Official Correct Choice:
Option C: \( (\text{CH}_3\text{COOH})=0.666\text{mol} \), \( (\text{C}_2\text{H}_5\text{OH})=0.666\text{mol} \), \( (\text{CH}_3\text{COOC}_2\text{H}_5)=1.333\text{mol} \), \( (\text{H}_2\text{O})=1.333\text{mol} \)
Concept:The value of \( K_c \) is a constant at a given temperature. If initial concentrations are doubled proportionally, the final equilibrium concentrations will also double to maintain the ratio of \( K_c \).
Formula:$$ K_c = \frac{[\text{Ester}][\text{Water}]}{[\text{Acid}][\text{Alcohol}]} = 4 $$
Solution:- Initial experiment started with 1 mol of each reactant and produced 0.666 mol of products.
- By adding another 1 mol of each reactant to the equilibrium mixture, the total moles put into the system is exactly 2 moles of each reactant.
- Since the system volume is unchanged and \( \Delta n = 0 \), this is identical to starting a fresh reaction with double the initial concentration (2 mol each).
- Because all stoichiometry is 1:1:1:1, doubling initial reactants exactly doubles the final equilibrium concentrations to maintain \( K_c = 4 \).
- New Reactants = \( 0.333 \times 2 = 0.666 \text{ mol} \).
- New Products = \( 0.666 \times 2 = 1.333 \text{ mol} \).
Why other options are incorrect:Other options represent asymmetrical shifts which violate the 1:1 stoichiometry of the balanced chemical equation.
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