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Chapter 15: Problem 15

An important environmental and physiological reaction is the formation of carbonic acid, \(\mathrm{H}_{2} \mathrm{CO}_{3}(\mathrm{aq})\) from carbon dioxide and water. Write the equilibrium constant expression for this reaction in terms of activities. Convert that expression into an equilibrium constant expression containing concentrations and pressures.

Short Answer

Expert verified
The equilibrium constant expression in terms of activities is \[K = \frac{a(H_{2}CO_{3})}{a(CO_{2})a(H_{2}O)}\] , and in terms of concentrations and pressures it is \[K = \frac{[H_{2}CO_{3}]}{P(CO_{2})[H_{2}O]} \]

Step by step solution

01

Understand the Reaction and Write it Down

The reaction is between carbon dioxide (CO2) and water (H2O) to form carbonic acid (H2CO3). The reaction can be written as: \[CO_{2(g)} + H_{2O(l)} \longleftrightarrow H_{2}CO_{3(aq)} \]
02

Write the Equilibrium Constant Expression in Terms of Activities

The equilibrium expression in terms of activities can be written based on the law of mass action. For the reaction mentioned above, the equilibrium constant expression can be written as: \[K = \frac{a(H_{2}CO_{3})}{a(CO_{2})a(H_{2}O)}\] where \(K\) is the equilibrium constant and \(a(Y)\) represents the activity of species \(Y\) in the reaction.
03

Convert the Equilibrium Constant Expression into Concentration and Pressure Terms

The activity of a species in solution is approximately equal to its molar concentration, and the activity of a gas is approximately equal to its partial pressure. Therefore, the equilibrium constant expression can be converted into concentration and pressure terms as follows: \[K = \frac{[H_{2}CO_{3}]}{P(CO_{2})[H_{2}O]}\] where \([Y]\) represents the concentration of species \(Y\), and \(P(Y)\) represents the partial pressure of species \(Y\) in the reaction.

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Most popular questions from this chapter

Equilibrium is established in a 2.50 L flask at \(250^{\circ} \mathrm{C}\) for the reaction $$\mathrm{PCl}_{5}(\mathrm{g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \quad K_{\mathrm{c}}=3.8 \times 10^{-2}$$ How many moles of \(\mathrm{PCl}_{5}, \mathrm{PCl}_{3},\) and \(\mathrm{Cl}_{2}\) are present at equilibrium, if (a) 0.550 mol each of \(\mathrm{PCl}_{5}\) and \(\mathrm{PCl}_{3}\) are initially introduced into the flask? (b) \(0.610 \mathrm{mol} \mathrm{PCl}_{5}\) alone is introduced into the flask?

The following is an approach to establishing a relationship between the equilibrium constant and rate constants mentioned in the section on page 660 \(\bullet\)Work with the detailed mechanism for the reaction. \(\bullet\) Use the principle of microscopic reversibility, the idea that every step in a reaction mechanism is reversible. (In the presentation of elementary reactions in Chapter \(14,\) we treated some reaction steps as reversible and others as going to completion. However, as noted in Table \(15.3,\) every reaction has an equilibrium constant even though a reaction is generally considered to go to completion if its equilibrium constant is very large.) \(\bullet\) Use the idea that when equilibrium is attained in an overall reaction, it is also attained in each step of its mechanism. Moreover, we can write an equilibrium constant expression for each step in the mechanism, similar to what we did with the steady-state assumption in describing reaction mechanisms. \(\bullet\)Combine the \(K_{\mathrm{c}}\) expressions for the elementary steps into a \(K_{\mathrm{c}}\) expression for the overall reaction. The numerical value of the overall \(K_{c}\) can thereby be expressed as a ratio of rate constants, \(k\) Use this approach to establish the equilibrium constant expression for the overall reaction, $$ \mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g}) $$ The mechanism of the reaction appears to be the following: Fast: \(\quad \mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{I}(\mathrm{g})\) Slow: \(\quad 2 \mathrm{I}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})\)

A sample of pure \(\mathrm{PCl}_{5}(\mathrm{g})\) is introduced into an evacuated flask and allowed to dissociate. $$ \mathrm{PCl}_{5}(\mathrm{g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) $$ If the fraction of \(\mathrm{PCl}_{5}\) molecules that dissociate is denoted by \(\alpha,\) and if the total gas pressure is \(P\) show that $$ K_{\mathrm{p}}=\frac{\alpha^{2} P}{1-\alpha^{2}} $$

A mixture of \(1.00 \mathrm{g} \mathrm{H}_{2}\) and \(1.06 \mathrm{g} \mathrm{H}_{2} \mathrm{S}\) in a 0.500 Lflask comes to equilibrium at \(1670 \mathrm{K}: 2 \mathrm{H}_{2}(\mathrm{g})+\mathrm{S}_{2}(\mathrm{g}) \rightleftharpoons\) \(2 \mathrm{H}_{2} \mathrm{S}(\mathrm{g}) .\) The equilibrium amount of \(\mathrm{S}_{2}(\mathrm{g})\) found is \(8.00 \times 10^{-6}\) mol. Determine the value of \(K_{p}\) at 1670 K.

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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