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Chapter 15: Problem 60

The reaction of an organic acid with an alcohol, in organic solvent, to produce an ester and water is commonly done in the pharmaceutical industry. This reaction is catalyzed by strong acid (usually $\left.\mathrm{H}_{2} \mathrm{SO}_{4}\right)$. A simple example is the reaction of acetic acid with ethyl alcohol to produce ethyl acetate and water: $$\begin{aligned} \mathrm{CH}_{3} \mathrm{COOH}(\mathrm{solv})+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(\mathrm{solv}) & \mathrm{CH}_{3} \mathrm{COOCH}_{2} \mathrm{CH}_{3}(\text { solv })+\mathrm{H}_{2} \mathrm{O}(\text { solv }) \end{aligned}$$ where "(solv)" indicates that all reactants and products are in solution but not an aqueous solution. The equilibrium constant for this reaction at \(55^{\circ} \mathrm{C}\) is 6.68 . A pharmaceutical chemist makes up $15.0 \mathrm{~L}\( of a solution that is initially \)0.275 \mathrm{M}$ in acetic acid and \(3.85 \mathrm{M}\) in ethanol. At equilibrium, how many grams of ethyl acetate are formed?

Short Answer

Expert verified
At equilibrium, approximately 1746.40 grams of ethyl acetate are formed in the reaction of acetic acid and ethanol in organic solvent with an initial solution of $0.275\, \mathrm{M}$ acetic acid and $3.85\, \mathrm{M}$ ethanol, a total volume of $15.0\, \mathrm{L}$, and an equilibrium constant of $6.68$.

Step by step solution

01

Write down the equilibrium expression

For the given reaction: $$\mathrm{CH}_{3} \mathrm{COOH} + \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COOCH}_{2} \mathrm{CH}_{3} + \mathrm{H}_{2}\mathrm{O}$$ The equilibrium constant expression can be written as: $$K = \frac{[\mathrm{CH}_{3}\mathrm{COOCH}_{2}\mathrm{CH}_{3}][\mathrm{H}_{2}\mathrm{O}]}{[\mathrm{CH}_{3}\mathrm{COOH}][\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{OH}]}$$
02

Set up an ICE table

Let's create an ICE (Initial, Change, Equilibrium) table to keep track of the changes in concentrations of the reactants and products: \[\begin{array}{c|ccc} & [\mathrm{CH}_{3}\mathrm{COOH}] & [\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{OH}] & [\mathrm{CH}_{3}\mathrm{COOCH}_{2}\mathrm{CH}_{3}] & [\mathrm{H}_{2}\mathrm{O}] \\ \hline \text{Initial} & 0.275 & 3.85 & 0 & 0 \\ \text{Change} & -x & -x & +x & +x \\ \text{Equilibrium} & 0.275-x & 3.85-x & x & x \\ \end{array}\]
03

Substitute values into the equilibrium expression

We are given the equilibrium constant for this reaction at \(55^{\circ} \mathrm{C}\), which is \(K = 6.68\). Substitute the values from the ICE table into the equilibrium expression and solve for x. $$6.68 = \frac{x \times x}{(0.275-x)(3.85-x)}$$
04

Solve for x

Since the value of K is large, we may assume that the value of x is small compared to the initial concentrations of the reactants. So, we can simplify the expression as follows: $$6.68 \approx \frac{x^2}{(0.275)(3.85)}$$ Now, we solve for x: $$ x^2 = 6.68 \times (0.275)(3.85)$$ $$x \approx 1.321$$
05

Find the amount of ester formed at equilibrium

Now that we have x, we can find the equilibrium concentration of ethyl acetate: $$[\mathrm{CH}_{3}\mathrm{COOCH}_{2}\mathrm{CH}_{3}]_\text{eq} = x = 1.321 \,\mathrm{M}$$ To find the amount of ethyl acetate formed in grams, we need to multiply the equilibrium concentration by the volume of the reaction and then by the molar mass of ethyl acetate. $$ \text{Amount of ethyl acetate (grams)} = [\mathrm{CH}_{3}\mathrm{COOCH}_{2}\mathrm{CH}_{3}]_\text{eq} \times V \times M $$ where \(V = 15.0 \,\mathrm{L}\) (given volume of the reaction) \(M = 88.11 \,\mathrm{g/mol}\) (molar mass of ethyl acetate) $$\text{Amount of ethyl acetate (grams)} = 1.321 \, \mathrm{M} \times 15.0 \, \mathrm{L} \times 88.11 \, \mathrm{g/mol}$$ $$\text{Amount of ethyl acetate (grams)} \approx 1746.40 \, \mathrm{g}$$ At equilibrium, approximately 1746.40 grams of ethyl acetate are formed.

