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Chapter 15: Problem 97

In organic synthesis many reactions produce very little yield, that is \(K \ll 1 .\) Consider the following hypothetical reaction: \(\mathrm{A}(\mathrm{aq})+\mathrm{B}(\mathrm{aq}) \longrightarrow \mathrm{C}(\mathrm{aq}), K=1 \times 10^{-2}\) We can extract product, \(\mathrm{C}\), from the aqueous layer by adding an organic layer in which \(\mathrm{C}(\mathrm{aq}) \longrightarrow \mathrm{C}(\mathrm{or})\), \(K=15 .\) Given initial concentrations of \([\mathrm{A}]=0.1 \mathrm{M}\) \([\mathrm{B}]=0.1,\) and \([\mathrm{C}]=0.1,\) calculate how much \(\mathrm{C}\) will be found in the organic layer. If the organic layer was not present, how much C would be produced?

Short Answer

Expert verified
If the organic layer is present, the final concentration of C in the organic layer is \(1.5 \times 10^{-3}\)M. Without the organic layer, the amount of C produced would be \(1 \times 10^{-4}\) M.

Step by step solution

01

Set up the first reaction

Here, we need to establish the equilibrium expression for the first reaction and also calculate the equilibrium concentrations of A, B, and C in the aqueous layer. The equilibrium expression for reaction A(aq) + B(aq) -> C(aq) is given as \(K = \frac{[C]}{[A]*[B]}\). Given that \(K = 1 \times 10^{-2}\) and the initial concentrations of A, B, and C are all 0.1M, plugging in these values we get \([C] = K \times [A] \times [B] = 1 \times 10^{-2} \times 0.1 \times 0.1 = 1 \times 10^{-4} M\). Thus, the equilibrium concentration of C in the aqueous layer is \(1 \times 10^{-4}\) M.
02

Set up the second reaction

For the second reaction, where C(aq) -> C(or), an organic layer is added that causes C to transfer from the aqueous layer to the organic layer. The equilibrium expression for this reaction is given as \(K = \frac{[C_{or}]}{[C_{aq}]}\). Rearranging this expression gives \([C_{or}] = K \times [C_{aq}]\). The equilibrium constant for this reaction is given as 15. Therefore, plugging in the values we get \([C_{or}] = 15 \times [1 \times 10^{-4} M] = 1.5 \times 10^{-3} M\). Thus, the concentration of C in the organic layer is \(1.5 \times 10^{-3}\) M.
03

Calculate the amount of C without the organic layer

Without the organic layer present, the distribution of C will only be determined by the first reaction (A(aq) + B(aq) -> C(aq)). Using the same process as in Step 1 and since the concentrations of A and B remain as 0.1M, the amount of C produced will be \([C] = K \times [A] \times [B] = 1 \times 10^{-2} \times 0.1 \times 0.1 = 1 \times 10^{-4} M\).

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Most popular questions from this chapter

Nitrogen dioxide obtained as a cylinder gas is always a mixture of \(\mathrm{NO}_{2}(\mathrm{g})\) and \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g}) .\) A \(5.00 \mathrm{g}\) sample obtained from such a cylinder is sealed in a \(0.500 \mathrm{L}\) flask at \(298 \mathrm{K}\). What is the mole fraction of \(\mathrm{NO}_{2}\) in this mixture? $$\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{g}) \quad K_{\mathrm{c}}=4.61 \times 10^{-3}$$

The following reaction is used in some self-contained breathing devices as a source of \(\mathrm{O}_{2}(\mathrm{g})\) $$\begin{aligned} 4 \mathrm{KO}_{2}(\mathrm{s})+2 \mathrm{CO}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{K}_{2} \mathrm{CO}_{3}(\mathrm{s}) &+3 \mathrm{O}_{2}(\mathrm{g}) \\\ K_{\mathrm{p}} &=28.5 \mathrm{at} 25^{\circ} \mathrm{C} \end{aligned}$$ Suppose that a sample of \(\mathrm{CO}_{2}(\mathrm{g})\) is added to an evacuated flask containing \(\mathrm{KO}_{2}(\mathrm{s})\) and equilibrium is established. If the equilibrium partial pressure of \(\mathrm{CO}_{2}(\mathrm{g})\) is found to be \(0.0721 \mathrm{atm},\) what are the equilibrium partial pressure of \(\mathrm{O}_{2}(\mathrm{g})\) and the total gas pressure?

In the human body, the enzyme carbonic anahydrase catalyzes the interconversion of \(\mathrm{CO}_{2}\) and \(\mathrm{HCO}_{3}^{-}\) by either adding or removing the hydroxide anion. The overall reaction is endothermic. Explain how the following affect the amount of carbon dioxide: (a) increasing the amount of bicarbonate anion; (b) increasing the pressure of carbon dioxide; (c) increasing the amount of carbonic anhydrase; (d) decreasing the temperature.

What effect does increasing the volume of the system have on the equilibrium condition in each of the following reactions? (a) \(\mathrm{C}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g})\) (b) \(\mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CaCO}_{3}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (c) \(4 \mathrm{NH}_{3}(\mathrm{g})+5 \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\)

The following reaction represents the binding of oxygen by the protein hemoglobin (Hb): $$\mathrm{Hb}(\mathrm{aq})+\mathrm{O}_{2}(\mathrm{aq}) \rightleftharpoons \mathrm{Hb}: \mathrm{O}_{2}(\mathrm{aq}) \quad \Delta H<0$$ Explain how each of the following affects the amount of \(\mathrm{Hb}: \mathrm{O}_{2}:\) (a) increasing the temperature; (b) decreasing the pressure of \(\mathrm{O}_{2} ;\) (c) increasing the amount of 6 hemoglobin.
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