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Chapter 13: Problem 48

Ethyl acetate is synthesized in a nonreacting solvent (not water) according to the following reaction: $$\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} \rightleftharpoons \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{C}_{2} \mathrm{H}_{5}+\mathrm{H}_{2} \mathrm{O} \quad K=2.2$$ For the following mixtures (a-d), will the concentration of $\mathrm{H}_{2} \mathrm{O}$ increase, decrease, or remain the same as equilibrium is established? a. $\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{C}_{2} \mathrm{H}_{5}\right]=0.22 M,\left[\mathrm{H}_{2} \mathrm{O}\right]=0.10 M$ $\quad\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right]=0.010 M,\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right]=0.010 M$ b. $\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{C}_{2} \mathrm{H}_{5}\right]=0.22 M,\left[\mathrm{H}_{2} \mathrm{O}\right]=0.0020 M$ $\quad\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right]=0.0020 M,\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right]=0.10 M$ c. $\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{C}_{2} \mathrm{H}_{5}\right]=0.88 M,\left[\mathrm{H}_{2} \mathrm{O}\right]=0.12 M$ $\quad\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right]=0.044 M,\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right]=6.0 M$ d. $\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{C}_{2} \mathrm{H}_{5}\right]=4.4 M,\left[\mathrm{H}_{2} \mathrm{O}\right]=4.4 M$ $\quad\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right]=0.88 M,\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right]=10.0 M$ e. What must the concentration of water be for a mixture with $\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{C}_{2} \mathrm{H}_{5}\right]=2.0 M,\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right]=0.10 M,$ and \(\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right]=5.0 M\) to be at equilibrium? f. Why is water included in the equilibrium expression for this reaction?

Short Answer

Expert verified
For the mixtures (a-d), the concentration of \(\mathrm{H}_{2}\mathrm{O}\) changes as follows: a) Decrease b) Decrease c) Decrease d) Remain the same For part (e), the concentration of water must be 0.55 M for the mixture to be at equilibrium. In part (f), water is included in the equilibrium expression because it is a product of the reaction in a nonreacting solvent, and its concentration can change affecting the position of equilibrium.

Step by step solution

01

Calculate the reaction quotient (Q) for each mixture

For mixture (a): \[ Q = \frac{[0.22] [0.10]}{[0.010] [0.010]} = 22 \] For mixture (b): \[ Q = \frac{[0.22] [0.0020]}{[0.0020] [0.10]} = 22 \] For mixture (c): \[ Q = \frac{[0.88] [0.12]}{[0.044] [6.0]} = 4.0 \] For mixture (d): \[ Q = \frac{[4.4] [4.4]}{[0.88] [10.0]} = 2.2 \]
02

Compare Q with K for each mixture to find the fate of water concentration

Mixture (a): Since \(Q > K\), the reaction will shift to the left (towards reactants), hence the concentration of \(\mathrm{H}_{2}\mathrm{O}\) will decrease. Mixture (b): Since \(Q > K\), the reaction will shift to the left (towards reactants), hence the concentration of \(\mathrm{H}_{2}\mathrm{O}\) will decrease. Mixture (c): Since \(Q > K\), the reaction will shift to the left (towards reactants), hence the concentration of \(\mathrm{H}_{2}\mathrm{O}\) will decrease. Mixture (d): Since \(Q = K\), the reaction is at equilibrium, hence the concentration of \(\mathrm{H}_{2}\mathrm{O}\) will remain the same.
03

Determine the concentration of water for the given mixture in part (e)

For reactants \(\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{C}_{2}\mathrm{H}_{5}\right] = 2.0 M, \left[\mathrm{CH}_{3} \mathrm{CO}_{2}\mathrm{H}\right] = 0.10 M\), and \(\left[\mathrm{C}_{2} \mathrm{H}_{5}\mathrm{OH}\right] = 5.0 M\), we will use the given equilibrium constant (K) to find the concentration of \(\mathrm{H}_{2}\mathrm{O}\): \[ K = \frac{[2.0] [\mathrm{H}_{2}\mathrm{O}]}{[0.10] [5.0]} \] Now, we simply need to find \([\mathrm{H}_{2}\mathrm{O}]\) by solving this equation: \[ [\mathrm{H}_{2}\mathrm{O}] = \frac{K [0.10] [5.0]}{[2.0]} = 0.55 M \] So, the concentration of water must be 0.55 M for the given mixture to be at equilibrium.
04