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Most popular questions from this chapter

Suppose that the gas-phase reactions \(A \longrightarrow B\) and $B \longrightarrow A\( are both elementary processes with rate constants of \)2.5 \times 10^{-2} \mathrm{~min}^{-1}\( and \)2.5 \times 10^{-1} \mathrm{~min}^{-1}$, respectively. (a) What is the value of the equilibrium constant for the equilibrium $\mathrm{A}(g) \rightleftharpoons \mathrm{B}(g) ?(\mathbf{b})$ Which is greater at equilibrium, the partial pressure of A or the partial pressure of B?

The protein hemoglobin (Hb) transports \(\mathrm{O}_{2}\) in mammalian blood. Each Hb can bind \(4 \mathrm{O}_{2}\) molecules. The equilibrium constant for the \(\mathrm{O}_{2}\) binding reaction is higher in fetal hemoglobin than in adult hemoglobin. In discussing protein oxygen-binding capacity, biochemists use a measure called the \(P 50\) value, defined as the partial pressure of oxygen at which \(50 \%\) of the protein is saturated. Fetal hemoglobin has a \(\mathrm{P} 50\) value of \(2.53 \mathrm{kPa},\) and adult hemoglobin has a P50 value of \(3.57 \mathrm{kPa}\). Use these data to estimate how much larger \(K_{c}\) is for the aqueous reaction $4 \mathrm{O}_{2}(g)+\mathrm{Hb}(a q) \rightleftharpoons\left[\mathrm{Hb}\left(\mathrm{O}_{2}\right)_{4}(a q)\right]\( in a fetus, compared to \)K_{c}$ for the same reaction in an adult.

The following equilibria were measured at \(823 \mathrm{~K}\) : $$\begin{array}{l} \mathrm{CoO}(s)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{Co}(s)+\mathrm{H}_{2} \mathrm{O}(g) \quad K_{c}=67 \\ \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \quad K_{c}=0.14 \end{array}$$ (a) Use these equilibria to calculate the equilibrium constant, \(K_{c},\) for the reaction $\mathrm{CoO}(s)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Co}(s)$ \(+\mathrm{CO}_{2}(g)\) at \(823 \mathrm{~K}\). (b) Based on your answer to part (a), would you say that carbon monoxide is a stronger or weaker reducing agent than \(\mathrm{H}_{2}\) at \(T=823 \mathrm{~K} ?(\mathbf{c})\) If you were to place \(5.00 \mathrm{~g}\) of \(\mathrm{CoO}(s)\) in a sealed tube with a volume of \(250 \mathrm{~mL}\) that contains \(\mathrm{CO}(g)\) at a pressure of $101.3 \mathrm{kPa}\( and a temperature of \)298 \mathrm{~K},$ what is the concentration of the CO gas? Assume there is no reaction at this temperature and that the CO behaves as an ideal gas (you can neglect the volume of the solid). (d) If the reaction vessel from part (c) is heated to $823 \mathrm{~K}$ and allowed to come to equilibrium, how much \(\operatorname{CoO}(s)\) remains?

Water molecules in the atmosphere can form hydrogenbonded dimers, \(\left(\mathrm{H}_{2} \mathrm{O}\right)_{2} .\) The presence of these dimers is thought to be important in the nucleation of ice crystals in the atmosphere and in the formation of acid rain. (a) Using VSF.PR theory, draw the structure of a water dimer, using dashed lines to indicate intermolecular interactions. (b) What kind of intermolecular forces are involved in water dimer formation? (c) The \(K_{p}\) for water dimer formation in the gas phase is 0.050 at $300 \mathrm{~K}\( and 0.020 at \)350 \mathrm{~K}$. Is water dimer formation endothermic or exothermic?

The following equilibria were attained at \(298 \mathrm{~K}:\) $$\begin{array}{c} \mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{AgCl}(s) \quad K_{c}=5.6 \times 10^{9} \\ \mathrm{Ag}^{+}(a q)+2 \mathrm{NH}_{3}(a q) \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}(a q) \\ K_{c}=1.6 \times 10^{7}\end{array}$$ Based on these equilibria, calculate the equilibrium constant $\text { for } \mathrm{AgCl}(s)+2 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}(a q)+\mathrm{Cl}^{-}(a q)$ at \(298 \mathrm{~K}\).
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