The ester, ethyl acetate is formed by the reaction between ethanol and acetic acid and equilibrium is represented as : \(\mathrm{CH}_{3} \mathrm{COOH}_{(\mathrm{l})}+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}_{(\mathrm{l})} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5(\mathrm{aq})}+\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})}\) (a) Write the concentration ratio (reaction quotient), \(Q_{\mathrm{e}}\), for this reaction. Note that water is not in excess and is not a solvent in this reaction. (b) At \(293 \mathrm{~K}\), if one starts with \(1.00\) mole of acetic acid and \(0.180\) of ethanol, there is \(0.171\) mole of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant. (c) Starting with \(0.500\) mole of ethanol and \(1.000\) mole of acetic acid and maintaining it at \(293 \mathrm{~K}, 0.214\) mole of ethyl acetate is found after some time. Has equilibrium been reached?

Short Answer

Expert verified
The reaction quotient Qe is given by \[Q = \frac{[\text{{ethyl acetate}}] \times [\text{{water}}]}{[\text{{acetic acid}}] \times [\text{{ethanol}}]}\]. The equilibrium constant K at 293 K is calculated as \[K = \frac{[0.171]^2}{[0.829] \times [0.009]}\]. Without calculating, it is uncertain if equilibrium has been reached for part (c), but comparison of the calculated Q with K will provide the answer.

Step by step solution

01

Writing the Reaction Quotient

According to the balanced chemical equation, the reaction quotient, Q, is given by the formula \[Q = \frac{[\text{{Products}}]}{[\text{{Reactants}}]} = \frac{[\text{{ethyl acetate}}] \times [\text{{water}}]}{[\text{{acetic acid}}] \times [\text{{ethanol}}]}\].
02

Calculate the Equilibrium Constant

The equilibrium constant, K, is calculated using the equilibrium concentrations of the reactants and products. The expression for K is the same as Q, but at equilibrium. Given the equilibrium concentrations: \[K = \frac{[0.171] \times [\text{{moles of water at equilibrium}}]}{[1.00 - 0.171] \times [0.180 - 0.171]}\].We are not provided with the equilibrium concentration of water, so we assume it is proportional to the amount of ethyl acetate formed. Thus, we have the same number of moles for water as ethyl acetate: 0.171. Now we can calculate K.
03

Solve for the Equilibrium Constant

Substitute the given values into the equilibrium constant expression: \[K = \frac{[0.171] \times [0.171]}{[1.00 - 0.171] \times [0.180 - 0.171]} = \frac{[0.171]^2}{[0.829] \times [0.009]}\].Calculate the K value by carrying out the multiplications and division.
04

Has Equilibrium Been Reached?

Assuming the equilibrium constant K from part (b), calculate Q for the given concentrations in part (c): \[Q = \frac{[0.214] \times [\text{{moles of water at non-equilibrium state}}]}{[1.000 - 0.214] \times [0.500 - 0.214]}\].Compare this Q with the calculated K from step 3 to determine if equilibrium has been reached. If Q equals K, then the system is at equilibrium. If Q does not equal K, the system has not reached equilibrium.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Reaction Quotient (Q)
The reaction quotient, often denoted as Q, is a crucial concept in understanding chemical reactions, particularly when assessing the progress of a reaction towards equilibrium. In a chemical process, different compounds may react to form products, and at any given point before reaching equilibrium, the concentrations of reactants and products can vary. Q allows us to express this ratio mathematically, taking the form of the equilibrium expression but using initial or non-equilibrium concentrations of the chemical species involved.

For the reaction involving ethyl acetate, the reaction quotient would be expressed as:
$$Q = \frac{[\text{ethyl acetate}] \times [\text{water}]}{[\text{acetic acid}] \times [\text{ethanol}]}$$
Here, the square brackets imply the concentration of each component. This ratio helps in determining the direction in which a reaction will shift to achieve equilibrium. If Q is less than the equilibrium constant (K), the reaction will proceed towards forming more products. Conversely, if Q is greater, the reaction will shift towards forming more reactants.
Equilibrium Constant (K)
The equilibrium constant, or K, is a fixed number for a given chemical reaction at a specific temperature, representing the ratio of concentrations of products to reactants at equilibrium. This value is a measure of the propensity of a reaction to occur. The equilibrium constant has the same form as the reaction quotient, Q, but is calculated only at the point when the system has reached equilibrium—when the rates of the forward and reverse reactions are equal and concentrations of reactants and products no longer change with time.

For the specific reaction involving acetic acid and ethanol, the equilibrium constant K is given by:
$$K = \frac{[\text{ethyl acetate}] \times [\text{water}]}{[\text{acetic acid}] \times [\text{ethanol}]}$$
at equilibrium conditions. By defining this constant, we can predict the extent of a reaction, how much product will be formed from specific initial amounts of reactants, and whether a certain set of concentrations represents an equilibrium state by comparing the calculated Q to the known K. If a reaction mixture's Q value is different from K, the system is not at equilibrium and will shift to minimize this difference.
IIT-JEE Physical Chemistry
The Indian Institutes of Technology Joint Entrance Examination (IIT-JEE) is a highly competitive examination in India, aimed at selecting candidates for premier engineering institutions. Physical chemistry is a significant part of the IIT-JEE syllabus, encompassing the study of macroscopic and particulate phenomena in chemical systems in terms of the principles, practices, and concepts of physics. This includes an extensive understanding of chemical equilibrium.

Students preparing for IIT-JEE must grasp the subtleties of equilibrium constants, reaction quotient, and related principles to solve problems efficiently. Such knowledge is not just important for theory-based questions but also for numerical problems where the calculation of K values and comparison with Q can help determine the direction of a reaction or composition of the equilibrium mixture. A problem such as the formation of ethyl acetate from acetic acid and ethanol showcases an application of these concepts and the level of understanding required for the examination.
Le Chatelier's Principle
Le Chatelier's Principle provides a qualitative method to predict the change in direction of a chemical reaction in response to shifts in external conditions, such as concentration, volume, and temperature. According to this principle, if a dynamic equilibrium is disturbed by altering the conditions, the system will adjust itself to counteract the effect of the disturbance and re-establish equilibrium.

In terms of the ethyl acetate reaction, if we were to increase the concentration of acetic acid or ethanol, Le Chatelier's Principle predicts that the equilibrium would shift to the right to favor product formation. This shift could be quantified by recalculating the reaction quotient Q and comparing it with the equilibrium constant K. Conversely, if the concentration of either the product ethyl acetate or water is increased, the reaction would shift to the left to produce more reactants. Students studying for competitive exams such as the IIT-JEE must not only understand Le Chatelier's Principle conceptually but also be able to apply it to predict the outcome of changes in reaction conditions.

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Most popular questions from this chapter

State which one is homogeneous or heterogeneous? (a) \(\quad \overline{\text { C }_{\text {Diamond }}} \rightleftharpoons \bar{C}_{\text {graphite }}\) (b) \(\mathrm{H}_{2} \mathrm{O}_{(\mathrm{s})}=\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})}\) (c) \(\mathrm{N}_{2(\mathrm{~g})}+3 \mathrm{H}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{NH}_{3(\mathrm{~g})}\) (d) \(\quad \mathrm{MgCO}_{3(\mathrm{~s})} \rightleftharpoons \mathrm{MgO}_{(\mathrm{s})}+\mathrm{CO}_{2(\mathrm{~g})}\) (e) \(\mathrm{PCl}_{3(\mathrm{~g})}+\mathrm{Cl}_{2(\mathrm{~g})}=\mathrm{PCl}_{5(\mathrm{~g})}\)

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