Explain why water is included in the equilibrium expression

Water is included in the equilibrium expression for this reaction because it is one of the products of the reaction, and its concentration in the nonreacting solvent can change due to the reversibility of the reaction. By including water in the equilibrium expression, it allows accurate prediction and assessment of the position of equilibrium depending on the initial concentrations of the reactants and products. In some reactions, water is excluded from the equilibrium expressions when it is a solvent. However, in this case, the reaction takes place in a non-aqueous (not water) solvent, so water should be considered in the equilibrium expression.

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Most popular questions from this chapter

At a particular temperature, \(K=3.75\) for the reaction $$\mathrm{SO}_{2}(g)+\mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g)+\mathrm{NO}(g)$$ If all four gases had initial concentrations of \(0.800 M,\) calculate the equilibrium concentrations of the gases.

Consider the reaction $\mathrm{A}(g)+2 \mathrm{B}(g) \rightleftharpoons \mathrm{C}(g)+\mathrm{D}(g)\( in a \)1.0-\mathrm{L}$ rigid flask. Answer the following questions for each situation \((\mathrm{a}-\mathrm{d}) :\) i. Estimate a range (as small as possible) for the requested substance. For example, [A] could be between 95\(M\) and 100\(M .\) ii. Explain how you decided on the limits for the estimated range. iii. Indicate what other information would enable you to narrow your estimated range. iv. Compare the estimated concentrations for a through d, and explain any differences. a. If at equilibrium \([\mathrm{A}]=1 M,\) and then 1 mole of \(\mathrm{C}\) is added, estimate the value for \([\mathrm{A}]\) once equilibrium is reestablished. b. If at equilibrium \([\mathrm{B}]=1 M,\) and then 1 mole of \(\mathrm{C}\) is added, estimate the value for \([\mathrm{B}]\) once equilibrium is reestablished. c. If at equilibrium \([\mathrm{C}]=1 M,\) and then 1 mole of \(\mathrm{C}\) is added, estimate the value for \([\mathrm{C}]\) once equilibrium is reestablished. d. If at equilibrium \([\mathrm{D}]=1 M,\) and then 1 mole of \(\mathrm{C}\) is added, estimate the value for \([\mathrm{D}]\) once equilibrium is reestablished.

Consider the reaction $\mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{C}(g)+\mathrm{D}(g) . \mathrm{A}$ friend asks the following: “I know we have been told that if a mixture of A, B, C, and D is at equilibrium and more of A is added, more C and D will form. But how can more C and D form if we do not add more B?” What do you tell your friend?

At a particular temperature, \(K=1.00 \times 10^{2}\) for the reaction $$\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \leftrightharpoons 2 \mathrm{HI}(g)$$ In an experiment, 1.00 mole of \(\mathrm{H}_{2}, 1.00\) mole of \(\mathrm{I}_{2},\) and 1.00 mole of \(\mathrm{HI}\) are introduced into a \(1.00-\mathrm{L}\) container. Calculate the concentrations of all species when equilibrium is reached.

The following equilibrium pressures were observed at a certain temperature for the reaction $$\begin{array}{c}{\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)} \\ {P_{\mathrm{NH}_{3}}=3.1 \times 10^{-2} \mathrm{atm}} \\\ {P_{\mathrm{N}_{2}}=8.5 \times 10^{-1} \mathrm{atm}} \\\ {P_{\mathrm{H}_{2}}=3.1 \times 10^{-3} \mathrm{atm}}\end{array}$$ Calculate the value for the equilibrium constant \(K_{\mathrm{p}}\) at this temperature. If $P_{\mathrm{N}_{2}}=0.525 \mathrm{atm}, P_{\mathrm{NH}_{3}}=0.0167 \mathrm{atm},\( and \)P_{\mathrm{H}_{2}}=0.00761 \mathrm{atm},$ does this represent a system at equilibrium?
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