-COOH)= 0.666 mol.dm-3(CH3-CH2-OH) = 0.666 mol.dm-3(CH3-CH2-O-CO-CH3)= 1.333 mol.dm-3(H2O)= 1.333 mol.dm-3

  • A. (CH3-COOH)= 0.333 mol.dm-3(CH3-CH2-OH) = 1.333mol.dm-3(CH3-CH2-O-CO-CH3)= 1.666 mol.dm-3(H2O)= 1.666 mol.dm-3
  • B. (CH3-COOH)= 1.333 mol.dm-3(CH3-CH2-OH) = 0.333mol.dm-3(CH3-CH2-O-CO-CH3)= 0.666 mol.dm-3(H2O)= 0.666 mol.dm-3
  • C. (CH3-COOH)= 0.666 mol.dm-3(CH3-CH2-OH) = 0.666 mol.dm-3(CH3-CH2-O-CO-CH3)= 1.333 mol.dm-3(H2O)= 1.333 mol.dm-3
  • D. (CH3-COOH)= 0.333 mol.dm-3(CH3-CH2-OH) = 0.333mol.dm-3(CH3-CH2-O-CO-CH3)= 1.333 mol.dm-3(H2O)= 1.333 mol.dm-3

Explanation

Change in concentration of of CH3COOH is 1-0.33 = 0.66 mol.dm-3Change in concentration of of CH3CH2OH is also 1-0.33 = 0.66 mol.dm-3Where as change in conc of CH3CH2OCOCH3 is 1 + 0.333 = 1.333 moldm-3And change in conc for H2O is also 1 + 0.333 = 1.333 mol.dm-3

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About Le Chatelier's Principle

Le Chatelier's principle predicts how an equilibrium responds when concentration, pressure or temperature changes. Questions apply it to reversible reactions, identify the direction of shift and distinguish genuine equilibrium changes from the effect of a catalyst, which changes the rate of reaching equilibrium but not its position.

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